UNL MATH DAY 2013 PROBE I SOLUTIONS v1111
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Question 1. A bottle containing 200 vitamin tablets weighs 180g. The same bottle containing 150 tablets weighs
165g. How much does the empty bottle weigh in grams?
(A) 103
(B) 115
C 120
(D) 125
(E) 146
Answer. If the bottle weighs B and a tablet weighs t, then B + 200t = 180 and B + 150t = 165. Subtracting, we
have 50t = 15, so t = 0.3. Then B = 180 − 200(0.3) = 180 − 60 = 120.
Question 2. What is the sum of all positive integers such that
A 64
(B) 16
7n − 50
is an integer?
n−8
(C) 56
(D) 44
(E) 50
6
7n − 50
= 7+
, which is an integer only if n−8 divides 6, meaning n−8 equals one of −6,−3,−2,−1,1,2,3,
Answer.
n−8
n−8
or 6, so n = 2, 5, 6, 7,9, 10, 11, or 14. The sum of these numbers n is 64.
Question 3. A cylindrical can with a radius of 6 inches is filled with 2 inches of water. When a smaller cylindrical
can 4 inches tall is placed inside of the first with its bottom lying on the bottom of the first, the water in the first
can rises a further inch to 3 inches. What is the radius of the smaller can?
√
√
√
B
12
(C) 15
(D) 4
(E) 18
(A) 3
Answer. The volume of water present is π(6)2 · 2 = 72π. If the small cylinder has radius r, then 72π = 3π(62 − r2 ),
since this is the volume
of a 3 inch tall column of water with the small cylinder removed. Solving for r yields
√
3πr2 = 36π, so r = 12.
√
√
Question 4. For what value(s) of x does x2 + 2x − 1 = x2 − 2x + 1 = 1 ?
(A)
5
6
B there are no real solutions
(C)
1
2
(D) 1
(E)
1 5
,
2 6
Answer. Squaring, we need 1 = x2 + 2x − 1 = x2 − 2x + 1 = (x − 1)2 . So either x − 1 = 1, so x = 2 (but then
x2 + 2x − 1 = 7), or x − 1 = −1, so x = 0 (but then x2 + 2x − 1 = −1). So there are no real solutions.
Question 5. What does the following sum equal?
√
1
1
1
1
√ +√
√ + ··· + √
√
√ + ··· + √
n+1+ n
2+ 1
3+ 2
49 + 48
7
(D) √
(E) 7
50
√
√
√
√
1 √
Answer. Multiplying tops and bottoms by n + 1 − n, we have √n+1+
= n + 1 − n. So our sum can be
n
√ √
√ √
√
√
√
√
√
√
√ √
written as ( 2− 1)+( 3− 2)+· · ·+√ 49− √48, or, in the opposite order, 49− 48+ 48− 47+· · ·+ 2− 1.
All of the middle terms cancel, leaving 49 − 1 = 7 − 1 = 6.
(A) 5
(B)
49
√
7 + 48
C 6
Question 6. A group of 20 students take an exam, and have an average score of 75. If the alphabetically first 8
students have an average of 90, what is the average score of the remaining 12 students?
(A) 55
(B) 60
C 65
(D) 70
(E) 75
Answer. The total of everyone’s score is (20)(75)=1500. The first 8 account for 8(90) = 720 of this total, leaving
780 for the remaining 12. These twelve then have an average of 780/12 = 65 .
Question 7. In the figure at right, the angles meeting at the center are
all equal, and the small triangles have their remaining vertices lying on the circle of radius 1 and the large triangles
have their remaining vertices lying on the circle of radius
2. What do the areas of the triangles add up to?
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(A) 3
(C) 6
(D)
√
13 2
3
E
√
15 3
4
Answer. All of the triangles are equilateral, and the large ones have side lengths 2 while the small ones have √
side
√
√
√
1√
3
.
3 · 2 = 3 and
lengths 1. They then have altitudes of lengths 3 and 3/2, respectively, and so have areas
2
4
√
√
√
3
15 3
The total area is therefore 3 3 + 3
=
.
4
4
Question 8. A package requires $1.53 in postage to mail. If only 5-cent and 8-cent stamps are available, what is
the smallest number of stamps that can be used to exactly cover the postage?
(A) 14
B 21
(C) 23
(D) 24
(E) 27
Answer. We want to solve 5a + 8b = 153 with a + b as small as possible. Since 5(8) = 8(5), we can replace any 8
5’s with 5 8’s and reduce the sum. So we need to subtract 5’s from 153 until we find a multiple of 8, and use exactly
that many 5’s. Since 153 − 5(5) = 128 = 8(16) (and no smaller multiple of 5 works), we use 16 8’s and 5 5’s, for a
total of 21 stamps.
Question 9. What is the area of the region in the x-y plane satisfying the inequalities xy ≤ 0 and x + 1 ≤ y ≤ x + 2?
(A) 1
B
3
2
(C) 2
(D)
5
2
(E) 3
Answer. The region lies in the second quadrant, and consists of the triangle with sides along the axes and the
line y = x + 2, with the triangle with sides along the axes and y = x + 1 removed. The larger triangle has area
1
1
3
1
(2)(2) = 2, and the smaller has area (1)(1) = , so the region in question has area .
2
2
2
2
Question 10. The polynomial x5 − 15x4 + 3x3 + 595x2 − 1404x − 1980 has five real roots. What is the sum of these
roots?
A 15
(B) 10
(C) 5
(D) 0
(E) −5
Answer. Writing the polynomial as a product (x − a)(x − b)(x − c)(x − d)(x − e), then when multiplying this out
the coefficient of x4 , −15, is equal to −(a + b + c + d + e). So the sum of the roots is equal to 15.
Question 11. In the figure at right a tower is made by stacking cubes of
side length 1 on top of one another against the walls in the
corner of a room. Some of the cubes used are not visible.
How many cubes in all were used?
(A) 10
(B) 15
(C) 18
D 20
(E) 24
Answer. From top to bottom, the k-th level has k more blocks than the previous one. So the levels have 1, 3, 6,
and 10 blocks, for a total of 20.
Question 12. If x + 3x + 9x + 27x = y +
(A) 1
(B)
40
27
y
y y
+ + , what is the ratio x/y?
3 9 27
(C)
27
40
D
1
27
(E) 27
Answer. If we multiply the right-hand side by 27, we get 27y + 9y + 3y + y = 40y. So the equation reads
1
x
1
(40y), so =
.
27
y
27
40x =
Question 13. A triangle in the plane has vertices at the points (2, 5), (2, 10), and a third point listed below. For
which third vertex does the resulting triangle have the smallest area?
(A) (4, 11)
1
√
(B) 4 2
(B) (0, 6)
(C) (11, 4)
2
(D) (7, 7)
E (1, −3)
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Answer. The triangle has a vertical side running between (2, 5) and (2, 10), of length 5. The coresponding altitude
of the triangle is the horizontal line running from the third vertex to this vertical line, with length the absolute value
of the difference between the x-coordinate of the third vertex and 2. The area will be smallest when this distance is
smallest; this occurs for the point (1, −3), with altitude of length 1.
Question 14. What is the sum 7 + 14 + 21 + · · · + 700 ?
(A) 34350
B 35350
(C) 36350
(D) 37350
(E) 38350
Answer. The sum is 7 times 1 + 2 + · · · + 50 + 51 + · · · + 100. This is 50 pairