ANSWERS
In addition to the answer, we have provided a difficulty rating for each problem. Our scale is 1-7, with 7 being the most
difficult. These are only approximations, and how difficult a problem is for a particular student will vary. Below is a general guide to the ratings:
Difficulty 1/2/3 - One concept; one- to two-step solution; appropriate for students just starting the middle school curriculum.
4/5 - One or two concepts; multistep solution; knowledge of some middle school topics is necessary.
6/7 - Multiple and/or advanced concepts; multistep solution; knowledge of advanced middle school topics and/or problem-solving strategies is necessary.
Warm-Up 1
Answer
Workout 1
Difficulty
Answer
Difficulty
1.200
(1)
6.88
(2)
31. 3.6
(2)
36. 7.2
(3)
2.59
(2)
7.7
(2)
32. 28
(2)
37. 32.6
(4)
3.1080
(3)
8.277
(2)
33. 64.5
(3)
38. 5
(4)
4.31.68
(2)
9.13
(3)
34. 1.45
(5)
39. 1 × 109
(4)
5.8
(3)
10. −800
(4)
35. 13
(2)
40. 12
(4)
Warm-Up 2
Answer
Workout 2
Answer
Difficulty
Difficulty
11. 1
(1)
16. 20
(5)
41.
6
(3)
46.
16.4
(3)
12. 2
(2)
17. 6
(2)
42.
19.2
(4)
47.
0.8
(2)
13. 12
(2)
18. 113
(3)
43.
5 or 5.00
(3)
48.
40
(3)
14. 5/3
(3)
19. 225
(4)
44.
42.9
(4)
49.
72
(4)
15. 36,300
(3)
20. 16/47
(4)
45.
2.70
(4)
50.
1*
(3)
Warm-Up 3
Answer
Warm-Up 4
Difficulty
Answer
21. 4
(2)
26. 81
(4)
51. 50
22. 223,200
(2)
27. 7.5
(4)
52. 18
23. 1/6
(4)
28. 9
(3)
24. 20
(4)
29. 107
25. 1/3
(5)
30. 10
Difficulty
(2)
56. 5940
(3)
(2)
57. 55
(4)
53. red
(2)
58. 48
(4)
(5)
54. 8
(5)
59. 550
(5)
(4)
55. 259
(4)
60. 20
(4)
1
3
* The plural form of the units is always provided in the answer blank, even if the answer appears to require
the singular form of the units.
MATHCOUNTS 2015-2016
57
Workout 4
Warm-Up 5
Answer
Answer
Difficulty
Difficulty
61. 34
(4)
66. 15
(3)
91.
15
(3)
96.
1091
(5)
62. 24
(3)
67. 1/5
(4)
92.
0.9
(4)
97.
3
(3)
63. 25
(3)
68. Monday
(3)
93.
7.5
(4)
98.
30.8
(3)
64. 75 or 75.00 (3)
69. 6
(4)
94.
3.04
(4)
99.
55
(2)
65. 432
70. 15
(4)
95.
157.5
(4)
100.7
(4)
(4)
Warm-Up 7
Warm-Up 6
Answer
Difficulty
Answer
71. 2
(4)
76. −27
72. 15
(3)
77. 328
73. 60
(3)
74. 64,000
75. 88
Difficulty
(4)
199
101.(3)
17/14
106.(3)
(4)
14
102.(3)
25
107.(6)
78. 3
(4)
1.25
103.(3)
226
108.(3)
(5)
79. 511/512
(4)
40
104.(4)
64
109.(3)
(3)
80. 60 or 60.00 (3)
26
105.(3)
1
8
5
8
Workout 3
Answer
10
110.(3)
Warm-Up 8
Difficulty
Answer
Difficulty
81.
25
(3)
12
86.
(5)
111.121
(3)
116.−144
(4)
82.
4.05
(3)
96
87.
(5)
112.12
(4)
117.3
(3)
83.
240
(4)
6.83
88.
(4)
113.6
(4)
118.14
(5)
84.
5.6
(4)
1.75
89.
(5)
114.5850
(4)
119.5
(4)
85.
0.025
(3)
24
90.
(3)
115.5/41
(4)
120.9
(5)
58
MATHCOUNTS 2015-2016
Warm-Up 10
Workout 5
Answer
Answer
Difficulty
Difficulty
121.4.7
(3)
126.26
(4)
151.7
(4)
156.17
(3)
122.−68.75
(4)
127.46,656
(5)
152.50
(3)
157.29,400
(4)
123.8.6
(4)
128.4
(5)
153.47
(4)
158.36
(5)
124.0.79
(5)
129.11/10
(3)
154.29
(4)
159.17
(3)
125.65
(3)
130.25
(4)
155.20
(5)
160.√145
(4)
Warm-Up 11
Workout 6
Answer
Difficulty
Answer
Difficulty
131.20
(3)
136.8.4
(5)
161.60
(4)
166.7
132.6/25
(4)
137.48
(5)
162.45
(3)
138.0.84
(2)
163.36
(5)
167.2π + 12
(5)
or 12 + 2π
139.5.7
(5)
164.625
(5)
140.6
(3)
165.2016
(5)
133. 2 − √3
(5)
or −√3 + 2
134.11.1
(5)
135.√3
(5)
Warm-Up 9
Answer
(5)
168.3
(4)
169.16
(4)
170.90
(4)
Workout 7
Answer
Difficulty
Difficulty
141.19
(4)
146.12
(5)
171.−5.6
(5)
176.6
142.1008
(4)
147.23
(4)
172.19.80
(4)
143.45
(3)
148.15
(4)
173.55.2
(5)
177. 10π − 4
(5)
or −4 + 10π
144.10
(4)
149.23
(4)
174.0.17
(5)
145.78
(5)
150.22
(4)
175.2014
(5)
MATHCOUNTS 2015-2016
(5)
178.678
(4)
179.9
(5)
180.5/12
(5)
59
Workout 8
Answer
Difficulty
181.36
(3)
186.9.2
(5)
182.2048
(4)
187.93
(4)
183.22√6
(6)
188. 66
(4)
221.(2)
855
226.(3)
10
184.√38
(5)
189.1275
(4)
222.(3)
56
227. 14
185.547
(4)
190.145
(6)
223.(3)
27
228.(3)
100
224.(3)
87
229.(4)
64
225.(4)
13
230.(4)
54
Counting Stretch
Answer
Difficulty
(3)
Warm-Up 12
Answer
Difficulty
191.1/24
(4)
196.5
(5)
192.20
(3)
197.7 or 7.00
(3)
193.4/3
(5)
198.54
(5)
194.135
(4)
199.1/7
(4)
195.33/20
(4)
200.−3
(4)
Warm-Up 13
Answer
1
Difficulty
201.9 3
(3)
206.243
(5)
202.24
(3)
207.3/7
(3)
203.16
(4)
208.126
(5)
204.1/2
(4)
209.(6, 2)
(5)
205.192.5
(5)
210.875/64
(6)
Area Stretch
Answer
(3)
232.321
(2)
233.7000
(3)
234.32
(3)
235.4π − 8
(4)
or −8 + 4π
236.4 + 8π
or 8π + 4
(5)
237.128 + 32√2 (6)
or 32√2 + 128
238.357/4
(6)
239.9/4
(7)
240.240/37
(6)
Modular Arithmetic
Stretch
Warm-Up 14
Answer
Difficulty
231.92
Difficulty
Answer
Difficulty
211.20
(4)
216.90
(5)
241.March
(3)
246.2
(4)
212.191/20
(4)
217.1
(4)
242.11:00 p.m.
(3)
247.3
(4)
213.157
(4)
218.13/34
(4)
243.110
(4)
248.83
(5)
214.64
(3)
219.11
(4)
244.8
(5)
249.61
(6)
215.31
(4)
220.124
(5)
245.5
(4)
250.59
(6)
60
MATHCOUNTS 2015-2016
MATHCOUNTS Problems Mapped to
Common Core State Standards (CCSS)
Forty-three states, the District of Columbia, four territories and the Department of Defense Education Activity
(DoDEA) have adopted the Common Core State Standards (CCSS). As such, MATHCOUNTS considers it
beneficial for teachers to see the connections between the 2015-2016 MATHCOUNTS School Handbook
problems and the CCSS. MATHCOUNTS not only has identified a general topic and assigned a difficulty level
for each problem but also has provided a CCSS code in the Problem Index (pages 62-63). A complete list of the
Common Core State Standards can be found at www.corestandards.org.
The CCSS for mathematics cover K-8 and high school courses. MATHCOUNTS problems are written to align
with the NCTM Standards for Grades 6-8. As one would expect, there is great overlap between the two sets
of standards. MATHCOUNTS also recognizes that in many school districts, algebra and geometry are taught in
middle school, so some MATHCOUNTS problems also require skills taught in those courses.
In referring to the CCSS, the Problem Index code for each of the Standards for Mathematical Content for grades
K-8 begins with the grade level. For the Standards for Mathematical Content for high school courses (such as
algebra or geometry), each code begins with a letter to indicate the course name. The second part of each code
indicates the domain within the grade level or course. Finally, the number of the individual standard within that
domain follows. Here are two examples:
•
6.RP.3 → Standard #3 in the Ratios and Proportional Relationships domain of grade 6
•
G-SRT.6 → Standard #6 in the Similarity, Right Triangles and Trigonometry domain of Geometry
Some math concepts utilized in MATHCOUNTS problems are not specifically mentioned in the CCSS. Two
examples are the Fundamental Counting Principle (FCP) and special right triangles. In cases like these, if a
related standard could be identified, a code for that standard was used. For example, problems using the FCP
were coded 7.SP.8, S-CP.8 or S-CP.9 depending on the context of the problem; SP → Statistics and Probability
(the domain), S → Statistics and Probability (the course) and CP → Conditional Probability and the Rules of
Probability. Problems based on special right triangles were given the code G-SRT.5 or G-SRT.6, explained
above.
There are some MATHCOUNTS problems that either are based on math concepts outside the scope of the
CCSS or based on concepts in the standards for grades K-5 but are obviously more difficult than a grade
K-5 problem. When appropriate, these problems were given the code SMP for Standards for Mathematical
Practice. The CCSS include the Standards for Mathematical Practice along with the Standards for Mathematical
Content. The SMPs are (1) Make sense of problems and persevere in solving them; (2) Reason abstractly and
quantitatively; (3) Construct viable arguments and critique the reasoning of others; (4) Model with mathematics;
(5) Use appropriate tools strategically; (6) Attend to precision; (7) Look for and make use of structure and
(8) Look for and express regularity in repeated reasoning.
MATHCOUNTS 2015-2016
61
PROBLEM INDEX
62
6.RP.3
6.RP.3
6.NS.1
6.RP.3
7.NS.3
7.RP.3
6.EE.7
6.RP.3
6.RP.3
6.RP.3
6.RP.3
7.RP.2
9 (3) 4.OA.4
32 (2) 7.NS.2
39 (4) 8.EE.2
51 (2) 7.NS.3
64 (4) 7.NS.3
68 (3) 7.NS.3
69 (4) SMP
81 (3) F-BF.2
93 (4) SMP
113 (4) SMP
119 (3) 6.NS.3
129 (3) 7.NS.3
130 (4) SMP
131 (3) SMP
144 (4) SMP
165 (5) 7.NS.2
166 (5) 8.EE.1
185 (4) SMP
213 (4) SMP
218 (4) 7.SP.7
220 (5) SMP
Modular Arithmetic Stretch*
8
(2)
6.EE.3
10
(4)
6.EE.7
23
(4)
8.EE.7
34
(5)
F-IF.2
35
(2)
A-REI.4
38
(4)
8.F.3
43
(3)
A-CED.2
57
(4)
8.EE.8
61
(4)
8.EE.8
71
(4)
A-CED.1
74
(4)
A-SSE.3
86
(5)
A-CED.2
99
(1)
F-IF.2
108
(3)
6.EE.2
112
(4)
6.EE.2
122
(4)
8.EE.8
125
(3)
F-IF.2
128
(5)
8.EE.8
134
(5)
8.EE.7
140
(3)
F-IF.2
141
(4)
8.F.1
147
(4)
A-SSE.3
148
(4)
4.OA.4
168
(4)
A-SSE.3
171
(5)
8.F.3
182
(4)
F-IF.2
195
(4)
8.F.3
196
(5)
A-SSE.2
200
(4)
A-SSE.3
203
(5)
F-IF.2
206
(3)
A-SSE.1
211
(4)
N-RN.2
Problem Solving (Misc.)
(2)
(3)
(3)
(4)
(3)
(4)
(4)
(3)
(3)
(3)
(3)
(4)
Algebraic Expressions & Equations
4
18
36
42
56
60
94
103
106
109
192
204
General Math
Number Theory
Proportional Reasoning
It is difficult to categorize many of the problems in the MATHCOUNTS School Handbook. It is very
common for a MATHCOUNTS problem to straddle multiple categories and cover several concepts.
This index is intended to be a helpful resource, but since each problem has been placed in exactly one
category and mapped to exactly one Common Core State Standard (CCSS), the index is not perfect.
In this index, the code 9 (3) 7.SP.3 refers to problem 9 with difficulty rating 3 mapped to CCSS
7.SP.3. For an explanation of the difficulty ratings refer to page 57. For an explanation of the CCSS
codes refer to page 61.
7
11
12
48
66
101
116
126
150
156
159
162
187
188
189
190
197
198
207
216
1
2
14
22
73
80
82
91
105
110
111
138
155
202
(2)
(1)
(2)
(3)
(3)
(3)
(4)
(4)
(4)
(3)
(3)
(3)
(4)
(4)
(4)
(6)
(3)
(5)
(3)
(5)
(1)
(2)
(3)
(2)
(3)
(3)
(3)
(3)
(3)
(3)
(3)
(2)
(5)
(3)
SMP
SMP
SMP
SMP
SMP
SMP
SMP
SMP
SMP
SMP
SMP
SMP
S-CP.9
SMP
S-CP.9
SMP
SMP
SMP
SMP
SMP
4.OA.2
7.NS.3
6.NS.1
4.OA.2
4.OA.2
N-Q.1
7.NS.3
A-CED.1
7.NS.3
SMP
7.NS.3
6.NS.3
SMP
SMP
MATHCOUNTS 2015-2016
(2)
(4)
(4)
(4)
(4)
(4)
(4)
(4)
(5)
(5)
(5)
(4)
F-BF.2
F-BF.2
F-BF.2
F-BF.2
F-BF.2
F-LE.2
F-BF.2
F-LE.2
F-LE.2
F-BF.2
F-BF.2
F-BF.2
MATHCOUNTS 2015-2016
G-GMD.3
G-GMD.3
G-GMD.3
7.G.6
7.G.6
G-GMD.3
8.G.9
8.G.9
G-GMD.3
G-GMD.3
G-GMD.4
G-GMD.3
154
(4)
6.G.3
158
(5)
A-SSE.3
177
(5)
7.G.6
209
(5)
8.G.1
210
(6)
8.G.8
217
(4)
8.G.8
29
31
37
59
75
95
143
170
194
201
219
6
33
54
104
118
149
(5)
(2)
(4)
(5)
(3)
(4)
(3)
(4)
(4)
(3)
(4)
(2)
(3)
(5)
(4)
(4)
(4)
6.RP.3
6.RP.3
7.RP.3
7.RP.3
7.G.6
6.RP.3
6.RP.3
7.RP.3
6.RP.3
7.RP.3
6.RP.3
6.SP.5
6.SP.5
F-BF.2
6.SP.5
6.SP.5
6.SP.5
Logic
(3)
(4)
(3)
(5)
(5)
(5)
(3)
(4)
(5)
(5)
(4)
(3)
21
53
78
97
135
152
169
(2)
(2)
(4)
(3)
(5)
(3)
(4)
8.G.3
SMP
SMP
SMP
G-SRT.6
SMP
SMP
Percents & Fractions
Solid Geometry
Coordinate Geometry
Measurement
28
45
50
63
65
89
90
92
124
186
205
214
16
27
46
77
83
100
121
136
172
176
193
212
(5)
(4)
(3)
(4)
(4)
(4)
(3)
(5)
(4)
(5)
(5)
(4)
6.EE.7
7.RP.3
7.NS.3
7.NS.3
6.RP.3
7.NS.3
6.RP.3
7.RP.3
7.RP.3
6.RP.3
A-CED.2
7.NS.3
Probability, Counting & Combinatorics
17
24
30
76
79
114
142
153
163
175
179
215
(3) 7.G.5
(2) 7.G.4
(3) 7.G.6
(4) G-C.2
(4) 7.G.6
(3) SMP
(2) G-C.2
(2) 6.NS.1
(4) 7.G.6
(4) 8.G.8
(3) 7.G.5
(3) 7.G.4
(5) 7.G.6
(4) 7.NS.3
(5) 8.G.7
(3) 7.G.4
(3) 8.G.8
(6) 8.G.5
(3) G-SRT.5
(4) G-C.2
(5) G-SRT.6
(5) 7.G.5
(5) G-SRT.6
(5) G-SRT.6
(4) 8.G.7
(4) A-SSE.3
(5) 8.G.7
(5) G-SRT.6
(5) G-SRT.6
(4) G-SRT.6
(6) 7.G.6
(5) 8.G.7
Stretch*
Statistics
Plane Geometry
Sequences, Series & Patterns
3
13
15
19
40
41
47
52
58
70
72
85
87
88
96
98
102
107
120
123
133
137
139
146
160
161
164
167
174
178
183
184
Area
5 (3) S-CP.9
20 (4) 7.SP.8
25 (5) 7.SP.3
26 (4) S-CP.9
44 (4) 7.SP.8
49 (4) S-CP.9
55 (4) 7.SP.8
62 (3) S-CP.9
67 (4) 7.SP.8
84 (4) 7.SP.7
115 (4) 7.SP.8
117 (3) 7.SP.7
127 (5) S-CP.9
132 (4) 7.SP.8
145 (5) S-CP.9
151 (4) S-CP.9
157 (4) S-CP.9
173 (5) 7.SP.8
180 (5) 7.SP.8
181 (3) 4.OA.4
191 (4) 7.SP.8
199 (4) 7.SP.8
208 (5) S-CP.9
Counting Stretch*
63
SOLUTIONS
The solutions provided here are only possible solutions. It is very likely that you or your students will come up with
additional—and perhaps more elegant—solutions. Happy solving!
Warm-Up 1
1. Each time Franco types MATHCOUNTS, he types 7 consonants and 3 vowels, which is 7 − 3 = 4 more consonants each time. He will type a
total of 50 × 4 = 200 more consonants than vowels.
2. Vicki finishes in 28 minutes 47 seconds. If Candace finished in 29 minutes 47 seconds, she would finish 1 minute, or 60 seconds, later than
Vicki. As it turns out, Candace actually finishes a second faster than that, so Vicki finishes 59 seconds ahead of Candace.
3. One way to find the measure of an interior angle of a regular polygon is to first find the measure of an exterior angle. Since the
octagon is regular, the exterior angle must have measure 360 ÷ 8 = 45 degrees. The interior angle is the supplement of 45,
which is 180 − 45 = 135 degrees. The sum of all 8 interior angles is 135 × 8 = 1080 degrees.
135º 45º
4. There are 16 ounces in one pound, so there are 1.5 × 16 = 24 ounces in a pound and a half. Each ounce of Vera’s favorite coffee blend
costs $1.32, so 24 ounces will cost 24 × 1.32 = $31.68.
5. There are 2 possible ends for Abby (A), 2 ways that Bea (B) and Jaclyn (J) can be next to each other and 2 choices for Lily (L) − either on the
other end or between Abby and the pair of girls, Bea and Jaclyn. Here are the 2 × 2 × 2 = 8 ways they all can be seated: A B J L, A J B L, A L B J,
A L J B, B J L A, J B L A, L B J A, L J B A.
6. Let’s arrange Maurice’s first five scores from least to greatest: 80, 84, 92, 99, 100. We are told that the median is 90 when a sixth score is
included in the ordered list. The missing score must go between 84 and 92 in the ordered list, and 90 must be halfway between the missing score
and 92. Since 92 is 2 more than the 90, the missing score must be 2 less than the 90, or 90 − 2 = 88.
1
2
Figure 1
1
2
1
2
4
3
Figure 2
5
3
6
4
Figure 3
7
7. One line divides the plane into two regions (Fig. 1). A second line that intersects the first line divides
each of those regions into two regions, creating four regions (Fig. 2). A third line that intersects the first
two lines divides three of these regions into two regions each (Fig. 3). That brings the total number of
regions determined by three lines in a plane to 7 regions, as shown.
8. There is no need to solve for x. Because 3x + 160 is 5 more than 3x + 155, the value of 3x + 160 is 272 + 5 = 277.
9. The prime factorization of 72 is 23 × 32. So the factors of 72 are the products of one these powers of 2: 20, 21, 22, 23 and one of these powers of
3: 30, 31, 32. Because Ash’s secret number is not a multiple of 3, the power of 3 must be 30. The only possible factors, therefore, are 1, 2, 4 and 8.
Also, since Ash’s secret number is not a prime number, it cannot be 2. The requested sum is 1 + 4 + 8 = 13.
1 3
2
× × n = 20 + × n
2 4
5
10. Translating this statement to algebra with n as the unknown number, we can write and solve an equation as shown.
2n
3n
= 20 +
5
8
Warm-Up 2
15 n = 800 + 16 n
−800 = n
11. The least possible sum of digits on a 12-hour digital clock is 1. This sum occurs at both 1:00 and 10:00.
12. The first and fourth digits of a 4-digit palindrome must be the same, and the second and third digits must be the same. Since both 1 and 2 must
be used, we must start with either 12 or 21, leading to the 2 palindromes 1221 and 2112.
13. The circumference of a circle is π times its diameter, so the diameter of this circle must be 24 cm. Since the diameter of a circle is twice its
radius, this circle must have radius 24 ÷ 2 = 12 cm.
14. The distance from one-half to 2.25 is 9/4 − 2/4 = 7/4. We want to divide this distance into thirds, and two-thirds of 7/4 is 2/3 × 7/4 = 14/12 =
7/6. We want the point that is 7/6 greater than one-half, so we add to get 3/6 + 7/6 = 10/6 = 5/3.
15. The length of the field is three times its width, so the perimeter of the field is eight times its width since P = 2l + 2w = 2(3w) + 2w = 6w + 2w
= 8w. The width is 880 ÷ 8 = 110 meters, which means the length must be 110 × 3 = 330 meters. It follows, then, that the area of the field is
110 × 330 = 36,300 m2.
MATHCOUNTS 2015-2016
65
16. When the price was increased by x percent in April, the new price was 1 + x/100 times the original price of d dollars, or (1 + x/100) × d.
When this price was decreased by x percent in May, the new price for May was 1 − x/100 times the price in April, or (1 − x/100) × (1 + x/100)
× d. We are told that this final price is equal to 4 percent less than the original price, or (1 − 4/100) × d = (96/100) × d = (24/25) × d. So we have
(1 + x/100) × (1 − x/100) × d = (24/25) × d. Dividing both sides by d, we get (1 + x/100) × (1 − x/100) = 24/25, and we see that the original price
of d dollars doesn’t matter when determining x. Multiplying through on the left-hand side of the equation, we get 1 − (x/100)2 = 24/25, so (x/100)2
= 1/25 and x/100 = 1/5. Since it must be positive, the value of x is 20.
17. As the number of rows increases, the total number of Ms is given by the sequence 1, 3, 6, 10, 15, 21, .... These are the well-known triangular
numbers, which come up often in many different contexts. The sixth triangular number, 21, is the first one that is divisible by 7. So the least number of
rows required for the total number of Ms to be divisible by 7 is 6 rows.
18. Since Josh mowed one-third of the 2000-ft2 lawn in 18 minutes, it would take him 18 × 3 = 54 minutes to mow all 2000 ft2. At this rate, he must
mow 200 ft2 every 5.4 minutes. Since 4200 ÷ 200 = 21, it would take him 21 × 5.4 = 113.4 minutes to mow the 4200-ft2 lawn. That’s 113 minutes
to the nearest whole number.
19. An angle inscribed in a circle is half the degree measure of the arc it intercepts. Since segment XZ is a diameter of the circle, the arc
intercepted by angle XYZ is arc XZ (not shown), a semicircle with degree measure 180. It follows that XYZ measures 90 degrees. We can use the
Pythagorean Theorem, which tells us that the sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of
the hypotenuse. For the right triangle in question, (XY)2 + (YZ)2 = (XZ)2, so the value of (XY)2 + (YZ)2 is 152 = 225 units2.
20. The last contestant must select a ball with a number greater than 67 to receive a cash prize. Since number 83 already has been selected,
there are 100 − 67 − 1 = 32 numbers that could win out of 100 − 6 = 94 numbers. The probability that the last contestant will win a cash prize is
32/94 = 16/47.
Shift
1
Shift
2
19
19
91
21. As the figure shows, 4 shifts must be performed to return 19 to its original orientation.
Shift
3
91
Warm-Up 3
Shift
4
19
22. There is an increase of 155 × 60 = 9300 people per hour. So every day, there are 24 × 9300 = 223,200 more people on Earth.
23. If 4 times 1/x equals 24, then 1/x has to equal 6. That means the reciprocal of x is 6, so x must be 1/6.
24. The sequence of units digits for the first row, which repeats for every 10 multiples of 7, is 7, 4, 1, 8, 5, 2, 9, 6, 3, 0. The sequence of units digits
for the second row, which repeats for every 10 multiples of 13, is 3, 6, 9, 2, 5, 8, 1, 4, 7, 0. Notice that every fifth term of these two sequences is
the same value, either 5 or 0. Since Maximo wrote down two rows of 100 numbers, the same units digit occurred in corresponding terms in the two
rows 100 ÷ 5 = 20 times.
25. Let’s suppose that the two adjacent squares are in the same row. Then there are 5 possible pairs in each of 6 rows, for 30 possible outcomes.
In row 1 and row 6, there is only 1 way to select a pair of adjacent squares of which one is shaded. In rows 2, 3, 4 and 5, there are 2 ways to do this.
That’s 1 + 2 + 2 + 2 + 2 + 1 = 10 out of 30 ways to select two adjacent squares that are different colors in the same row. The same reasoning
applies to selecting adjacent squares in the same column, so the probability is (10 + 10)/(30 + 30) = 20/60 = 1/3.
26. One billion is written 1,000,000,000. Integers less than 1,000,000,000 contain 1, 2, 3, 4, 5, 6, 7, 8 or 9 digits. If the integer is classified as a
very round number, the first digit must be the only nonzero digit. There are 9 options for that digit. Therefore, 9 × 9 = 81 integers less than one billion
are very round numbers.
27. Five percent of a number is 1/20 of the number, so the original number must be 3/8 × 20 = 15/2 = 7.5.
28. The volume V of a right rectangular prism is the product of its length l, its width w and its height h. For this particular prism, we are told that
V = 108 cm3, l = 4 cm and w = 3 cm. Thus, we have 108 = 4 × 3 × h, so h must be 108 ÷ 12 = 9 cm.
29. Note that Chanel’s speed is 2/3 Marcus’s speed, so she will type 2 pages for every 3 pages that Marcus types. This means that Chanel will type
2/5 of the 40 pages, or 16 pages, while Marcus will type the other 24. Chanel’s 16 pages contain 16 × 240 = 3840 words. So at a rate of 36 words
per minute, it will take her 3840 ÷ 36 = 106 2/3 minutes. Marcus will complete his 24 pages during this same time period, so all 40 pages of notes
will be typed in 106 2/3 ≈ 107 minutes.
30. The table shows how many cars the salesman will sell on each of his first five days
on the job and the total number of cars he will have sold since he began. It appears that
the daily total forms the sequence 20, 21, 22, 23, 24,…, and the cumulative total
forms the sequence (21) − 1, (22) − 1, (23) − 1, (24) − 1,…. So to generalize, on the nth day he will sell 2(n – 1) cars. And, including the cars he sells
on the nth day, he will have sold a total of (2n) − 1 cars. On the tenth day he will sell 29 = 512 cars, and he will have sold a total of (210) − 1 =
1024 − 1 = 1023 cars. So it will take him 10 days to sell at least 1000 cars.
Day
Daily Total
Cumulative Total
66
1st
1
1
2nd
2
3
3rd
4
7
4th
8
15
5th
16
31
MATHCOUNTS 2015-2016
Workout 1
31. There are 12 × 10 = 120 months in 10 years. Deshawn can expect each fingernail to grow 120 × 3 = 360 mm during that time. For her
10 fingers, she can expect the total fingernail growth to be 360 × 10 = 3600 mm, or 3600 ÷1000 = 3.6 m.
32. Of the 25 prime numbers less than 100, the 7 primes with a units digit of 3 are 3, 13, 23, 43, 53, 73 and 83. That’s 7/25 = 0.28 = 28%.
33. Since there are more girls than boys in Mrs. Abel’s class, the girls’ average height will have more of an impact on the overall average than the
boys’ average height. We calculate the “weighted” average as follows: (16 × 63.7 + 14 × 65.4)/(16 + 14) ≈ 64.5 inches.
34. Using the formula given, P = 5 + 1.25x, we calculate the total cost for downloading 25 applications to be P = 5 + 1.25(25) = 36.25. The
average cost per application was 36.25/25 = $1.45.
35. The prime factorization of 299 is 13 × 23. The factors of 299 are 1, 13, 23 and 299. There is only one factor pair that works, since there are
more than 2 and fewer than 20 members in the gaming club. The gaming club must have 13 members, who each paid $23 dollars for the book.
36. First, let’s convert 12 feet to 12 × 12 = 144 inches. Next, since the construction of the staircase must follow the local building code, we divide
144 by the 7.5, and get 19.2. So any staircase Jamee constructs must have at least 20 steps. A staircase with 20 steps would have steps of the
maximum allowable height, which is 144 ÷ 20 = 7.2 inches.
37. The ratio of the thickness of the pages of book two to that of book one must be 11 to 10. Therefore, the thickness of the pages in book two
is 11/10 of 14 mm, or 15.4 mm. Since the thickness of book two, including the front and back covers, is 17 mm, the covers must be 17 − 15.4 =
1.6 mm thick. The thickness of book one, including the front and back covers, is 14 + 1.6 = 15.6 mm. Therefore, the total thickness of the two books
stacked must be 15.6 + 17 = 32.6 mm.
38. The point (0, b), the intersection of the line and the y-axis, is called the y-intercept. The line containing (−3, 14) and (4, −7) has slope
(14 − (−7))/(−3 − 4) = 21/(−7) = −3. Since (0, b) is on the line, the slope between it and (4, −7) also must be −3. Therefore, (b − (−7))/(0 − 4) =
−3. Simplifying and solving for b, we have (b + 7)/(−4) = −3 → b + 7 = 12 → b = 5.
39. Since we are looking for a multiple of 10, we will start with 2 × 5, which equals 10. Now, to get the greatest value possible, we’ll need to raise
this quantity to some power. The greatest value we can have using the remaining 2 and 3 also will be a power. Our options are 23 = 8 and 32 = 9.
2
Since 32 is greater, we have (2 × 5) 3 , which is 1 × 109, expressed in scientific notation.
40. Since the altitude to the 18-cm side of the triangle is 10 cm, the triangle has area (1/2) × 18 × 10 = 90 cm2. If h is the altitude to the 15-cm
side of the triangle, when we multiply (1/2) × 15 × h, the product also must equal 90 cm2. So (15/2)h = 90 and h = 12 cm.
Workout 2
A
6
H1
G
B
5
2
F
3
4
C
D
41. The octagon shown has eight vertices, labeled A through H. From A, we can draw diagonals to every vertex of the octagon
except A, B and H. As the figure shows, ABC, ACD, ADE, AEF, AFG and AGH are the 6 triangles that are formed.
E
42. Since 8/12 < 4/5, Alicia’s frame fits the full width of 12 inches. But the frame has some extra height. To find the frame’s height, h, we set up a
proportion: 4/5 = h/12. Cross-multiplying, we have 5h = 48 and h = 48/5 inches. Therefore, the frame exceeds the height of the picture by 48/5 −
8 = 48/5 − 40/5 = 8/5 inches, and the area of the extra space is 8/5 × 12 = 96/5 = 19.2 in2.
43. Let’s use the letters b and d to represent the prices of a burger and a dog, respectively. Translating the given information to algebra, we get the
equations 4b + 2d =16.40 and 6b + 4d = 26.40. In the second equation, there are two more burgers and two more dogs than there are in first. The
difference is 2b + 2d = 10, and dividing by 2, we see that b + d = $5.
44. Among the seventh and eighth graders, there are 61 + 54 = 115 who play soccer, 116 + 103 = 219 who don’t play soccer, and 115 + 219
= 334 students in all. If two students are chosen at random, the probability that neither student plays soccer is (219/334) × (218/333) ≈ 0.429 =
42.9%.
45. Using the formula for the volume of a cylinder, V = π × r2 × h, we determine that the volume of the smaller container is π × 12 × 4 = 4π in3.
The volume of the larger container is π × 32 × 10 = 90π in3. The volume of the second container is 90π ÷ (4π) = 22.5 times the volume of the first
container, so the second container would cost 22.5 times as much to fill with soda. That would be 22.5 × 0.12 = $2.70.
46. It took 10% less time to create each decoration after the first than it did to create the previous one, which means it took exactly 90% of the
time it took to create the previous one. The total time it took Kim to create the first five decorations, then, was 4 + (0.9)(4) + (0.9)2(4) + (0.9)3(4) +
(0.9)4(4) = 4(1 + 0.9 + 0.92 +0.93 + 0.94) = 16.3804 ≈ 16.4 minutes.
MATHCOUNTS 2015-2016
67
47. The length of string used is the circumference of each circle, which equals 2π times the circle’s radius. Lily’s string has length 26 cm, so her
circle has radius 26/(2π). Willow’s string has length 31 cm, so her circle has radius 31/(2π). Subtracting, we see that the absolute difference
between these two radii is (31 − 26)/(2π) = 5/(2π) ≈ 0.8 cm.
48. Since the result has 0 in the ones place, B + G = 10, and a 1 is carried into the tens column. The thousands and hundreds places of the result
are 1 and 0, respectively, so we know that nothing is carried to the hundreds column when we add 1 + A + A. Since 1 + 2A = 9, it follows that
2A = 8 and A = 4. Therefore, the value of A × (B + G) is 4 × 10 = 40.
49. For the sum of any two adjacent integers to be odd, one must be even and one must be odd. When arranging the integers 1 through 6 in a row,
there are 3 × 2 × 1 = 6 ways we can place 1, 3 and 5 in the first, third and fifth positions. For each of these possibilities, there are 6 ways to place
the 2, 4 and 6 in the second, fourth and sixth positions. So far, that’s 6 × 6 = 36 arrangements. There are another 36 arrangements if 1, 3 and 5 are
placed in the second, fourth and sixth positions with 2, 4 and 6 in the first, third and fifth positions. That’s a total of 72 ways in all.
50. A cube of volume of 8 units3 has edges of length 2 units. The largest sphere that will fit inside this cube has a diameter equal to the length of the
cube’s edge, and the sphere has a radius of 1 unit.
Warm-Up 4
51. These factors are very close to values rounded to the nearest tenth, so we expect the actual product to be really close to the product of the
rounded factors. Rounding each factor to the nearest tenth, we would get the product 2.5 × 4 × 5 = 50. The actual product will be just a hair less
than this. So the smallest integer greater than the actual product is 50.
Y
52. A rectangle of length l and width w has area l × w. For this particular rectangle, we have l = 2 1/2 = 5/2
1
and w = 7 1/3 = 22/3. Thus, the rectangle has an area of 5/2 × 22/3 = 5 × 11/3 = 55/3 = 18 3 m2.
53. If we consider the circles in the order numbered here, we can determine exactly which
5
color must be placed at each vertex. Circle 1 must be colored yellow since we are told that
the circles at the other two vertices of this triangle are colored blue and red. Then circle 2 will be
colored red, and circle 3 will be colored yellow. Circle 4, then, will be colored blue, which leaves
circle 5, the one marked with , to be colored red, as shown.
R
R
B
3
2
4
B
Y
R
B
1
B
Y
R
54. Solution 1: The absolute difference between the first and second terms in an arithmetic sequence is the absolute value of the common
difference d. We are told that the mean of the first five terms is 27 and the mean of the first eight terms is 39. Since the mean of the first five terms
is equal to the middle term, it follows that the value of the third term is 27. The first eight terms of the sequence can be written 27 − 2d, 27 − d, 27,
27 + d, 27 + 2d, 27 + 3d, 27 + 4d, 27 + 5d. The mean of the first eight terms is equal to the mean of the fourth and fifth terms, so we can write the
equation [(27 + d) + (27 + 2d)]/2 = 39. Multiplying both sides by 2 and solving for d, we see that (27 + d) + (27 + 2d) = 78 → 54 + 3d = 78 →
3d = 24 → d = 8.
Solution 2: The absolute difference between the first and second terms in an arithmetic sequence is the absolute value of the common difference
d. Let a1, a2, a3, a4, a5, a6, a7 and a8 be the first eight terms of the arithmetic sequence. We are told that the mean of the first five terms is 27 and
the mean of the first eight terms is 39. Since the mean of the first five terms is equal to the middle term, it follows that a3 = 27. In addition, the sum
of these five terms is 27 × 5 = 135 and the sum of the first eight terms is 39 × 8 = 312. The difference 312 − 135 = 177 must be the sum of the
terms a6, a7 and a8. Their mean, 177 ÷ 3 = 59, must equal a7. Since a3 + 4d = a7, we have 27 + 4d = 59. Solving for d, we get 4d = 32 and d = 8.
55. There are 366 − (9 × 12) = 258 double-digit days, including February 29th. If there were 259 people with no two having the same birthday,
then at least one of them must have a single-digit birthday.
56. The car and every point on it travel 3000 − 30 = 2970 feet each way, for a round-trip distance of 2 × 2970 = 5940 feet.
57. Let x be the amount that Fancy Caterers charges per guest and y be the fixed delivery fee. We can write the following equation: 25x + y =
2(10x + y). Simplifying, we have 25x + y = 20x + 2y, so 5x = y. This means that the delivery fee is equal to the cost for 5 guests. We also have the
following equation: nx + y = 2(25x + y). Simplifying, we have nx + y = 50x + 2y, so nx − 50x = y. Factoring x on the left-hand side yields (n − 50)x
= y. Because 5x = y, it follows that n − 50 = 5 and n = 55.
D
58. Solution 1: Since mABC = 120 degrees, it follows that the mBCD = 180 − 120 = 60 degrees by the properties of alternate interior angles,
and mDAB = 120 degees. If we drop the perpendiculars from A and B to G and F on side CD, as shown, we create
A
B
rectangle ABFG and congruent right triangles BCF and ADG. Since side FG is opposite side AB, we know that FG = AB =
18 cm and CF = DG = (22 − 18)/2 = 4/2 = 2 cm. Since BCF and ADG are 30-60-90 right triangles with a shorter leg of
G
F C length 2 cm, hypotenuses BC and AD must have length 2 × 2 = 4 cm. Therefore, the perimeter of trapezoid ABCD is
18 + 4 + 22 + 4 = 48 cm.
Solution 2: Since mABC = 120 degrees, it follows that the mBCD = 180 − 120 = 60 degrees by the properties of alternate
interior angles. Let E be on side CD so that DE = 4 cm and EC = 18 cm. When we draw segment AE, which is parallel to
segment BC, we create parallelogram ABCE and equilateral triangle ADE, as shown. It follows, then, that DE = AD = EA =
D
BC = 4 cm. Therefore, the perimeter of trapezoid ABCD is 18 + 4 + 22 + 4 = 48 cm.
68
A
B
60º
60º
60º
4 E
18
60º
MATHCOUNTS 2015-2016
C
59. Since Kaci’s jogging speed is 60% greater than her walking speed, it must be 160% of her walking speed, or 8/5 times as fast. We can use
this 8 to 5 ratio in several ways. It means her jogging time for one mile is 5/8 of her walking time and her walking time is 8/5 of her jogging time. The
extra 5 1/2 minutes it takes her to walk one mile must be the extra 8/5 − 5/5 = 3/5 times her jogging time. If we multiply the 5 1/2 = 11/2 minutes
by 5/3, we will get the full jogging time of 11/2 × 5/3 = 55/6 minutes. This is 55/6 × 60 = 550 seconds.
60. Rewrite the sentence as an equation with the percents expressed as fractions and the unknown percent as x. Then (30/100) × (80/100) × 20 =
(x/100) × (60/100) × 40 → (3/10) × (4/5) × 20 = (x/100) × (3/5) × 40 → 24/5 = 24x/100 → x = (24/5) × (100/24) → x = 20.
Warm-Up 5
61. Let n, d and q represent the number of nickels, dimes and quarters, respectively. Then the total value of the coins, $2.90, or 290 cents, can
be written as 5n + 10d + 25q. Also, we know that n = 1 + 6q and d = 4q. Substituting for n and d in the equation 5n + 10d + 25q = 290 yields
5(1 + 6q) + 10(4q) + 25q = 290 → 5 + 30q + 40q + 25q = 290. When we combine like terms and solve for q, we get 95q = 285 → q = 3. Now
that we know there are 3 quarters, we can determine that there are 1 + 6 × 3 = 19 nickels and 4 × 3 = 12 dimes. Zoey has a total of 3 + 19 + 12 =
34 coins.
62. We can use the Fundamental Counting Principle, which basically says that the total number of ways for a series of events to occur equals the
product of the number of ways each event can occur. Since there are 3 choices of entrée, 4 choices of drink and 2 choices of dessert, the number
of combinations of one entrée, one drink and one dessert is 3 × 4 × 2 = 24 combinations. Therefore, it would take 24 days to try every combination.
63. The volume of the clay in the original cube is 10 × 10 × 10 = 1000 cm3. Each rectangular prism has volume 2 × 4 × 5 = 40 cm3. There is
enough clay to make 1000 ÷ 40 = 25 prisms.
64. If Roberto spent one more dollar, he would get 16 coupons, but he only gets 15 coupons for the first $600 he spent at the restaurant. The total
dollar value of those coupons is 15 × 5 = $75.
A
12
D
9
H
9
9
3
E
F I G
B 65. The length, width and height of the medium box are each twice that of the small box. So the medium box should contain 2 × 2 × 2
= 8 times as many golf balls as the small box. That’s 8 × 12 = 96 golf balls. The large box, then, should contain 3 × 3 × 3 = 27 times
as many golf balls as the small box. That’s 27 × 12 = 324 golf balls. If Jake orders one box of each size, he will receive 12 + 96 + 324
C = 432 golf balls.
66. When the cube is formed, the face labeled 5 is opposite the face labeled 1. That means the remaining four numbers must be on the faces
adjacent to the face labeled 1. The sum of these numbers is 2 + 3 + 4 + 6 = 15.
67. Solution 1: There are 6C2, read “six choose two,” ways to select two numbers from the set. That’s 6C2 = 6!/(2! × (6 − 2)!) = 6!/(2! × 4!) =
(6 × 5)/(2 × 1) = 30/2 = 15 ways. Since the product of two numbers is odd only when both numbers are odd, and there are only 3C2 ways to select
two odd numbers from the subset of odd numbers {1, 3, 5}, it follows that there are only 3C2 = 3!/(2! × (3 − 2)!) = 3!/(2! × 1!) = 3!/2! = 3 ways to
get an odd product. Therefore, the probability that the product of two numbers randomly selected from this set is odd is 3/15 = 1/5.
Solution 2: The product is odd only if each factor is odd. The probability that the first chosen number is odd is 3/6 = 1/2. The probability that the
second chosen number, then, is odd is 2/5. Therefore, the probability that the product is odd is (1/2) × (2/5) = 1/5.
68. Beginning with tomorrow, which is Saturday, the days of the week occur in the following 7-day cycle: Saturday, Sunday, Monday, Tuesday,
Wednesday, Thursday, Friday. Since 101 divided by 7 is 14 remainder 3, it follows that day 7 × 14 = 98 will be a Friday. Three days later, which is
101 days from today, will be a Monday.
69. In the base b number 431, the 1 is in the b0 place, 3 is in the b1 place and 4 is in the b2 place, so in base 10, 4b2 + 3b1 + 1b0 = 163. At this
point, we could solve a quadratic equation, or we can simply determine the greatest value of b such that 4b2 < 163, or b2 < 40.75. Since the base b
number contains the digit 4, we also know that b > 4. It follows, then, that b = 5 or b = 6. If b = 5, we get 4b2 + 3b1 + 1b0 = 4(52) + 3(51) + 1(50)
= 100 + 15 + 1 = 116. So b ≠ 5. To verify that b = 6, we substitute to get 4b2 + 3b1 + 1b0 = 4(62) + 3(61) + 1(60) = 144 +18 + 1 = 163. Thus,
Thomas must be using base 6.
y
70. Solution 1: The ant has crawled a total of 1 + 3 + 5 = 9 units to the right and a total of 2 + 4 + 6 = 12 units up. We can draw
a right triangle on the grid, as shown. Using the Pythagorean Theorem, we can find the length d of the hypotenuse, which is the ant’s
distance from the origin. We have 92 + 122 = d2 and 81 + 144 = d2. Since d2 = 225, it follows that d = √225 = 15 units.
12
10
d
8
6
4
Solution 2: You may have recognized that the legs of the right triangle are in the ratio 9/12 = 3/4, the same as those of a 3-4-5 right
triangle. We can deduce that d/12 = 5/4, so 4d = 60 and d = 15 units.
2
0
2
4
6
8 10
Warm-Up 6
71. Let the integer be n. Then we have 2n + 3 = n2 − 5 → n2 − 2n − 8 = 0. Solving the quadratic equation by factoring gives us (n − 4)(n + 2) =
0, so n − 4 = 0 and n = 4, or n + 2 = 0 and n = −2. The sum of these integers is 4 + (−2) = 2.
MATHCOUNTS 2015-2016
69
x
C
72. Since triangle ABC is isosceles, mABC = mACB = (180 − 30) ÷ 2 = 75 degrees. Angle BCD is the complement of
ACB; therefore, it has degree measure 90 − 75 = 15 degrees.
A
75º
30º
75º
15º
B
D
73. Each of 15 teams play 8 games, so we might think there are 15 × 8 = 120 games. But this counts separately the game of team A versus team B
and the game of team B versus team A. Dividing by 2, we see that the number of games that take place during the season is 60 games.
74. Let n be the cube root of Laree’s current mileage. Then n3 + 4921 = (n + 1)3. Let’s first expand the right-hand side to get n3 + 4921 =
n3 + 3n2 + 3n + 1. Combining like terms, we get 3n2 + 3n − 4920 = 0. We can simplify further by dividing by 3, which gives us n2 + n − 1640 = 0.
This quadratic equation can be solved by factoring since 40 and 41 have a difference of 1 and their product is 1640. We have (n − 40)(n + 41) = 0,
so n − 40 = 0 and n = 40, or n + 41 = 0 and n = −41. Since Laree’s mileage must by positive, n = 40 and Laree’s current mileage must be 403 =
64,000 miles. To check this, we add 64,000 + 4921 = 68,921 = 413.
75. Let x be the width and y be the height of the upper left small rectangle. Because the area of the upper right small rectangle is 24, its width must
be 24/y. Similarly, the height of the lower left small rectangle must be 15/x. This makes the area of the lower right small rectangle (24/y) × (15/x)=
360/xy = 360/40 = 9. The total area is 40 + 24 + 15 + 9 = 88 units2.
76. In an arithmetic sequence, the difference between consecutive terms is constant. Subtracting the first term from the second term, we get
(18x − 3) − (17x + 20) = x − 23. Subtracting the second term from the third term, we get d = (20x + 1) − (18x − 3) = 2x + 4. Setting these two
expressions equal to each other and solving, we have x − 23 = 2x + 4, so x = –27.
77. Donovan takes 500 mg of the medication at 8:00 a.m. and by 12:00 noon, he has 500 × 1/4 = 125 mg of effective medication in his body.
He then takes another 500 mg at 12:00 noon and by 4:00 p.m., he has (500 + 125) × 1/4 = 625/4 = 156 1/4 mg of effective medication left in
his body. He then takes another 500 mg at 4:00 p.m. Since 1/4 of the effective medication remains after 4 hours, it follows that twice that amount,
or 1/2 of the medication, must be left after 2 hours. So at 6:00 p.m., the amount of medication in Donovan’s body will be (500 + 156 1/4) × 1/2 =
1
2625/4 × 1/2 = 328 8 mg.
78. Ron should take 3 marbles to leave Martin with 36 marbles. To explain this, we will work backward. At the end of the game, Ron wants to leave
1 marble so Martin will have to take the last marble. Thus, Martin will lose. To ensure that he can leave Martin with 1 marble, Ron should leave 6
marbles before that. By doing so, Ron can sure that 5 marbles are removed between his turn and Martin’s final turn. For example, if Martin takes
1 marble, Ron takes 4 marbles (or vice-versa). Or, if Martin takes 2, Ron takes 3 (or vice-versa). Therefore, as long as Ron leaves Martin with a
number that is 1 more than a multiple of 5, Ron will completely control the game. Try it on an unsuspecting friend!
79. The common denominator of these powers of 2 is 512. The sum of the fractions is (256 + 128 + 64 + 32 +16 + 8 + 4 + 2 + 1)/512 =
511/512.
80. We want to find combinations that satisfy the equation 25 ≤ 3p + 4s, where p is the number of pizzas and s is the number of subs. Mrs. Lowe
can get 3 pizzas for 9 students and 4 subs for 16 students. The total cost would be 3 × 8 + 4 × 9 = 24 + 36 = $60. By way of comparison, she
could get 2 pizzas and 5 subs for $61, 7 pizzas and 1 sub for $65, 9 pizzas for 9 × 8 = $72 and have 2/3 of a pizza left over, or she could get 7
subs for 7 × 9 = $63 and have 3/4 of a sub left over.
Workout 3
81. The units digits of the powers of 3, which repeat every four terms, are 3, 9, 7 and 1. Starting with 33, every fourth power of 3 has a units digit of
7. Therefore, in the sequence 31, 32, 33, …, 3100 , which has 100 terms, 100/4 = 25 terms have a units digit of 7.
82. The rider must pay $2.40 for the first quarter of a mile, leaving 10 − 2.40 = $7.60 to spend on additional mileage. At $0.40 per additional
fifth mile, that’s enough to travel another 7.60/0.40 = 19 fifths of a mile. The total distance a rider can travel is 1/4 + 19/5 = (5 + 76)/20 = 81/20 =
4.05 miles.
83. In the 15 minutes from 11:15 a.m. to 11:30 a.m., 80 − 35 = 45% of the video game was downloaded, with another 100 − 80 = 20% remaining
to be downloaded. Since it takes 15 minutes to download 45%, to determine the number of minutes t it will take the other 20% to download, we can
set up the following proportion: 15/45 = t/20. Cross-multiplying and solving for t, we have 45t = 20 × 15 = 300, so t = 300/45 = 20/3 = 6 2/3
minutes, or 6 minutes 40 seconds. So a = 6, b = 40 and a × b = 6 × 40 = 240.
84. Each time Nick selects a widget at random, the probability is 1/70 that he selects a widget with a defect and 69/70 that he selects a widget
without a defect. There are five possible outcomes when Nick randomly selects four widgets. He might select 4, 3, 2, 1 or 0 widgets with a defect.
We want the sum of the probabilities of the first four cases. Since the probabilities of all five cases must have a sum of 1, let’s determine the
probability that he selects zero defective widgets and subtract that value from 1. (This technique is sometimes called complementary counting.) The
value we want is 1 − (69/70)4 ≈ 0.056 = 5.6%.
85. The circumference of a coin with a diameter of 1.205 inches is 1.205 × π inches. The 150 ridges must be spaced (1.205 × π)/150 ≈ 0.025 inch
apart.
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MATHCOUNTS 2015-2016
86. Let M and J represent Margo’s and Joy’s current ages. From the statements given, we can write the following equations: M = 2(J − 30) =
2J − 60 and J = 3(M − 20) = 3M − 60. If we substitute 2J − 60 for M in the equation J = 3M − 60, we get J = 3(2J − 60) − 60. Simplifying and
solving for J yields J = 6J − 180 − 60 and 5J = 240, so J = 48. Joy is currently 48 years old. Substituting 48 for J in the equation M = 2J − 60, we
get M = 2(48) − 60 = 96 − 60 = 36. Margo is currently 36 years old. Therefore, Joy was 48 − 36 = 12 years old when Margo was born.
87. The diagonals of a rhombus are perpendicular and bisect one another, thereby dividing the rhombus into four right triangles. The diagonals have
a difference of 4 inches, so the two legs of each right triangle must differ by 2 inches. If we call the shorter leg x then the longer leg will be x + 2.
Some students may immediately recognize that these are 6-8-10 triangles. But those who don’t can now use the Pythagorean Theorem to determine
the side lengths of each right triangle. We have x2 + (x + 2)2 = 102 → x2 + x2 + 4x + 4 = 100 → 2x2 + 4x − 96 = 0 → x2 + 2x − 48 = 0. Factoring,
we get (x + 8)(x − 6) = 0. So x + 8 = 0 and x = −8, or x − 6 = 0 and x = 6. Since the side length of the triangle must be positive, we have a
shorter leg of length 6 inches and a longer leg of length 6 + 2 = 8 inches. Therefore, the rhombus has area 4 × ((1/2) × 8 × 6) = 96 in2.
88. When the value of the stock increases by 10%, the new value of the stock is 110% of the old value. We can calculate the new value directly by
multiplying by 1.1. Three successive increases of 10% are equivalent to multiplying by 1.13 = 1.331, which amounts to an increase of 33.1%. This is
better than a 30% increase. Unfortunately, when this value dropped by 30% on Friday, the stock retained only 70% of its value. So Octavio’s stock
was left with 0.7 × 1.331 = 0.9317 or 93.17% of his original amount. This represents a decrease of 6.83%.
89. The volume of a cylinder equals the area of its base times its height, or V = πr2h. Since the circumference of a circle equals 2 × π × r, we have
2 = 2πr and r = 1/π for the smaller size Tootie Frootie. Its volume is π × (1/π)2 × 1.25 = 1.25/π in3. The volume of the larger size Tootie Frootie is
twice this amount, which is 2.5/π in3. Therefore, 2.5/π = π × r2 × 3.25 and 2.5 = 3.25 × (πr)2, so r2 = 2.5/(3.25 × π2) → r = (1/π)√(2.5/3.25) inches.
Therefore, the larger size Tootie Frootie has circumference 2 × π × (1/π)√(2.5/3.25) = 2√(2.5/3.25) ≈ 1.75 inches.
90. The total amount of dirt that must be hauled away is 32 × 15 × 8 = 3840 cubic feet. The truck holds a maximum of 6 cubic yards, and each
cubic yard is 3 × 3 × 3 = 27 cubic feet. So the truck carries 6 × 27 = 162 cubic feet of dirt per load. Dividing, we get 3840/162 ≈ 23.7 truckloads.
Therefore, it will take 24 truckloads to haul away the dirt.
Workout 4
91. Solution 1: Let’s work backward to solve this problem. After the five days, Laurie had twice as many dimes as quarters, so because she started
and ended with a total of 30 coins, she must have ended with 20 dimes and 10 quarters. After four days, she would have had 19 dimes and
11 quarters; after three days, 18 dimes and 12 quarters; after two days, 17 dimes and 13 quarters; and after the first day, 16 dimes and 14 quarters.
So she must have started with 15 dimes and 15 quarters.
Solution 2: We also can solve this algebraically with a system of equations. Let d and q represent the initial numbers of dimes and quarters in the
jar, respectively. From the information provided, we can write the following equations: d + q = 30 → d = 30 − q, and d + 5 = 2(q − 5) → d + 5 =
2q ­− 10. Substituting 30 − q for d in the equation d + 5 = 2q ­− 10 gives us 30 − q + 5 = 2q − 10. Simplifying and solving for q yields 3q = 45,
so q = 15 quarters.
92. The area of the garden is the sum of the areas of a 12-by-8 foot rectangle and the area of a circle of diameter 8 feet. The garden has area
(12 × 8) + (π × 42) = 96 + 16π ft2. To fill the garden with topsoil uniformly to a depth of 2 inches = 1/6 foot, the total volume of topsoil needed is
(1/6)(96 + 16π) = 16 + (8/3)π ft3. There are 3 × 3 × 3 = 27 ft3 in one cubic yard, so to fill the garden as described, the volume of topsoil needed is
(16 + (8/3)π)/27 ≈ 0.9 yd3.
93. In the proportion a/x = x/b, if we replace a, x and b with m, 9 and m + 3.3, respectively, we have m/9 = 9/(m + 3.3). Cross-multiplying yields
m(m + 3.3) = 81 → m2 + 3.3m − 81 = 0. Using the quadratic formula, for a = 1, b = 3.3 and c = −81, we have m = (−b ± √(b2 − 4ac))/(2a) =
(−3.3 + √(3.32 − 4(1)(−81)))/(2(1)) = 7.5.
94. Let r represent the rate, in miles per hour, that Oscar walked for the first 40 minutes, or 2/3 hour. Then he walked at a rate of r + 1 mi/h for
10 minutes, or 1/6 hour. Because Distance = Rate × Time, the distance he walked is r × (2/3) + (r + 1) × (1/6). Setting this equal to 2.7 and
multiplying the equation by 6 gives us 4r + r + 1 = 16.2, so 5r = 15.2 and r = 3.04 mi/h.
95. Since Allen can paint the entire fence by himself in 9 hours, he paints 1/9 of the fence every hour. Similarly, since Asanji can paint the entire
fence by himself in 7 hours, then he paints 1/7 of the fence every hour. Working together, they paint 1/9 + 1/7 = (7 + 9)/63 = 16/63 of the fence
every hour. In other words, together they can paint the entire fence in 63/16 hours, or 63/16 × 60 = 236.25 minutes. Therefore, it will take them
236.25 × (2/3) = 157.5 minutes to paint 2/3 of the fence.
A
60
0
45°





96. After two hours, traveling at a speed of 300 mi/h, plane A will have flown 300 × 2 = 600 miles, and traveling at a speed of 250 mi/h,
plane B will have flown 250 × 2 = 500 miles. The figure shows the two routes on a coordinate grid. Notice that the route taken by plane A
is at a 45-­degree angle with both the x- and y-axes and forms the hypotenuse of a 45-45-90 right triangle. Similarly, the route taken by
plane B is at a 30-degree angle with the x-axis and at a 60-degree angle with the y-axis and forms the hypotenuse of a 30-60-90 right
triangle. Using properties of 45-45-90 right triangles, we see that plane A is 600 ÷ √2 = 300√2 miles north and 300√2 miles
45°
60°
east of the starting point. Using properties of 30-60-90 right triangles, we see that plane B is 500 ÷ 2 = 250 miles south and
0
50
250√3 miles west of the starting point. So the distance between the two planes is the length of the hypotenuse of a right
triangle
30°
B
with legs of length 300√2 + 250√3 miles and 300√2 + 250 miles. We can use the Pythagorean Theorem to determine
250√3
300√2
this distance. So the planes are √((300√2 + 250√3)2 + (300√2 + 250)2) ≈ 1091 miles apart.















MATHCOUNTS 2015-2016
71
300√2
250
97. In the first figure, the sum of the bottom and top numbers is 8 + 1 = 9, and the difference between the bottom and top numbers is 8 − 1 = 7.
In the second figure, the same relationships hold true, since 7 + 5 = 12 and 7 − 5 = 2. In the third figure, for these same relationships to hold, it
must be true that x + (−3) = 0 and x − (−3) = 6. Solving either of these equations yields x = 3.
98. Since the answer is requested in inches, let’s first convert the diameter of 122 feet 8 inches to 122 × 12 + 8 = 1472 inches. The circumference
of a circle with this diameter is 1472π inches. If the chef cuts 150 equal slices, the distance between slices along the circumference would be
1472π ÷ 150 ≈ 30.8 inches.
99. Since g(−4) = (−4)2 − 2(−4) + 5 = 16 + 8 + 5 = 29, and f(10) = 3(10) − 4 = 30 − 4 = 26, then g(−4) + f(10) = 29 + 26 = 55.
100. Since the reservoir declines by 10% each year, it retains 90% of the amount it had the previous year. After two years, the amount of water is
0.9 × 0.9 = 0.81 of what it was, or 81%. After three years, it’s 0.93 = 0.729 or 72.9% of what it was. We are looking for the exponent n such that
0.9n is less than 0.5 for the first time. We see that 0.96 ≈ 0.531, but 0.97 ≈ 0.478. Therefore, the minimum number of whole years it will take for the
water to be less than half its original amount is 7 years.
Warm-Up 7
101. Each audience member has a total of 1 + 2 + 3 = 6 points to assign to the three finalists. That’s 6 × 100 = 600 points to be awarded. Since
the winner must have fewer points than each of the other finalists, but we’re asked to find the winner’s greatest possible point total, the point totals
for the three finalists, without ties, should be as close to each other as possible. The distribution of points would be as follows: 201 points for third
place, 200 points for second place and 199 points for first place.
102. The obvious segment lengths are 1, 2, 3 and 4 units. Less obvious, however, are the lengths of the segments whose endpoints are not in the
same row or in the same column. The figures below show these additional segments.
2
1
1
5
1
2
1
10
3
1
17
4
2
13
2 2 2
2
3
2
2 5
4
3
3 2
3
3
5
4
4
4 2
4
Notice that each of these segments is the hypotenuse of a right triangle. Therefore, the 10 different lengths of the segments can be determined
using the Pythagorean Theorem. In all, segments can be drawn having 4 + 10 = 14 distinct lengths.
103. The surface area of a 1-inch cube is 6 × 1 × 1 = 6 in2, while the surface area of a 1-foot cube is 6 × 12 × 12 = 864 in2, which is 144 times
as much. Therefore, the amount of time it takes for a 1-inch cube of ice to melt will be 1/144 of the 3 hours, or the 3 × 60 = 180 minutes, it takes a
1-­foot cube of ice to melt. So it will take 180/144 = 5/4 = 1.25 minutes for a 1-inch cube of ice to melt.
104. Since the 24 students have an average score of 88, the sum of their scores is 24 × 88 = 2112. When the four lowest scores are excluded,
the sum of the other scores is 20 × 94 = 1880. The four lowest scores must have a sum of 2112 − 1880 = 232, giving the four lowest scores an
average of 232 ÷ 4 = 58. Similarly, when the four highest scores are excluded, the sum of the other scores is 20 × 86 = 1720. The four highest
scores must have a sum of 2112 − 1720 = 392, with an average of 392 ÷ 4 = 98. The absolute difference between these two averages is 98 − 58
= 40.
105. We are told that each time the ball strikes the ground, it bounces back to a height one-fourth of its previous height. Initially, the ball travels
down 16 feet. The first four times the ball strikes the ground, it bounces back to the following heights, in feet: 4, 1, 1/4, 1/16. But with each bounce
up, it travels the same distance back down, so that by the time the ball strikes the ground for the fifth time, it will have traveled a total of 16 + 4 + 4
5
+ 1 + 1 + 1/4 + 1/4 + 1/16 + 1/16 = (256 + 2(64) + 2(16) + 2(4) + 2(1))/16 = 426/16 = 26 8 feet.
106. Based on the formula Distance = Rate × Time, we know that the distance the jet travels east equals its rate R1 times 3 1/2 = 7/2 hours. But
this is equivalent to the distance the jet travels west, which equals its rate R2 times 4 1/4 = 17/4 hours. We have (7/2)R1 = (17/4)R2. So the ratio of
the jet’s speed going east to its speed going west is R1/R2 = (17/4)/(7/2) = (17/4) × (2/7) = 17/14.
107. We know that angle BAC is acute, meaning it has a degree measure between 0 and 90. We also know that triangle ABC has integer side
lengths. If the measure of angle BAC were 90 degrees, the length of side AC would be √(162 − 152) = √(256 − 225) = √31, which has a value
between √25 = 5 and √36 = 6. It follows, then, that AC ≥ 6 since anything less would result in angle BAC being obtuse. Also AB + BC = 15 + 16
= 31. That means AC ≤ 30 since anything greater would result in ABC not being a triangle. This property of triangles is called the Triangle Inequality.
Therefore, side AC can have any integer length from 6 to 30 units. That’s a total of 25 possible integer lengths.
108. The operation # gives the sum of the quotient and the product of two numbers. Starting with the expression within parentheses, we have 6 # 2
= 6/2 + 6 × 2 = 3 + 12 = 15. Now we evaluate 15 # 15 to get 15/15 + 15 × 15 = 1 + 225 = 226.
109. Twenty percent of 60 minutes is 1/5 × 60 = 12 minutes, so the time must be 4:12 p.m. From 1:00 p.m. to 4:12 p.m. is 3 hours 12 minutes =
3 × 60 + 12 = 180 + 12 = 192 minutes. Since there are 5 hours = 5 × 60 = 300 minutes from 1:00 p.m. to 6:00 p.m., the percent of time that had
elapsed by 4:12 p.m. is 192/300 = 64/100 or 64%.
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MATHCOUNTS 2015-2016
7 dollars
2
1
1
1
0
0
0
0
0
0
3 dollars
0
2
1
0
5
4
3
2
1
0
1 dollar
1
2
5
8
0
3
6
9
12 15
110. The table shows the 10 combinations of coins that can be used to pay exactly 15 dollars in Woodington.
Warm-Up 8
111. Initially, only the committee chairperson knew the theme. At 8:00 a.m. he or she told the three other committee members, making the total 1 +
3 = 4 people who knew the theme. Then each of those three committee members told 3 other people, so that by 9:00 a.m., 1 + 3 + 9 = 13 people
knew the theme. There appears to be a pattern. At 8 a.m. 31 = 3 new people were told, and by 9 a.m. 32 = 9 new people were told. Therefore, the
total number of people who knew the theme by 11:00 a.m. was 1 + 3 + 9 + 27 + 81 = 121 people.
112. Since |(3 ▲ x) − 8| = 2, it follows that 3 ▲ x = 6 or 3 ▲ x = 10. If 3 ▲ x = 6, then x = −3 or x = 9. And if 3 ▲ x = 10, then x = −7 or x = 13.
Thus, the sum of all numbers x that satisfy the given equation is −3 + 9 + (− 7) + 13 = 12.
113. The prime factorization of 300 is 22 × 3 × 52. That means among the prime factorizations of 150 and m, there must be at least one each of 22,
31 and 52, with 21 and 51 also being possible factors. The prime factorization of 150 is 2 × 3 × 52, which includes 31 and 52, but not 22. Therefore, 22
must be part of the prime factorization of m. The possible prime factorizations for m, then, are 22, 22 × 3, 22 × 5, 22 × 3 × 5, 22 × 52 and 22 × 3 × 52.
Therefore, the 6 possible integer values for m are 4, 12, 20, 60, 100 and 300.
114. On the 30th day, Carrick must have read 50 + 29 × 10 = 340 words. The total numbers of words Carrick read in 30 days is 50 + 60 + 70 +
…+ 320 + 330 + 340. This is an arithmetic series, which has a value equal to the sum of the first and last terms times the number of terms divided
by 2. That’s (340 + 50) × 30 ÷ 2 = 5850 words.
115. Bailey, Cooper and Dallas have totals of 21, 14 and 19 points, respectively. Currently, Austin has 17 points and needs 5 or more points to win.
The common suit for Austin’s cards is diamonds. So far, the Ace, King, Queen, 9 and 8 of diamonds have been dealt. That means the Jack, 10, 7, 6,
5, 4, 3 and 2 of diamonds are among the 52 − 11 = 41 cards left in the deck from which Austin’s final card will be dealt. Austin only will win if dealt
the 5, 6, 7, 10 or Jack of diamonds. The probability of that happening is 5/41.
116. If the set contained only positive numbers, to get the least possible value of (l × m)/n we would minimize the quantity l × m while maximizing n.
However, this set contains both positive and negative numbers. If we can get the result to be negative, then maximizing |l × m| and minimizing |n| will
result in the least possible value of (l × m)/n. Doing so, we find that (2 × 12)/(−1/6) = −24 × 6 = –144 is the least value possible.
117. There are 3 × 2 × 1 = 6 ways to replace X, Y and Z with 3, 5 and 7. Only 3 of them make the inequality true. They are 0.35 < 0.7, 0.53 < 0.7
and 0.37 < 0.5.
118. Let’s assume we have the following list of eight integers: 8, 8, 8, 8, 8, 8, 8, 8. The mean, median and unique mode of this list are all 8. But we
also need the range of the numbers to be 8. Subtracting four from the first 8 and adding it to the last gives us the following: 4, 8, 8, 8, 8, 8, 8, 12.
We now have a range of 8, but let’s see if we can get a range of 8 with a greater maximum value than 12. One possibility is this list: 5, 6, 8, 8, 8, 8,
8, 13. The list that gives us the greatest maximum value is the following: 6, 6, 6, 8, 8, 8, 8, 14. Therefore, the greatest possible integer that could be
in the list is 14.
119. If x = 0.19 and y = 0.29, then xy = 0.0551, so the hundredths digit of the product could be as large as 5. On the other hand, x < 0.2 and
y <0.3, so xy < 0.2 × 0.3 = 0.06. Therefore, the hundredths digit cannot be larger than 5, and the greatest possible value of the hundredths digit of
the product is 5.
120. Draw segment HI through E and parallel to side AD, as shown. Since segments BF and AG bisect angles ABC and BAD, respectively,
triangles AHE and BHE are congruent 45-45-90 right triangles with AH = BH = EH = 9 units. Notice that similar triangles EFG and EAB are in the
ratio 1:3, so FG/AB = 1/3 and FG/18 = 1/3 → FG = 18/3 = 6 units. Thus, the area of triangle EFG is (1/2) × 6 × 3 = 9 units2.
Workout 5
121. Enlarging the area of the photo by 150% means that the new area is 250%, or 2.5, times the original area. The scale factor for the width and
the height will be the square root of 2.5, since each of them gets multiplied by this number separately. The original height was 3 inches, so the height
of the enlarged photo is 3 × √2.5 ≈ 4.7 inches.
122. If we let a and b represent the two numbers, we can write the equations a + b = 7 and a − b = 18. We can conclude that one of the numbers
is negative since their difference is greater than their sum. Adding these two equations yields a + b + a − b = 7 + 18 → 2a = 25 → a = 12.5.
Substituting this back into the first equation, we see that 12.5 + b = 7 → b = 7 − 12.5 → b = −5.5. Therefore, the product a × b = 12.5 × (−5.5) =
–68.75.
A
4
4 2G
B
r
D r
r
4 2 − 4F
123. Solution 1: The figure shows circle D inscribed in triangle ABC. Hypotenuse AC is 8 cm, so it follows that AB = BC = (8/√2) × (√2/√2)
= 8√2/2 = 4√2 cm. Since circle D is inscribed in triangle ABC, its center is the intersection of its three angle bisectors. Radius DE
E
coincides with the bisector of angle B, so E is the midpoint of side AC. Thus, AE = EC = 4 cm. When we draw radius DF perpendicular
to side BC and radius DG perpendicular to side AB, as shown, we create kites CEDF and AEDG with DE = DF = GB = DG = FB =
4
r and EC = FC = AE = AG = 4 cm. That means r = 4√2 − 4 cm. Using the area formula, we see the circle has area π × (4√2 − 4)2
C
4
≈ 8.6 cm2.
MATHCOUNTS 2015-2016
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Solution 2: By symmetry, AE = EC = 8/2 = 4, and because triangle BEC is an isosceles right triangle, EB = 4 also. Let r = DE = DF be the radius
of the circle. Triangle DFB is an isosceles right triangle, so DB = r√2. Therefore, 4 = EB = ED + DB = r + r√2 = r(1 + √2).Solving for r, we see that
r = 4/ (1 + √2), so the circle has area π × (4/(1 + √2))2 ≈ 8.6 cm2.
124. Because the conical tank and the cone of water in the tank are similar three-dimensional geometric figures, the ratio of the depth of the water
to the height of the tank must be the cube root of the ratio of the volume of the water to the volume of the tank. Since the ratio of the volumes is
3
given as 1/2, the ratio of the depth of the water to the height of the tank is ratio is √(1/2) ≈ 0.79.
125. Moesha’s 18th birthday comes 18 − 13 = 5 years after her 13th birthday, so t = 5. Substituting this value into the formula H(t) = 61 + 20.4t, we
see that Moesha should expect her height on her 18 birthday to be H(5) = 61 + 20.4(5) = 61 + 4 = 65 inches.
126. Let’s first consider the 2-digit perfect squares since there are only six of them. They are 16, 25, 36, 49, 64 and 81. The digit that
belongs in the cell in the top row, third column is the units digit of one 2-digit perfect square and the tens digit of another. The same is
true for the digit that belongs in the cell in the bottom row, first column. There are four possible pairs for each corner (16, 64), (36, 64),
(64, 49) and (81,16). This gives 16 possible ways to fill in all but the middle cell. Let’s try the pattern shown in the first grid, which uses
the smallest sum of digits for each corner (1 + 6 + 4 = 11). There is a 3-digit perfect square with a hundreds digit of 1 and a units digit
of 4; namely, placing a 4 in the center cell results in the perfect square 144. So this is one set of perfect squares that works, and these
digits have a sum of 1 + 6 + 1 + 4 + 4 + 6 + 4 = 26. Because the corner sums for (64, 49) and (81, 16) are 19 and 15, respectively,
we cannot do better by using either of these patterns. Because the corner sum for (36, 64) is 13, we cannot do better by using this
pattern in both corners. (There is such a grid that works, but its digit sum is 28.) The only other arrangement to try is to have (16, 64)
in one corner and (36, 64) in the other, as shown in the second grid. But here there is no middle digit that will create 3-digit perfect
squares in both the middle row and the middle column. We conclude that the minimum possible digit sum is 26.
1 6
1
4
6 4
1 6
3
4
6 4
127. For the top row of six squares, each square can be colored with one of three colors, so there are 36 = 729 ways to color the original top row.
When we rotate the rectangle 180 degrees, the colors for the original bottom row will appear in the new top row in the reverse arrangement they
were in for the original bottom row. Since each location must be a different color than it was before the rotation, each location in the original bottom
row must be colored one of the two colors not chosen for the reversed location in the original top row . Thus, there are 26 = 64 ways to color the
bottom row. That’s a total of 729 × 64 = 46,656 ways to color the squares as described.
128. Solution 1: Starting from the bottom of the table, let’s assign the letters a, b, c and d to represent the number of people in each category. We
know that there are 12 family members, so a + b + c + d = 12. Since the total cost for the family was $73 and we have the list of fees, we can write
the equation 10a + 7b + 3c + 0d = 73. Notice that if b = c, then 7b + 3c is a multiple of 10. If this multiple of 10 is added to 10a, then the total
would be a multiple of 10. There are only two ways we can get a dollar amount that ends in a 3. Either we have b − c = 9, since 7 × 9 = 63, or we
have c − b = 1, since 1 × 3 = 3. We are told that there are more “others” than seniors, which means a > b. This eliminates the first case and tells us
that c − b = 1, or c = b +1. Substituting b +1 for c (and ignoring d for the moment), we have 10a + 7b + 3(b + 1) = 73. This can be rewritten as
10(a + b) + 3 = 73, then 10(a + b) = 70 and finally a + b = 7. We want the maximum number of children, but we must have a > b, so we settle on
a = 4 and b = 3. This means c = 4 and d = 1. To confirm our results, we have 4 + 3 + 4 + 1 = 12 people, and the cost is 10 × 4 + 7 × 3 + 3 × 4
+ 0 × 1 = $73. The maximum is 4 children ages 3 to 12.
Solution 2: Let w, x, y and z represent the numbers of individuals 2 and under, from 3 to 12, 62 and older and all others, respectively. Based on the
information provided, we have the equations 0w + 3x + 7y + 10z = 73 and w + x + y + z = 12. Let’s assume x = 1. Then 7y + 10z = 70. This is
true only if y = 10 and z = 0, or if y = 0 and z = 7. But since we were told that z > y, only the second scenario is possible. Now let’s suppose x =
2. Then 7y + 10z = 67. This is true when y = 1 and z = 6. Let’s try x = 3. Now 7y + 10z = 64. This is true when y = 2 and z = 5. Now let x = 4. So
7y + 10z = 61. This is true when y = 3 and z = 4. We need not try x = 5 since higher values of x result in y > z. Therefore, the maximum possible
number of children ages 3 to 12 is 4 children.
129. If the two positive integers are x and y, then the sum of their reciprocals is (x + y)/(xy) = 11/(xy). To maximize this quotient, we want to minimize
the denominator, and this is easily seen to occur when the integers are 10 and 1. Therefore, the maximum possible sum is 11/10.
130. An integer that has the same number of even factors and odd factors must have exactly one factor of 2. Another way to say this is that they
are numbers that are multiples of 2, but not multiples of 4. The numbers are all 2 times an odd number. One out of every four consecutive positive
integers has this property (2, 6, 10, ...), so there are 100/4 = 25 such positive integers less than or equal to 100.
Workout 6
131. Consider, first, the equation a2 + b2 = c2. We recognize this as the Pythagorean Theorem, which has infinitely many positive integer solutions,
called Pythagorean Triples. Another theorem, Fermat’s Last Theorem, states that there are no positive integer solutions to equations of the form
an + bn = cn for n > 2. So a3 + b3 = c3 has no positive integer solutions. In this problem, however, we are allowing negative integers as well. We
need to determine if there are solutions when one or more of a, b or c is negative. If a < 0 and b < 0, then a3 < 0 and b3 < 0. This leads to a negative
value for c3 < 0 and, thus, c < 0 also. The equation (−a)3 + (−b)3 = (−c)3 → −a3 − b3 = −c3 is just the negative version of a3 + b3 = c3, which has
no positive integer solutions. Now let’s suppose that a is positive and b is negative, and write b = –b'. Then we essentially have a difference of two
cubes equal to another cube. Assume first that |a| > |b|, which implies that c > 0. Then the equation a3 + (−b')3 = c3 → a3 − b'3 = c3 → a3 = b'3 + c3,
and we already know that the sum of two positive cubes can never equal another cube. If |a| < |b| then a similar argument applies. This leaves only
solutions with zeros, sometimes referred to as “trivial solutions.” For example, if a = 5 and b = −5, we have 53 + (−5)3 = 0. To maximize the sum a +
b + c, we can choose a = 0, b = 10 and c = 10. That gives us 03 + 103 = 103, and the value of a + b + c = 0 + 10 + 10 = 20.
74
MATHCOUNTS 2015-2016
132. The 8 multiples of 6 from 1 to 50 are 6, 12, 18, 24, 30, 36, 42 and 48. The 6 multiples of 8 from 1 to 50 are 8, 16, 24, 32, 40 and 48. In
total, there are 12 different multiples of 6 and 8 (24 and 48 are multiples of both). Therefore, the probability that the card Matt draws at random is a
multiple of 6 or 8 is 12/50 = 6/25.
133. It is easy to see that the length of each segment in Figure 1 is 1 unit, and the total length is 2 units. In Figure 2, AD = BD = CD, and segments
AD, BD and CD each are a subset of a median of the triangle. Point D, the intersection of the three medians, is known as the centroid of triangle
ABC. A property of the centroid of a triangle is that a segment drawn from a vertex of the triangle to the centroid has a length that is 2/3 the length
of the median drawn from that vertex to the midpoint of the opposite side. In an equilateral triangle, the median drawn from any vertex is an altitude of
the triangle. Given that triangle ABC has side length 1 unit, by the properties of 30-60-90 right triangles, we know its altitude is √3/2 unit. Therefore,
AD = BD = CD = (2/3) × (√3/2) = √3/3 unit, and AD + BD + CD = 3 × (√3/3) = √3 units. The absolute difference between the sum of the lengths
of the two segments in Figure 1 and the sum of the lengths of the three segments in Figure 2 is 2 – √3 units.
134. Let s represent the sales tax rate expressed as a decimal. Then Erik paid 2s as a tipping rate. Alicia’s tipping rate was 2(1 + s)/10. Since the
tip amounts were the same, these two tipping rates must be equal. We need to solve the equation 2s = 2(1 + s)/10. Doing so, we have 10s = 1 + s
→ 9s = 1 → s = 1/9 ≈ 11.1%. Therefore the percent sales tax was approximately 11.1%.
1
135. Rectangle BCEF, shown here, has vertices that coincide with four vertices of hexagon ABCDEF. The measure of each
interior angle of a regular hexagon is 180 − (360 ÷ 6) = 120 degrees. If we draw the diagonal from A to D, we create four
congruent 30-60-90 right triangles. Consider, for example, triangle ABG with AB = 1 unit. By properties of 30-60-90 right
triangles, it follows, then, that AG = 1/2 unit and BG = √3/2 unit. In addition, BF = 2 × BG = 2 × √3/2 = √3 units. Now, since
BC = 1 unit, we calculate the area of rectangle BCEF to be 1 × √3 = √3 units2.
B
30°
A
60°
F
G
C
E
D
136. A uniform yearly percent of increase means the price increase was compounded from year to year over the 10-year period. So with starting
price p, ending price q and uniform yearly rate of increase r, we have q = p × (1 + r)10 or q/p = (1 + r)10. Since the price increased from $1.72 to
$3.84, we substitute to get 3.84/1.72 = (1 + r)10. Taking the 10th root of each side, we get 10√(3.84/1.72) = 1 + r → 10√(3.84/1.72) − 1 = r. Using a
calculator, we find that 10√(3.84/1.72) − 1 ≈ 0.084. That’s an annual percent increase of 8.4%.
A
E
57°
k
81°
99°
137. The sum of the angles in quadrilateral FEBG, shown here, must be 360 degrees. We already know that mEFG = 90 degrees
since lines j and k are perpendicular. Also, FEB is supplementary to 1, so mFEB = 180 − 81 = 99 degrees. Since ABCD is
a parallelogram, angles A and B also are supplementary, so mB = 180 − 57 = 123 degrees. So far, we have mEFG +
G
mFEB + mB = 90 + 99 + 123 = 312 degrees. Therefore, mFGB = 360 − 312 = 48 degrees. Since FGB and 2 are
2
vertical angles they are congruent and therefore, m2 = 48 degrees.
Items
Cost Per Item ($)
123°
48°
F
D
B
C
j
138. To find the shipping cost per item, we need to divide each total shipping cost by the number of items. The table shows
the shipping cost per item for each number of items. The least shipping cost a customer can pay per item is $0.84.
1
2
3
4
5
6
7
8
1.85 ÷ 1 = 1.85
2.15 ÷ 2 = 1.075 ≈ 1.08
2.65 ÷ 3 = 0.883 ≈ 0.88
3.35 ÷ 4 = 0.8375 ≈ 0.84
4.25 ÷ 5 = 0.85
5.35 ÷ 6 = 0.8916 ≈ 0.89
6.65 ÷ 7 = 0.95
8.15 ÷ 8 = 1.01875 ≈ 1.02
139. Solution 1: Since we are not given the side lengths for the rectangle, only their ratio, let’s assume
9.85 ÷ 9 = 1.094 ≈ 1.09
9
rectangle ABCD is a 5 × 7 rectangle, as shown. By the properties of 45-45-90 right triangles, we know
11.75 ÷ 10 = 1.175 ≈ 1.18
10
that isosceles
right
triangles
ABH
and
CDF
have
legs
of
length
7/√2
×
√2/√2
=
(7√2)/2
units.
Also
by
the
5
G
E
properties of 45-45-90 right triangles, isosceles right triangles ADE and BCG have legs of length 5/√2 × √2/√2 = (5√2)/2 units.
H
From the figure, we see that the side length of the shaded square is simply the difference between the lengths of legs AH and AE
C
D
of triangles ABH and ADE, respectively. So EH = (7√2)/2 − (5√2)/2 = (2√2)/2 = √2 units, and the shaded region has area (√2)2 =
2
2 units . The area of rectangle ABCD is 5 × 7 = 35 units2, and the shaded region represents 2/35 ≈ 0.057 = 5.7% of this area.
A
7
B
F
A
7
B
Solution 2: Again, we’ll assume rectangle ABCD is a 5 × 7 unit rectangle. Then by the properties of 45-45-90 right triangles,
isosceles right triangles ADE and BCG have legs of length 5/√2 × √2/√2 = (5√2)/2 units. Also by the properties of 45-45-90
5
G
E
right triangles, the altitudes from vertex E to side AD and from vertex G to side BC of triangles ADE and BCG, respectively, have
H
length 5/2 units. The combined length of these altitudes is 10/2 = 5 units. But since rectangle ABCD measures 7 units across,
C the shaded square must have diagonal length EG = 7 − 5 = 2 units. The area of a square is half the square of its diagonal length,
D
so the shaded square EFGH has area (1/2) × (2)2 = 2 units2. Rectangle ABCD has area 5 × 7 = 35 units2 so the shaded region
represents 2/35 ≈ 0.057 = 5.7% of the rectangle’s area.
F
140. To determine the value of the first expression, (10  11)  12, we evaluate what’s inside the parentheses and get 10  11 = 10 × 11 + 3 =
113. Then we evaluate 113  12 and get 113 × 12 + 3 = 1359. Next, we need to determine the value of the second expression, 10  (11  12),
by first evaluating what’s inside the parentheses to get 11  12 = 11 × 12 + 3 = 135. Then we evaluate 10  135 and get 10 × 135 + 3 = 1353.
The absolute difference between the two expressions is 1359 − 1353 = 6.
Warm-Up 9
141. We have f(g(−2)) = f(2(−2) + 5) = f(1) = 3(1)2 − 7 = −4. We also have g(f(−2)) = g(3(−2)2 − 7) = g(5) = 2(5) + 5 = 15. The absolute
difference between these values is |−4 − 15| = |−19| = 19.
142. Notice that the sums 1 + 2015 = 2016 and −2 −2014 = −2016 cancel one another, as do 3 + 2013 = 2016 and −4 −2012 = −2016.
The sums continue to cancel each other out until we get to 1007 + 1009 = 2016, to which we must add −1008 and get 2016 −1008 = 1008.
MATHCOUNTS 2015-2016
75
143. Ted and Fred are approaching each other at a constant speed of 30 + 30 = 60 mi/h. Initially they are 60 miles apart, so it will take them 1 hour
to meet, each man traveling half the distance, or 30 miles. At a rate of 45 mi/h, the carrier pigeon will have flown a total of 45 miles during that time.
0
1
y 144. The value of 11 in base 3 is 113 = 1 × 3 + 1 × 3 = 1 + 3 = 4. Now we calculate 3(113) = 3 × 4 = 12 in base 10, so x = 10.
(0, 2)
(1, 2)
(2, 2)
(0, 1)
(1, 1)
(2, 1)
(0, 0)
(1, 0)
(2, 0)
x
145. The coordinate grid shown here has the nine lattice points with integer coordinates (x, y) such that 0 ≤ x ≤ 2 and 0 ≤ y ≤ 2.
There are 9C4 = 9!/(5! × 4!) = 126 ways to choose 4 of these 9 lattice points. However, not all of these combinations can
represent vertices of quadrilaterals. In fact, any combination that includes three collinear lattice points from one of the 3 rows,
one of the 3 columns or one of the 2 diagonals won’t work. The 3 + 3 + 2 = 8 sets of three collinear lattice points are each
included in 6 different combinations of 4 lattice points. That’s a total of 8 × 6 = 48 combinations of 4 lattice points that won’t
work. We subtract these from our original calculation to get a total of 126 − 48 = 78 quadrilaterals.
146. The side length of the equilateral triangle is 6 units. D is the midpoint of base BC, so DC = 3 units. Those familiar with 30-60-90 right
triangles will know that altitude AD is 3√3 units, which can be confirmed with the Pythagorean Theorem. M is the midpoint of altitude AD, so MD =
3√3/2 units. We can use the Pythagorean Theorem to find MC since it is the hypotenuse of right triangle CDM. We have MC2 = (3√3/2)2 + 32 =
27/4 + 9 = (27 +36)/4 = 63/4 → MC = √(63/4) = (3√7)/2 units. So a = 3, b = 7, c = 2 and a + b + c = 3 + 7 + 2 = 12.
147. The difference of the perimeters of squares A and B is 28, so we can write 4y − 4x = 28, which simplifies to y − x = 7. The difference in their
areas is 161, so we can write y2 − x2 = 161. This difference of two squares can be factored into (y − x)(y + x) = 161. Substituting 7 for y − x, we
get (7)(y + x) = 161 and y + x = 161/7 = 23.
148. Four pairings of 2-tiles can be used to create four 4-tiles. Then use two pairings of 4-tiles to produce two 8-tiles. Pairing these two 8-tiles
results in a 16-tile. These same 4 + 2 + 1 = 7 pairings can be repeated to create another 16-tile. Finally, the pairing of two 16-tiles results in a
32-tile. In all, 7 + 7 + 1 = 15 pairings were needed.
149. Since 3 is the least of the five values and 22 is the greatest, there are three cases to consider for possible values of x.
Case 1: We have the five ordered integers 3, x, 7, 12, 22. In this case, the possible values of x are 4, 5 or 6, and the median of the set will be 7.
For each of these possible values of x, the mean will be greater than 7 so we can write the inequality (44 + x)/5 − 7 ≥ 2. Solving for x, we see that
(44 + x)/5 ≥ 9 → 44 + x ≥ 45 → x ≥ 1. Because 4, 5 and 6 all satisfy this condition all three are possible values for x.
Case 2: We have the five ordered integers 3, 7, x, 12, 22. In this case, the possible values of x are 8, 9, 10 and 11, and the median of the set will be
x. For these possible values of x, the mean will be greater than or equal to x, so we can write the inequality 44 + x)/5 − x ≥ 2. Solving for x, we see
that 44 + x − 5x ≥ 10 → −4x ≥ −34 → x ≤ 8 1/2. The only value that satisfies this condition is 8.
Case 3: We have the five ordered integers 3, 7, 12, x, 22. In this case, the possible values of x are 13, 14, 15, 16, 17, 18, 19, 20 and 21, and the
median will be 12. For each of these possible values of x, the mean will be between 11.4 and 13 so the absolute difference between the mean and
median will be less than 2 so none of these possible values of x satisfies the given conditions. The possible values for x, then, are 4, 5, 6 and 8. The
sum of these values is 4 + 5 + 6 + 8 = 23.
150. Recall that proper divisors of a number n include all positive factors of n except n itself. Let’s first consider positive integers with two digits. If
the tens digit is 1, a units digit of 2, 3, 4, 5, 6, 7, 8 or 9 results in a 2-digit divisive integer. Then, if the tens digit is 2, a units digit of 4, 6 or 8 results
in a 2-digit divisive integer. If the tens digit is 3, a units digit of 6 or 9 results in a 2-digit divisive integer. Finally, if the tens digit is 4, a units digit of 8
gives us a divisive integer. Thus, there are 8 + 3 + 2 + 1 = 14 divisive 2-digit positive integers. Next, let’s consider 3-digit integers. A hundreds digit
of 1 followed by any of the 2-digit divisive integers with a tens digit of 2, 3 or 4 results in a 3­-digit divisive integer. A hundreds digit of 2 followed
by 48 gives us the only other 3-digit divisive integer. That’s a total of 3 + 2 + 1 + 1 = 7 divisive 3-digit positive integers. Now we consider 4-digit
divisive integers. A thousands digit of 1 followed by 248 gives us the only 4-digit divisive positive integer. There is no way to form a divisive integer
with five digits, so that brings the grand total to 14 + 7 + 1 = 22 positive integers that can be classified as divisive.
Warm-Up 10
151. If there were only 5 players on Coach Washington’s team, he would have only 1 choice for the starting 5 players. If there were 6 players, he
would have 6 choices for the starting 5, since there are 6 options in which 1 player would be left out. If there were 7 players, the coach would have
21 choices for the starting 5 since there are 7C2 = 7!/(2! × 5!) = (7 × 6)/(2 × 1) = 42/2 = 21 options in which 2 players would be left out. So, the
coach needs at least 7 players in order to pick a different starting 5 for all 20 games.
152. The first figure is a single cube. The second is composed of (2 × 2 × 2) − (1 × 1) = 8 − 1 = 7 cubes. The third has (3 × 3 × 3) − [(1 × 1) +
(2 × 2)] = 27 − (1 + 4) = 22 cubes. If this pattern continues, the next figure will be composed of (4 × 4 × 4) − [(1 × 1) + (2 × 2) + (3 × 3)] =
64 − (1 + 4 + 9) = 50 cubes.
153. Solution 1: Consider the arithmetic sequence with the first term a and common difference d. The first seven terms can be written as a, a + d,
a + 2d, a + 3d, a + 4d, a + 5d, a + 6d. The sum of the second and seventh terms would be (a + d) + (a + 6d) = 2a + 7d. The sum of the fourth
and fifth terms would be (a + 3d) + (a + 4d) = 2a + 7d. Notice that these two sums are equal. We are told that the sum of the second and seventh
terms is 85. It follows, then, that the sum of the fourth term, 38, and the fifth term also is 85. So the fifth term must be 85 − 38 = 47.
Solution 2: Let x, y and d be the second term, seventh term and common difference, respectively. Then x + 2d = 38 and y = 38 + 3d. Adding these
two equations gives us x + y + 2d = 38 + 38 + 3d, so d = x + y − 76. Because we are told that x + y = 85, we have d = 9. Therefore, the fifth
term is 38 + 9 = 47.
76
MATHCOUNTS 2015-2016
y
154. Point B is the image of A after it is rotated 90° counterclockwise about O, and it has coordinates (−7, 3), as shown. Since the angle
of rotation, O, has degree measure 90, it follows that triangle ABO is a right triangle. It also is an isosceles triangle since AO = BO.
B
That means the area of triangle ABO is half the area of the square of side length AO = √(32 + 72) = √(9 + 49) = √58. Therefore, the
area of triangle ABO is (√58)2/2 = 58/2 = 29 units2.
A
7
O
3
155. We can rewrite this as x2 + y2 = 502. This is an equation for the circle centered at (0, 0) with radius 50 units. We are looking for points that lie
on this circle that have integer coordinates. The obvious ones are the points (50, 0), (0, 50), (−50, 0) and (0, −50). To find the remaining ordered
pairs, we’ll call to mind a few common Pythagorean Triples. First let’s consider a multiple of the 3-4-5 Pythagorean Triple, namely 30-40-50. This
gives us the points (30, 40), (30, −40), (−30, 40), (−30, −40), (40, 30), (40, −30), (−40, 30) and (−40, −30). Lastly we’ll consider 14-48-50, a
multiple of the 7-24-25 Pythagorean Triple. This gives us the points (14, 48), (14, −48), (−14, 48), (−14, −48), (48, 14), (48, −14), (−48, 14) and
(−48, −14). We’ve found a total of 4 + 8 + 8 = 20 ordered pairs that satisfy the given equation.
156. Observe the numbers 1 through 20 here written out in words. One way to
approach this problem would be to look at the letters in each number and narrow it
down to those numbers that contain letters with higher values. For example, several
numbers contain the letters V, W, X and Y, which have values of 22, 23, 24 and
25, respectively. We also might take note of how many letters these number have, since numbers with more letters are likely to have higher values.
In fact, this is a good technique, as SEVENTEEN has nine letters, more than any of the other numbers, and with a value of 19 + 5 + 22 + 5 + 14 +
20 + 5 + 5 + 14 = 109, 17 has the greatest value.
ONE
TWO
THREE
FOUR
FIVE
SIX
SEVEN
EIGHT
NINE
TEN
ELEVEN
TWELVE
THIRTEEN
FOURTEEN
FIFTEEN
SIXTEEN
SEVENTEEN
EIGHTEEN
NINETEEN
TWENTY
157. There are 5C2 = 5!/(3! × 2!) = (5 × 4)/(2 × 1) = 10 ways to choose the two 6th graders. There are 7C4 = 7!/(3! × 4!) = (7 × 6 × 5)/(3 × 2 × 1)
= 35 ways to choose the four 7th graders. There are 9C6 = 9!/(3! × 6!) = (9 × 8 × 7)/(3 × 2 × 1) = 84 ways to choose the six 8th graders. Therefore,
there are 10 × 35 × 84 = 29,400 ways to choose members of the Spirit Committee.
158. To determine the length of the base, we need to find the x-coordinates of the points where the parabola intersects the line y = −5. We can do
this by solving the quadratic equation 4x − x2 = −5, which can be rewritten as x2 − 4x − 5 = 0. Factoring we get (x − 5)(x + 1) = 0 so x − 5 = 0
and x = 5 or x + 1 = 0 and x = −1. Therefore, b = 5 − (−1) = 6. The x-coordinate of the vertex of the parabola is the midpoint of this base. Now we
need to determine the y-coordinate when x = 5 − (6/2) = 5 − 3 = 2. Substituting into the equation y = 4x − x2, we get y = 4(2) − 22 = 8 − 4 = 4.
Thus, the parabola has vertex (2, 4). It follows, then, its height is h = 4 − (−5) = 9. Using the formula A = (2/3)bh, we see the area of the shaded
region is A = (2/3)(6)(9) = 36 units2.
159. If we make a table of counting numbers and begin circling amounts that can be paid with coins in
8
1
2
3
4
5
6
7
9 10
the denominations of 5, 8 or 11 fizz, then—if we are careful and thorough—what’s left not circled are the 12 13 14 15 16 17 18 19 20 21
amounts that cannot be paid. As the table shows, 17 fizz is the greatest integer amount that cannot be
paid using a whole number of coins. We know there is no amount greater than 17 since we can make 18, 19, 20, 21 and 22 fizz and any amount
greater can be made by adding one or more 5-fizz coins to one of these amounts.
11
22
160. To get the longest possible diagonal, we will only fold the 8-inch side. Folding lengthwise in this manner is sometimes
referred to as a “hotdog fold.” As the figure shows, the three folds result in rectangles of size 4 × 12, 2 × 12 and 1 × 12.
The diagonal of the final 1 × 12 rectangle is √(12 +122) = √(1 + 144) = √145.
12
8
4
2
1
Warm-Up 11
161. Since we know the perimeter is 40 inches, we have 2n + n2 − 1 + n2 + 1 = 40 → 2n2 +2n − 40 = 0 → n2 + n − 20 = 0. If we solve this
quadratic equation by factoring, we get (n + 5)(n − 4) = 0, so n + 5 = 0 and n = −5, or n − 4 = 0 and n = 4. Since we are told that n is a positive
integer, it follows that n = 4, and the triangle has side lengths 2(4) = 8, (4)2 − 1 = 15 and (4)2 + 1 = 17 inches. The 17-inch side must be the
hypotenuse, so the two other sides are the legs. The area of the triangle, then, is (1/2) × 8 × 15 = 60 in2.
162. Let XYZ be a three-digit integer. If Y = 1, then X = 1, which results in the three-digit integer 110. If Y = 2, there are 2 possible values for X,
either 1 or 2, which results in the possible three-digit integers 121 and 220. If Y = 3, there are 3 possible values for X, 1, 2 or 3, which results in the
possible three-digit integers 132, 231 and 330. For each Y from 1 to 9, there are Y possible values for X and consequently Y possible three-digit
integers with Y in the tens place. Thus, there are 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 three-digit integers in which the tens digit is the sum of
the units digit and the hundreds digit.
163. Let’s call the first term a. Then the common difference is equal to 3a. Thus the first five terms of the sequence are a, 4a, 7a, 10a and 13a, and
their sum is 35a. Because a ≥ 1, the absolute difference between the first term and its square is a2 − a. We know that a2 − a = 35a. Solving for a
yields a2 − 36a = 0 → a(a − 36) = 0. The sequence contains positive integers, so a − 36 = 0 and a = 36.
164. Let l, w and d represent the rectangle’s length, width and diagonal length, respectively. The rectangle has area lw = 168 and perimeter
2(l + w) = 62 → l + w = 31 → l = 31 − w. Substituting for l in the area equation, we get (31 − w)w = 168 → w2 − 31w + 168 = 0 →
(w − 24)(w − 7) = 0, so w − 24 = 0 and w = 24, or w − 7 = 0 and w = 7. This means the rectangle is 24 inches by 7 inches. By the Pythagorean
Theorem we know the product of the diagonals is d2 = 72 + 242 = 49 + 576 = 625 in2.
MATHCOUNTS 2015-2016
77
x
165. Dividing, we see that the decimal equivalent of 2240/1111 = 2.0162, where the bar over the digits 0, 1, 6 and 2 indicates a sequence of four
repeating digits. Since 16/4 =4 R0, n16 = 2, which is the 4th repeating digit. Similarly, since 13/4 = 3 R1, n13 = 0, the 1st repeating digit. Since
10/4 = 2 R2, n10 = 1, the 2nd repeating digit. Since 7/4 = 1 R3, n7 = 6, the 3rd repeating digit. Therefore, the value of 1000n16 + 100n13 + 10n10
+ n7 is 1000 × 2 + 100 × 0 + 10 × 1 + 6 = 2016.
166. The expression 65 − 45 − 25 can be rewritten as 25(35 − 25 −1) = 25(243 − 32 − 1) = 25(210). We’ll factor 210 to see if there is a prime
factor greater than 2. Since 210 = 2 × 3 × 5 × 7, it follows that the greatest prime factor is 7.
167. Upon inspection, we notice that the figure is composed of 3/4 of square ABCD and 1/4 of a circle with radius AE. Radius AE is 1/2 the
diagonal length of a square of side length 4 inches. Therefore, AE = (1/2) × 4√2 = 2√2 inches. Thus, the figure has area 3/4 × 42 + 1/4 × π × (2√2)2
= 12 + 2π in2.
y
II
−4
168. As shown, the graph of the function f(x) = −x2 + 8x − 15 is a parabola that opens downward. The x-coordinate of the vertex of
III
a parabola given by the equation y = Ax2 + Bx + C is −B/(2A). So the vertex of the parabola has coordinates (4, 1), in the 1st
quadrant. The y-intercept is (0, −15), so the graph must cross the x-axis into the 4th quadrant. Substituting any value of x < 0 results
in a point in the 3rd quadrant. This is enough to conclude that the parabola will contain points in only 3 quadrants.
I
2
−2
2
4
6
8
10
−2
IV
−4
−6
−8
−10
−12
−14
169. Two of the L-pieces can be put together to make a 2 × 4 rectangle of 8 square units. The figure shows one
example of the entire 8 × 8 board can be covered using a total of 8 × 2 = 16 L-pieces.
−16
−18
170. Shelby walks an extra 10 steps, and it saves her 12 seconds. It follows, then, that the escalator moves 10 steps every 12 seconds. In the
72 seconds that Justin is on the escalator, it must move 60 steps. He walks and he walked 30 steps, so that’s 90 steps in all. Similarly, Shelby rides
50 steps in 60 seconds and walks another 40 steps, which also is 90 steps.
y=−
5
y=− x+3
4
Workout 7
S
171. First, solving the equation (−5/4)x + 3 = 14, we get −5x + 12 = 56 → −5x = 44 → x = −44/5, meaning
the point S(−8.8, 14) is on the line y = (−5/4)x + 3. Next, solving the equation (−5/4)x + 11 = 14, we get −5x +
44 = 56 → −5x = 12 → x = −12/5, meaning the point U(−2.4, 14) is on the line y = (−5/4)x + 11. As the figure
shows, the points S(−8.8, 14) and U(−2.4, 14) are on the line y = 14, with the point T(t, 14) located on that line,
midway between S and U. Point T will in fact be equidistant from the two lines. (The distance from T to a line is
defined to be the distance from T to the closest point on the line, which is obtained by dropping a perpendicular
segment from T to the line.) The x-coordinate of the midpoint of the segment with endpoints S(−8.8, 14) and
U(−2.4, 14) is (−8.8 −2.4)/2 = −5.6, which is the value of t.
172. The $0.99 difference must equal 20 − 15 = 5 percent of the absolute difference between the price of the
camera and the price of the printer. Dividing $0.99 by 0.05, we get an absolute difference of $19.80.
5
x + 11
4
y
16
T
U
14
y = 14
12
10
8
6
4
2
−10
−8
−6
−4
−2
2
173. Let W and L represent a win and a loss, respectively, by the Lemurs. These are the four possible scenarios in which the Lemurs win at least
two games: WWL, WLW, LWW, WWW. Since the Lemurs have a 60% chance of winning on their home field, the probability of a win in each of the
first and third games is 3/5, while the probability of a win in the second game is only 2/5. Let’s find the probability of each scenario.
WWL: (3/5) × (2/5) × (2/5) = 12/125 WLW: (3/5) × (3/5) × (3/5) = 27/125
LWW: (2/5) × (2/5) × (3/5) = 12/125 WWW: (3/5) × (2/5) × (3/5) = 18/125
Therefore, the overall probability is (12 + 27 + 12 + 18)/125 = 69/125 = 55.2%.
174. When we draw the radii from P and Q to S, equilateral triangle PSQ is created, as shown. The region bounded by arc SR and central angle
SQR is sector SR of the quarter-circle centered at Q. Sector SR has measure 30 degress and its area is 30/360 = 1/12 that of
T
R
S
a complete unit circle, or (1/12) × π × 12 = π/12 unit2. The shaded region, however, excludes the segment of the quarter-circle
centered at P that is bounded by arc SQ and chord SQ. The area of this segment is the area of sector SQ minus the area of
1
1
triangle PSQ. Sector SQ has measure 60 degrees so its area is 60/360 = 1/6 that of a complete unit circle, or (1/6) × π × 12 =
π/6 unit2. Using the formula for the area of an equilateral triangle, we see that triangle PSQ, with sides 1 unit in length, has area
P
Q
1
(12√3)/4 = √3/4 unit2. The area of segment SQ of the quarter-circle centered at P is π/6 − √3/4. Now we can calculate the area
of the shaded region to be π/12 − (π/6 − √3/4) ≈ 0.17 unit2.
175. The table shows the terms of the sequence from a2 through a10. Notice that a2, a4, a6, a8 and a10 are just the five even integers: 0, 2, 4, 6, 8. For
the terms a2, a4, a6, a8, a10, … it looks like ak, the kth term, is simply n − 2 when n is even. Based on this, we can conclude that the 2016th term of the
sequence is the 2016 − 2 = 2014.
n
an + 1
–an + 2n
78
1
a2
0
–2 + 2(2)
2
a3
4
0 + 2(2)
3
a4
2
–4 + 2(3)
4
a5
6
–2 + 2(4)
5
a6
4
–6 + 2(5)
6
a7
8
–4 + 2(6)
7
a8
6
–8 + 2(7)
8
9
a9 a10
10 8
–6 + 2(8) –10 + 2(9)
MATHCOUNTS 2015-2016
x
x
176. Since 60%, or 3/5, of the students in the original class are girls, the ratio of the number of girls in the class to the total number of students in
the class can be written as 3n/5n for some positive integer value of n. When g girls and 1 boy are added to the class, 75%, or 3/4, of the students
in the class are girls. Since the number of girls in the class increases by g and the total number of students increases by g + 1, that ratio becomes
(3n + g)/(5n + g + 1) = 3/4. Cross-multiplying, we get 4(3n + g) = 3(5n + g + 1) → 12n + 4g = 15n + 3g + 3 → g = 3n + 3 → g = 3(n + 1).
We see that g has to be a multiple of 3, and if we let n be the least positive integer 1, we get g = 3 × 2 = 6. To check this, we verify that 3 girls out
of 5 students is 60% to begin with. When 6 girls and 1 boy join the class, we have 9 girls out of 12 students, which is 75%, which confirms our
answer that the least possible value of g is 6.
177. The area inside the curve can be described as two quarter circles of radius 2 units plus two quarter circles of radius 4 units, minus a square of
side length 2 units that otherwise is counted twice. Combining the areas, we get 2 × (1/4) × π × 22 + 2 × (1/4) × π × 42 − 22 = 10π – 4 units2.
178. At each corner of the octagon, there is an isosceles right triangle. The legs of these triangles each have length (190 − 120) ÷ 2 =
70 ÷ 2 = 35 feet. By the properties of 45-45-90 right triangles, we know that each has a hypotenuse of length 35√2 feet. That means
the octagon has perimeter 4 × 120 + 4 × 35√2 = 480 + 140√2 feet ≈ 678 feet.
190 ft
120 ft
179. The absolute difference between the last term and the first term of this sequence is 77 − 13 = 64. In an arithmetic sequence, there is a
common difference, d, between consecutive terms. Since all the terms of the sequence are integers, the common difference d between consecutive
terms must be a factor of 64. The seven factors of 64 are 1, 2, 4, 8, 16, 32 and 64. Each of these factors is a possible value for d in a different
sequence that begins with 13 and ends with 77. For d = 1, the common difference is added 64 ÷ 1 = 64 times to complete this sequence of 65
terms. For d = 2, the common difference is added 64 ÷ 2 = 32 times to complete the sequence of 33 terms. If we continued for all possible values
of d, we would see that the possible values for the number of terms n are 65, 33, 17, 9, 5, 3 and 2. The median value is 9.
180. The divisors, or factors, of 20 are 1, 2, 4, 5, 10 and 20. These are the 6 values from which Will and Ian choose. We
have used a table to model the possible pairs of factors they can choose and have indicated the LCM for each pair. As the
table shows, of the 36 possible pairs that can result when Will and Ian each randomly choose a divisor of 20, only 15 have
an LCM of 20. That’s a probability of 15/36 = 5/12.
Workout 8
Ian
1
2
4
5
10
20
1
1
2
4
5
10
20
2
2
2
4
10
10
20
4
4
4
4
20
20
20
5
5
10
20
5
10
20
10
10
10
20
10
10
20
20
20
20
20
20
20
20
Will
181. The prime factorization of 2016 is 25 × 32 × 7. If we multiply any of the 6 powers of 2 from 20 through 25 by any of the 3 powers of 3 from 30
through 32, and then multiply that result by either 70 or 71, we will get a positive integer divisor of 2016. Since there are 6 × 3 × 2 = 36 combinations
of numbers we can multiply, it follows that 2016 has 36 positive integer divisors.
182. To solve this problem, we must go through three iterations of the function, beginning with f(2). The value of f(2) is 22 = 4. The value of f(4) is 44
= 256. The value of f(256) is 256256. We don’t need to find 256256, which we won’t calculate, since we just need to determine what it is when
expressed as 2k. The value of k is what we’re asked to find. Using the laws of exponents, we can rewrite 256256 as (28)256 = 28 x 256 = 22048. Thus, the
value of k is 2048.
183. In the figure at right, we introduce segment PW parallel to line m. Triangle POW is a right triangle with hypotenuse OW of
length 8 + 3 = 11 and leg PO of length 8 − 3 = 5. Leg PW must have length √(112 − 52) = √(121 − 25) = √96 = 4√6. The
area of triangle POW is (1/2) × 5 × 4√6 = 10√6. The area of rectangle BPWL is 3 × 4√6 = 12√6. Finally, the area of
quadrilateral BOWL is the sum of the areas of triangle POW and rectangle BPWL, which is 10√6 + 12√6 = 22√6 units2.
O
5
m
8
P
3
3
B
E
6
D
9
G
7
9
7
6
F
L
D' 184. Let’s rotate a copy of the triangle 180 degrees about point G so that we create a parallelogram as shown.
According to the parallelogram law, the sum of the squares of the lengths of all four sides of a parallelogram equals the
sum of the squares of the lengths of the two diagonals. If we let EF = x, then we have 62 + 92 + 62 + 92 = 142 + x2 →
234 = 196 + x2 → x2 = 38. So EF = √38 units.
185. The largest prime number in the 700s that uses distinct digits from 1 to 7 is 761, which, leaves us with the digits 2, 3, 4 and 5 with which to
construct 2 two-digit primes. It is not possible to do that given those four digits. In fact, none of the prime numbers in the seven hundreds would
allow for 2 two-digit primes. If we focus, instead, on the available digits for the ones place of our three prime numbers, we discover that we need to
use the 1, the 3, and the 7 in the ones place. As there are no primes with a ones digit of 2, 4, 5 or 6, we will need to use those for the other place
values in our three prime numbers. Since we found no three-digit prime that works in the 700s, we’ll try the 600s. That means the digits 2, 4 and 5
will be in the tens place of the three prime numbers. There are several primes that look promising, but we find that we are unable to make 2 two-digit
primes using the remaining digits. For example, 653 is a three-digit prime and we can make 21 and 47 or 27 and 41. Since neither 21 nor 27 is a
prime number, this can’t be a solution. It turns out that there are no three-digit primes in the 600s that work either. We’ll try the 500s next. The prime
number 563 uses the same digits as 653, thus giving us the same options for pairs of two-digit primes, which we’ve already determined won’t work.
When we get to 547, finally we are able to construct the primes 23 and 61 out of the other digits. Thus, the greatest possible value of the three-digit
prime number is 547.
186. Inside the can, the soda is in the shape of a truncated cylinder. Unlike a right cylinder, whose volume is the area of its base times its height,
for a truncated cylinder the volume is the area of its base times its average height. On the low side the height of the soda is 1/3 the can’s height,
and on the high side it is 1/2 the can’s height. The average of 1/3 and 1/2 is (1/3 + 1/2)/2 = 5/12, so the average height of the soda is 5/12 ×
4.5 = 1.875 inches. The base of the truncated cylinder formed by the soda is just a circle with radius 2.5 ÷ 2 = 1.25 inches and area π × 1.252 =
1.5625π. Therefore, the volume of soda in the can is 1.5625π × 1.875 ≈ 9.2 in3.
MATHCOUNTS 2015-2016
W
3
79
187. There are 10C3 = 10!/(7! × 3!) = (10 × 9 × 8)/(3 × 2 × 1) = 120 ways to choose three of the ten dots to be the vertices of a triangle. We need
to subtract the number of sets with 3 collinear dots. The figures in groups A, B and C show the 12 ways 3 collinear dots can be selected. We also
need to subtract the number of sets of 3 dots that make equilateral triangles. As illustrated in groups D, E and F, there are 15 ways to create an equilateral triangle with
each vertex at a point in the array. Nine are of the size shown in group D, 2 are the size
A
D
E
in group E and 4 are the size in group F. That leaves 120 − 12 − 15 = 93 distinct nonequilateral triangles.
B
C
F
188. Let a and b be the two numbers, and without loss of generality assume that a > b. We want a + b = n(a – b) with a – b as large as possible.
Thus we probably want n to be as small as possible. The smallest n could be is 2. In this case a + b = 2(a – b), which is equivalent to a = 3b, so
a – b = 2b. Because a ≤ 100, it follows that the largest b could be is 33. If we take b = 33 and a = 99, then (99 + 33)/(99 – 33) = 132/66 = 2.
This suggests that the greatest possible value of a – b is 99 – 33 = 66.
189. It is convenient to start Pascal’s Triangle with a zero row and include a “zeroth” number in each row. Thus, the numbers that are on the vertical
line of symmetry in Pascal’s triangle are actually the even rows, where there is an odd number of elements. The sum we seek is 0C0 + 2C1 + 4C2 +
6C3 + 8C4 + 10C5 + 12C6 = 1 + 2 + 6 + 20 + 70 + 252 + 924 = 1275.
190. The fraction a/(a + b) will always be written below and to the left of a/b and it is certainly less than 1. The fraction 31/4159 is one level below
and to the left of 31/4128. Notice that the denominator 4128 is 31 less than 4159. Since our new fraction is still less than 1, it must also be below
and to the left of a fraction that has 31 over a denominator that is 31 less than 4128. Since 4159 = 134 × 31 + 5, we will go up (and to the right)
134 levels until we get to the fraction 31/5. Since 31/5 is greater than 1, it has the (a + b)/b form and must be below and to the right of 26/5. Since
31 = 6 × 5 + 1, we will go up (and to the left) 6 levels until we get to the fraction 1/5. Finally, since 5 = 5 × 1 + 0, we will go up (and to the right) 4
levels to reach 1/1. We went up a total of 134 + 6 + 4 = 144 levels to reach Level 1, so the fraction 31/4159 is on Level 145 and n = 145.
Warm-Up 12
191. The only prime products possible when three dice are rolled are 2, 3 and 5. These products occur only when the number rolled on two of the
dice is 1 and a 2, 3 or 5 is rolled on the third die. Each of these 3 outcomes can occur in 3 ways, so the probability is (3 × 3)/(6 × 6 × 6) = 1/24.
192. The tree’s initial height was 8 feet, or 8 × 12 = 96 inches. After one year, its height was 96 × 7/6 = 112 inches. After two years, its height was
96 × 1.375 = 132 inches. Therefore, during the second year, the tree grew 132 − 112 = 20 inches.
193. Let h be the number of quarters initially showing heads, and let t be the number of quarters initially showing tails. After Kenyatta turns over 20%
= 2/10 of the quarters initially showing heads and 10% = 1/10 of the quarters initially showing tails, the number of quarters showing heads equals
the number of quarters showing tails.
10
10
h =
10
10
8
h
10
+
1
t
10
=
t =
9
t
10
+
2
h
10
As the figure shows, then 8/10 of the quarters initially showing heads combined with the 1/10 of the quarters initially showing tails, which now show
heads, is equal to the 9/10 of the quarters initially showing tails plus 2/10 of the quarters initially showing heads, which now show tails. We have
(8/10)h + (1/10)t = (2/10)h + (9/10)t. Combining like terms, we see that (8/10)h − (2/10)h = (9/10)t − (1/10)t → (6/10)h = (8/10)t → 6h = 8t →
h/t = 8/6 = 4/3 was the initial ratio of the number of quarters showing heads to the number of quarters showing tails.
194. If three machines fill 80 boxes in 2 hours, then the same three machine can fill 40 boxes in 1 hour and one machine can fill 40/3 boxes in
1 hour. Dividing 150 boxes by 40/3 boxes per hour, we get 150 ÷ 40/3 = 150 × 3/40 = 45/4 hours. This means it would take 45/4 hours for a
single machine to fill 150 boxes. If we have five machines, it should take only 1/5 as long, or (1/5) × (45/4) = 9/4 hours = 9/4 × 60 = 135 minutes.
195. The slope is rise over run. So for the run of 3 units, the rise along line m to point A is (3/4) × 3, and the rise along line n to point B is (1/5) × 3.
Therefore, the distance from A to B is (3/4 − 1/5) × 3 = (11/20) × 3 = 33/20 units.
196. To solve this problem without a calculator, it may be helpful to rationalize the denominator. Multiplying the numerator and denominator by the
conjugate of √102 − √98, we get 1/(√102 − √98) × (√102 + √98)/(√102 + √98) = (√102 + √98)/(102 − 98) = (√102 + √98)/4. Since √102 is
just a little more than 10 and √98 is just a little less than 10, the quantity √102 + √98 is very close to 20. Therefore, the integer closest in value to
1/(√102 − √98) is 20/4 = 5.
197. Suppose Victor buys 3 bottles of soda for a total of $3. Then, when he returns the empties for these 3 bottles of soda, he gets a bottle of soda
for free and buys 2 more for a total of $2. So far, Victor has paid $5 for 6 bottles of soda. When he returns the empty bottles for the last 3 sodas, he
again gets a free bottle of soda and buys 2 more bottles of soda for a total of $2. At this point, Victor will have spent $7 for 9 bottles of soda. Now
he simply returns the 3 empty bottles from his last 3 sodas to get a free bottle of soda. This free bottle of soda is his 10th, and he will have paid a
total of $7.
80
MATHCOUNTS 2015-2016
198. If the center circle is colored red, then there are 3 ways the circles along each diagonal can be colored red (R) or blue (B). They are RRB,
BRR and BRB. Therefore, there are 3 × 3 × 3 = 27 different ways to color the circles. Similarly, if the center circle is colored blue, the 3 ways the
circles along each diagonal can be colored are BBR, RBB and RBR. That’s another 27 ways the circles can be colored. Altogether, the circles can
be colored in 27 + 27 = 54 ways so that no three collinear circles are the same color.
199. There are 8C2 = 8!/(6! × 2!) = (8 × 7)/(2 × 1) = 56/2 = 28 ways to randomly select two people from the four couples. Since 4 of those result
in the selection of a married couple, the probability is 4/28 = 1/7.
200. Since the roots of Ax2 + Bx + C are 2/3 and 4, we have Ax2 + Bx + C = (x − 2/3)(x − 4)k, where k is some positive integer that makes A, B
and C relatively prime integers. Multiplying the binomials yields (x − 2/3)(x − 4) = x2 − (14/3)x + 8/3. To make the coefficients all integers, we must
choose k to be a multiple of 3. If k = 3, then we get 3x2 − 14x + 8 = 0. The greatest common factor of A, B and C is 1, as required, and A + B + C
is 3 − 14 + 8 = −3.
Warm-Up 13
1
1
201. Since Jenny averaged 10 1/3 minutes per mile for the first 4 miles, she has used 4 × 10 3 = 41 3 minutes so far. If she wants to finish 6 miles
1
2
2
in 1 hour = 60 minutes, then she needs to finish the last 2 miles in 60 − 41 3 = 18 3 minutes. To do this, she must run an average of 18 3 ÷ 2 =
9 13 minutes per mile.
202. Each hour, Clock B gains 15 + 15 = 30 minutes, or 1/2 hour, relative to Clock A. So the next instance when these two clocks will show the
same time will be when Clock B displays a time that is 12 hours ahead of Clock A. That will occur 12 ÷ (1/2) = 24 hours after the clocks are started.
203. We evaluate the expression in the innermost set of parentheses and work our way to the outermost set. We begin with f(19) = (19 − 3)/2 = 8.
Then we have f(8 + 1) = f(9) = (9 − 3)/2 = 3. Finally, we have f(3 + 1) = f(4) = 42 = 16.
204. Since the ratio of the area of trapezoid EBCD to triangle AED is 5 to 1, triangle AED must be 1/6 of the area of rectangle ABCD. That means
(1/2)(AE)(AD) = (1/6)(AB)(AD) and AE/AB = 2/6 = 1/3. It follows, then, that the ratio AE:EB is 1/2.
l
205. The diagram shows what Kendra will have left of the cardboard after four squares are
removed from the corners. To form the sides of the box, she will fold the cardboard along the
w dashed lines. So the length l and width w of the rectangle formed by the dashed lines will be
the length and width of the box, and the side length of the square removed from each corner
will be the height h of the box. We can use these measurements to determine the area of the
largest box Kendra can create. For example, if she cuts a square of side length 0.5 inch from
each corner, that results in a box with dimensions l = 16 − 2(0.5) = 15 inches, w = 12 − 2(0.5) = 11 inches and
h = 0.5 inch. This box has volume 15 × 11 × (0.5) = 82.5 in3. We can continue to increase the side length of the
square removed from each corner in increments of 0.5 inch until we reach the maximum volume. The table shows
the dimensions and volume of the boxes created when the corner square has side lengths 0.5, 1, 1.5, 2, 2.5 and
3 inches. We see that the greatest volume is 192.5 in3.
h
Height
(h)
Length
(l)
Width
(w)
Volume
(l × w × h)
0.5
15
11
82.5
1
14
10
140
1.5
13
9
175.5
2
12
8
192
2.5
11
7
192.5
3
10
6
180
3.5
9
5
157.5
4
8
4
128
4.5
7
3
94.5
5
6
2
60
5.5
5
1
27.5
6
4
0
0
206. Solution 1: We could expand the expression (a + b)5 by multiplying (a + b)(a + b)(a + b)(a + b)(a + b). Although it would take some time to
complete, we would see that the terms of this expanded fifth-degree polynomial are a5b0, a4b1, a3b2, a2b3, a1b4 and a0b5. Before all the like terms are
combined, we also would see that there is 5C0 = 1 a5b0 term, there are 5C1 = 5 a4b1 terms, 5C2 = 10 a3b2 terms, 5C3 = 10 a2b3 terms, 5C4 = 5 a1b4
terms and there is 5C5 = 1 a0b5 term. So (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5. This is an example of a method known as Binomial
Expansion. Using this method, we now replace a and b in this example with x and 2y, respectively. Doing so, we get (2x + y)5 = (1)(2x)5 + (5)(2x)4y
+ (10)(2x)3y2 + (10)(2x)2y3 + (5)(2x)y4 + (1)y5 = 32x5 + 80x4y + 80x3y2 + 40x2y3 + 10xy4 + y5. The sum of the coefficients, then, is 32 + 80 + 80
+ 40 + 10 + 1 = 243.
Solution 2: We could simply recognize that each term in the expansion will consist of a coefficient times a power of x times a power of y. If we let
x = y = 1, then we will get precisely the sum of the coefficients. Therefore, the sum of the coefficients is (2 × 1 + 1)5 = 35 = 243.
207. At least one dart lands on the bull’s-eye scoring 11 points. The player’s other two darts have an equal likihood of landing on the board scoring
7 points or scoring 11 points. The possible combinations of points scored for the three darts are 11 + 7 + 7 = 25 points, 11 + 11 + 7 = 29 points
and 11 + 11 + 11 = 33 points. The only combination that yields a prime number of points is when exactly two darts land on the bull’s-eye. There are
3 orders in which the three darts can be thrown if only one lands on the bull’s-eye. There also are 3 orders in which the three darts can be thrown if
two land on the bull’s-eye. There is only 1 way to get three bull’s-eyes. That’s a total of 7 possible outcomes when three darts are thrown. Since 3 of
these outcomes result in a total score of 29 points, the probability that the total number of points scored will be prime is 3/7.
208. We could make an organized list that starts with 1234, 1235, 1236 and ends with 6789. There are some interesting patterns in the list, but we
can determine the total number of integers without actually completing the list. We know that zero cannot be one of the digits in a four-digit number
that has digits in strictly ascending order, nor can there be any repeat digits. We just need to pick four digits from 1 to 9 and then arrange them in
the only way that makes them strictly ascending. Therefore, there are 9C4 = 9!/(5! × 4!) = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) = 126 integers composed
of four digits in strictly ascending order.
MATHCOUNTS 2015-2016
81
y
y = −x
209. As the figure shows, to get from P(−3, −2) to the line y = −x, we go up 2 units and right 3 units. From there, we
can go up 3 units and right 2 units to get to the reflection P'(2, 3) across the line y = −x. When P' is translated 4 units
right and 1 unit down, we arrive at P"(6, 2).
4
P'(2, 3)
2
−4
−2
P"(6, 2)
2
4
6
−2
P(−3, −2)
210. Let O be (0, 0). Then the y-axis divides right triangle ABC into right triangles BOC and AOB such that all
−4
three triangles are similar. The x-axis divides right triangle BCD into right triangles COD and BOC such that all three
triangles are similar. Finally, the y-axis divides right triangle CDE into right triangles DOE and COD such that all three triangles are similar. Therefore,
right triangles AOB, BOC, COD and DOE are similar. Triangle AOB is a 3-4-5 right triangle, with AB = 5. For triangles BOC and AOB we have
BO/AO = 3/4 → BO = (3/4) × AO. Each side length in triangle BOC is 3/4 the corresponding side length in triangle AOB. In fact, each side length
in triangle COD is 3/4 the corresponding side length in triangle BOC, and each side length in triangle DOE is 3/4 the corresponding side length
in triangle COD. It follows, then, that BC = (3/4) × 5 = 15/4 units, CD = (3/4) × (15/4) = 45/16 units and DE = (3/4) × (45/16) = 135/64 units.
Therefore, AB + BC + CD + DE = 5 + 15/4 + 45/16 + 135/64 = (320 + 240 + 180 + 135)/64 = 875/64 units.
Warm-Up 14
211. Using the laws of exponents, we can rewrite the expression 95x as (32)5x = 310x = (3x)10. Since 3x = y2, it follows that (3x)10 = (y2)10 = y20. Thus, z
= 20.
212. First, simplifying the fractions in the expression (3 + 6)/9 + (12 + 15)/18 + (21 + 24)/27 + (30 + 33)/36 + (39 + 42)/45 + (48 + 51)/54,
we get 1 + 3/2 + 5/3 + 7/4 + 9/5 + 11/6 = (60 + 90 + 100 + 105 + 108 + 110)/60 = 573/60 = 191/20.
213. Any number that leaves a remainder of 4 when divided by 9 will leave a remainder of 1 when divided by 3, so we can ignore the fact that there
is one band member left over when Namakshi arranges the band members in rows of 3. The numbers that are 4 more than a multiple of 9 are 13, 22,
31, 40, 49, 58, 67, and so on. We need a number that leaves a remainder of 2 when divided by 5, so its ones digit must be 2 or 7. The numbers in
this list that end in 2 will be 90 apart: 22, 112, 202, and so on; and those that end in 7 will also be 90 apart: 67, 157, 247, and so on. Dividing these
numbers by 7 shows that the smallest one that leaves a remainder of 3 is 157. The band has 157 members.
214. The 4 × 4 square must be a face of the finished cube. That means the cube has edge length 4 cm and volume 4 × 4 × 4 = 64 cm3.
215. Let x be the 100th term in the sequence. Then the 101st term is 7 + x and the 102nd term is 7 + 2x. We are told that the 102nd term is 69.
So we have 7 + 2x = 69. Solving for x, we get 2x = 62 and x = 31.
216. Let p1, n1 and d1 be the number of pennies, nickels and dimes in the left pocket, and similarly for p2, n2 and d2. We are told that p1 + n1 + d1 =
p2 + n2 + d2 and that p1 + 5n1 + 10d1 = p2 + 5n2 + 10d2. Subtracting these equations gives 4n1 + 9d1 = 4n2 + 9d2, which is equivalent to
4(n1 – n2) = 9(d2 – d1). If n1 = n2, then d1 = d2, and the pockets will have the same contents, in violation of the condition that the pockets contain
different numbers of pennies. Therefore, n1 – n2 must be a positive multiple of 9, and the smallest choices would be n1 = 9 and n2 = 0. Similarly, the
smallest choices for the numbers of dimes are d2 = 4 and d1 = 0. Substituting these values into the original equations shows that p2 = p1 + 5. The
smallest solution will have p1 = 0 and p2 = 5. Then the left pocket will contain no pennies, 9 nickels and no dimes; and the right pocket will contain
5 pennies, no nickels and 4 dimes. Each pocket contains 45 cents, and the total value is 90 cents.
217. Let’s start by determining the possible lengths of segments drawn between pairs of dots. Since we don’t yet know how to
1
2
3
5
4
C
E
B
D
A
label the dots with the letters A through E, the figure shows them labeled with the numbers 1 through 5. From least to greatest,
the possible lengths for segments drawn between pairs of dots in the grid are, in units, 1, √2, 2, √5, 2√2. Assume that the
longest segment, DE, has length 2√2 units. Then segment DE must have endpoints at dots 2 and 5. The shortest segment, AB,
has length 1 unit. Since D or E is at the dot labeled 2, it follows that segment AB must have endpoints at dots 3 and 4. That
leaves only one location for C, at the point labeled 1. Since segment CD needs to be shorter than segment DE but longer than
segment BC, D must be at the dot labeled 5, which means E is at the dot labeled 2. Since AB < BC < CD, or 1 < BC < √5,
segment BC could have length √2 units or 2 units. Of the possible locations for B, neither results in BC = 2 units. But when
B is located at the dot labeled 3, BC = √2 units. That leaves the final location, at the dot labeled 4, for A. As the second figure
shows, the length of segment CE is 1 unit. It can be verified that there is no arrangement of the labels in which segment DE has
length less than 2√2 units and satisfies the given conditions.
218. We’ve listed all the two-digit numbers that contain the digit 1 or the digit 9 and then circled the 13 that are prime.
10 11 12 13 14 15 16 17 18 19 21 29 31 39 41 49 51 59 61 69 71 79 81 89 90 91 92 93 94 95 96 97 98 99
Based on this, the probability that a two-digit number containing at least one 1 or 9 is prime is 13/34.
219. Alfonso, Benjamin and Carmen are walking at speeds of 2 × 60 = 120, 6 × 60 = 360 and 8 × 60 = 480 feet per minute, respectively. All
three start together, heading in the same direction, walking around the 440 × 3 = 1320-foot track. Notice that Benjamin is walking 3 times Alfonso’s
speed and Carmen is walking 4 times Alfonso’s speed, which means they each will have walked at least one full lap around the track before the three
of them meet again. Let t represent the number of minutes it takes all three friends to be at the same location on the track again at the same time.
Since Distance = Rate × Time, we know that after t minutes, Alfonso, Benjamin and Carmen will have walked 120t feet, 360t feet and 480t feet,
respectively. These distances all must be multiples of 1320 feet. If we find the GCF of 120, 360, 480 and 1320, we can determine the number of
minutes that will result in these three distances being multiples of 1320 feet. The prime factorizations are 120 = 23 × 3 × 5, 360 = 23 × 32 × 5, 480
= 25 × 3 × 5 and 1320 = 23 × 3 × 5 × 11. So the GCF is 23 × 3 × 5 = 120. Therefore, they all will be in the same location on the track again at the
same time after they have been walking 1320 ÷ 120 = 11 minutes.
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MATHCOUNTS 2015-2016
x
220. If we know the prime factorization of a number, we can calculate both the number of factors and the sum of the factors without having to list all
the factors. For example, the prime factorization of 54 is 21 × 33. There are 2 × 4 = 8 ways to create factors of 54, by multiplying one of the 2 powers
of two (1, 2) by one of the 4 powers of three (1, 3, 9, 27). So there are 8 factors of 54. Now the sum of the factors of a number is the product of the
sums of the powers of the primes in each list. For example, the sum of the factors of 54 is (1 + 2) × (1 + 3 + 9 + 27) = 3 × 40 = 120. Using this
method, we determine S(n) for n = 1 to 55, and we see that the greatest possible value is S(48) = (1 + 2 + 4 + 8 + 16) × (1 + 3) = 31 × 4 = 124.
Counting Stretch
221. There are 9 digits in the numbers from 1 through 9. There are 90 × 2 = 180 digits in the two-­digit numbers from 10 through 99. There are
200 × 3 = 600 digits among the 200 three-digit numbers from 100 to 299. Finally, there are 22 × 3 = 66 digits in the 22 three-digit integers from
300 to 321. So when writing down the integers 1 through 321, a total of 9 + 180 + 600 + 66 = 855 digits are written.
222. Consider the figure shown where each of the 16 divisions of the square has been labeled. The 8 divisions above the
diagonal from the lower left vertex to the upper right vertex are labeled A through H. The reflection of each of these division
across that diagonal are labeled A’ through H’. We will consider the figure in sections to ensure that we don’t miss counting
any of the triangles. The triangles in each section are identified by the labels of the divisions of the overall figure that compose
the triangle. First we examine each of the four “quadrants” (small squares in lower left, upper left, lower right and upper right
corners), then the four rectangles each composed of two of the square quandrants (left half, right half, top half, bottom half),
then the four Ls each composed of three of the of the square quadrants, and finally we’ll consider all four quandrants together
and look for any triangles that cover a portion of each quandrant.
Lower Left – 6
1
2
3
4
5
6
Upper Left - 6
7
8
9
10
11
12
A+B
A' + B'
A
B
A'
B'
Left Half – 7
27
28
29
30
31
32
33
A' + B' + B + D + F
A+B+C+D
B' + B + D + F
A+ C+E
B+D+F
B+D
A+C
Left/Bottom L - 4
45
46
47
48
13
14
15
16
17
18
Top Half - 2
Left/Top L - 3
Top/Right L - 1
36
37
52
A+B+C+D+E+F+G+H
B+D+F+G+H
B+D+F+G
C
A
G
H'
F
G'
D
F'
B
B' D'
E'
C'
A'
Upper Right - 8
19
20
21
22
23
24
25
26
C' + D'
E '+ F'
D' + F'
D'
E'
F'
Right Half - 2
D' + F' + G + G'
34
E' + F' + G' + H'
35
49
50
51
A + A' + B + B' + C + C ' + D + D'
A + B + B' + C + D + D'
A+B+C+D
B + B' + D + D'
Lower Right - 6
C+D
E+F
D+F
D
E
F
H
E
G+H
G' + H'
G + G'
H + H'
G
H
G'
H'
Bottom Half - 7
38
A + B + B' + D' + F'
D + F + G' + G
E+F+G+H
39
40
41
42
43
44
A' + B' + C' + D'
B' + B + D + F
A '+ C' + E'
B' + D' + F'
B' + D'
A' + C'
Right/Bottom L - 3
E + E' + F + F' + G + G' + H + H'
53
54
55
A' + B' + C' + D' + E' + F' + G' + H'
B' + D' + F' + G' + H'
B' + D' + F' + G'
In addition to the large triangle composed of B + B' + D + D’ + F + F’ + G + G’, there are a total of 56 triangles in the figure.
223. The 27 possible ways to give the knife, fork and spoon are shown in the table.
Position
2
Position 1
Knife
Knife, Fork
Knife, Spoon
Fork, Spoon
Fork, Spoon
Knife, Spoon, Fork
Position 3
Fork
Spoon
Fork
Knife
Knife
Distinct Ways to
Switch Positions
Spoon
6
6
6
6
6
3
224. We’ll start with the largest possible number of quarters, then the largest possible number of dimes, then nickels, and change nickels to pennies
one nickel at a time. When you can’t go any further, break one quarter, form the largest possible number of dimes, and repeat the process. Once you
see the pattern, you won’t have to list the remaining cases individually.
Quarters
Dimes
Nickels
Pennies
# Ways
2
1
1
1
2
0
0
7
2
3
2
2
7
1
12
4
0
17
4
0
2
1
2
2
3
1
7
3
0
12
4
2
3
17
2
2
12
5
1
17
0
22
Extending the pattern in the table for 1 quarter and 1 through 0 dimes gives us 7 ways with 1 dime and 9 ways with 0 dimes.
Extending the pattern to include all the numbers of quarters and dimes gives us:
2 quarters with 1 through 0 dimes: 2 + 4 = 6 ways
1 quarter with 4 through 0 dimes: 1 + 3 + 5 + 7 + 9 = 25 ways
0 quarters with 6 through 0 dimes: 2 + 4 + 6 + 8 +10+ 12+ 14 = 56 ways
The total number of ways to make 670 is then 6 + 25 + 56 = 87.
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225. The following table lists all sums of scores for 6 turns:
Same score 6 times
One score 5 times and another once
One score 4 times and another twice
One score 4 times and the other 2 scores once each
One score 3 times and another score 3 times
One score 3 times, one twice, and one once
Each score twice
18,42, 66
22, 26, 38, 46, 58, 62
26, 34, 34, 50, 50, 58
30, 42, 54
30, 42, 54
34, 38, 38, 46, 46, 50
42
With duplicates crossed off, this leaves scores of: 18, 22, 26, 30, 34, 38, 42,46, 50, 54, 58, 62, and 66 = 13 scores.
226. The desired numbers must be divisible by 5 x 17 = 85. The smallest such 3-digit number is 170 (85 x 2) and the largest is 935 (85 x 11). The
number of 3-digit numbers divisible by 85 is then 11 - 2 + 1 = 10.
227. Since 7 is prime and 10 = 2 x 5, no number with a factor of 2, 5, or 7 is relatively prime to 7 or 10. All the other primes under 40, plus the
number 1, are relatively prime to 7 and 10. Thisgives 1, 3,11,13,17,19, 23, 29, 31, and 37, or 10 numbers. In addition, products of the above 9
primes that are still under 40 are also relatively prime to 7 and 10. This gives the additional four numbers 9, 27, 33 and 39, for a total of 10 + 4 =
14 numbers.
228. The two-digit palindromes are 11, 22, 33,44, 55, 66, 77, 88, and 99 = 9 palindromes. Three-digit palindromes are in the form 101, 111, 121,
202, 292, ... There are 10 in each of the 100s, 200s, 300s, 400s, 500s, 600s, 700s, 800s and 900s. That’s a total of 90 three-digit palindromes.
Finally, 1001 is the last palindrome less than 1009. Total number of palindromes, then, is 9 + 90 + 1 = 100 palindromes.
R1 O R2
O
T
O
R3 O R4
229. Label the cells as shown. The number of ways starting with R1 include two ways to get to the T and two ways to get from there
to R1, R2, R3, or R4. This equals (2 × 2) × 4 different endpoints for 16 different paths. Using the same reasoning, there are
1 2
3
16 different paths that start with each of R2, R3 and R4. So the total number of paths is 16 x 4 = 64 paths.
T
230. Per the above additions, the number of ways to get from P to Ais 3, the number of ways to get from Ato T is 6, and the
number of ways to get from Tto His 3. Therefore, the total number of paths that make PATH is 3 x 6 x 3 = 54.
A
Area Stretch
P
1
3
1
2
1
3
6
1
2 1
3
1
1
1
H
1
1
231. The area of the original square sheet of paper was 102 = 100 in2. From each corner, Norm cut an isosceles right triangle of area (1/2) × 2 × 2
= 2 in2. Each triangle is 2/100 = 2% of the original square’s area. Together, all four triangles account for 4 × 2 = 8% of the total area. That means
that the octagon that remains after the four corners are removed accounts for 100 – 8 = 92% of the area of the original square.
232. We can decompose the area of the office floor into two rectangles: one with dimensions 12 feet by 24 feet, and one with dimensions 3 feet by
11 feet (the upper-left corner). The areas of these rectangles are 288 ft2 and 33 ft2, respectively. So the total area of the office floor is 288 + 33 =
321 ft2.
233. The area enclosed by the track is composed of a rectangle, whose dimensions are 100 meters by 50 meters, and two semicircles, each with
radius 25 meters. The area of the rectangle is 5000 square meters, and the two semicircles can be combined to form a circle of area π × 252 =
625π m2. So the total enclosed area is 5000 + 625π ≈ 7000 m2.
234. We can decompose this pentagon into two trapezoids: one with vertices at (0, 0), (2, 0), (2, 10) and (0, 6), and one with vertices at (0 0),
(−2, 0), (−2, 10) and (0, 6). These two trapezoids are mirror images of each other across the y-axis and, therefore, are congruent. So we’ll compute
the area of the one on the right and multiply the area by 2. The area of the right trapezoid is (1/2) × 2 × (10 + 6) = 16, so the area of the entire
pentagon is 2 × 16 = 32 units2.
235. The shaded region is called a segment of the circle. To find the area of this segment, we first draw radii to each endpoint of the segment. We
see that the segment in question is the area inside a sector of the circle that lies outside an isosceles right triangle. Since the sector is one-fourth of
the circle, the area of the sector is (1/4) × π × 42 = 4π units2. The area of the right triangle is (1/2) × 4 × 4 = 8 units2. So the area of the segment is
4π – 8 units2.
236. The total area covered by the two discs can be decomposed into a half of the smaller disc, an isosceles right triangle, and three-quarters of
the larger disc. The smaller disc has radius 2 inches, so half of the disc will have area (1/2) π × 22 = 2π in2. The right triangle has legs of length
2√2 inches, so its area is (1/2) × (2√2)2 = 4 in2. The larger disc has radius 2√2 inches, so three-quarters of the disc will have area (3/4) × π × (2√2)2
= 6π in2. Thus, the total area covered by the two discs is 2 + 4 + 6 = 4 + 8π in2.
237. Because the octagon has reflection symmetry across AE and CG, the quadrilateral ACEG is a square. This square has diagonals of length
16 units, so the side length of the square is 8√2 units. So half of a side of the square has length 4√2 units, and so by the Pythagorean Theorem,
the distance from vertex B to the side of the square is √(62 – (4√2)2) = √4 = 2 units. So we can decompose the octagon into a square of area
(8√2)2 = 128 units2, and four isosceles triangles, each with base length 8√2 units and height 2 units. Therefore, the area of the octagon is
128 + 4 × (1/2) × 8√2 × 2 = 128 + 32√2 units2.
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MATHCOUNTS 2015-2016
238. Because the triangles AEQ and PCB are congruent and isosceles, the angles EAQ, DPQ, DQP and CBP are all congruent. Therefore,
triangle DPQ is similar to triangles EAQ and CPB. So triangle DPQ is a dilation of triangle EAQ. Let k < 1 be the scale factor of this dilation. We
can view the perimeter of ABCDE as the perimeter of triangle AEQ plus the perimeter of triangle BCP minus the perimeter of triangle DPQ. So
the perimeter is 32 + 32 – 32k = 64 – 32k. Since the perimeter is known to be 52, we have 64 – 32k = 52 → 32k = 12 → k = 3/8. So the area
of triangle DPQ is (3/8)2 times the area of triangle EAQ. By the Pythagorean Theorem, the height of triangle EAQ is 8, so the area of the triangle
is 48 units2. So the area of pentagon ABCDE is the area of triangle EAQ plus the area of triangle BCP minus the area of triangle DPQ, which is
48 + 48 – (3/8)2(48) = 357/4 units2.
B
239. We mark the points of intersection of the two triangles P, Q and R as shown. We can find the area of the overlapping region
by subtracting the area of triangle PBM from the area of triangle RBQ. We know that triangle PBM is similar to triangle ABC. So
since BM = 5/2, PB = (5/4) × (5/2) = 25/8. We also know that MP = 3/4 × 5/2 = 15/8. Since MB’ = 5/2, PB’ = (5/2) – (15/8)
= 5/8. Since triangle PB’Q is also similar to triangle ABC, we have PQ = 3/5 × 5/8 = 3/8. So BQ = BP + PQ = 25/8 + 3/8 =
7/2, and thus QR = 3/4 × 7/2 = 21/8. So the area of triangle RBQ is (1/2) × (7/2) × (21/8) = 147/32. The area of triangle PBM
P
is (1/2) × (5/2) × (15/8) = 75/32. So the area of the overlapping region is 147/32 – 75/32 = 72/32 = 9/4 units2.
B'
A'
M
R
Q
C
240. Since ABC is a right triangle, we recognize the leg lengths AC = 10 BC = 24 as a derivative of the 5-12-13 Pythagorean Triple. Thus, AB =
26. Let k be the distance from X to AB. Then the distance from X to AC is k/2, and the distance from X to BC is k/4. Draw a segment from X to each
vertex of triangle ABC, dividing the triangle into triangles AXB, BXC and CXA. The areas of these three triangles must add up to the area of triangle
ABC, which is (1/2) 10 × 24 = 120 units2. So since the shortest distance from X to each side of ABC is an altitude of one of the three smaller
triangles, and we can write the equation (1/2) × 26 × k + (1/2) × (k/4) × 24 + (1/2) (k/2) × 10 = 120. Solving for k, we get (37k)/2 = 120 → 37k =
240 → k = 240/37 units.
Modular Arithmetic Stretch
241. Every 12 months we will arrive back at the same month we started therefore this is a problem modulo 12. A multiple of 12 close to the 152
would be 12 × 12 = 144. Since 152 is 8 more than this we can write 152 º 8 (mod 12), and 8 months from July will be March.
242. Since we want to know the time and whether it is AM or PM our modulus will be 24, not 12. A multiple of 24 close to 255 is 24 × 10 = 240.
So we can write 245 º 15 (mod 24). Fifteen hours from 8:00 a.m. will be 11:00 p.m.
243. Since the track is 400 meters around Jennie will return back to the point she started at every 400 meters – this is our modulus.
5310 º 110 (mod 400). Jennie’s distance from her starting point when she finished will be 110 meters.
244. As we start computing powers of 2, the last digit of each power depends only on the last digit of the previous power. This is because every
positive integer can be written in the form 10a + b for nonnegative integers a and b, where b <10. Then 2(10a + b) = 20a + 2b has the same last
digit as 2b. Thus the powers of 2 end in 2, 4, 8, 6, 2, 4, …. We see that the pattern repeats with period 4 and 2015 = 4(503) + 3, so the last digit
of 22015 is the same as the last digit of 23 = 8.
245. Note that 122 = 11 × 11 + 1 and 71 = 11 × 6 + 5 therefore, 122 º 1 (mod 11) and 71 º 5 (mod 11) Instead of multiplying 122 times 71
and then finding the answer it is easier to use modular arithmetic first. The answer will be 122 × 71º 1 × 5 º 5 (mod 11).
246. We can use the same approach as the previous problem. Since we are looking at a modulus of 5 it is easy to find the remainder of large digit
numbers. We know that multiples of 5 will end in either a 5 or a 0 so we only need to look at the last digit to determine the remainder. 5981 × 8162 × 476 º 1 × 2 × 1º 2 (mod 5). So the remainder will be 2.
247. Here what we are really solving for is the product of 29 times 51 (mod 12) = 29 × 51 º 5 × 3 º 3 (mod 12). So Jon will have 3 extra donuts.
248. Since 7 and 11 are both prime numbers they must be relatively prime. The least common multiple will then be 7 × 11 = 77. If we want a
remainder of 6 when divided by 7 and 11 we simply add 6 to their least common multiple. Since 83 º 6 (mod 7) and 83 º 6 (mod 11), our answer
is 83.
249. We are looking for a number that leaves a remainder of 1 when divided by 3, 4, 5, or 6, and hence is 1 more than a multiple of the LCM of 3, 4,
5 and 6, which is 60. Since 1 is too small and 121 is too big, she must have 61 pencils.
250. Same idea as #249, but one less than a multiple of 60. The answer is 59 pencils.
MATHCOUNTS 2015-2016
C'
A
85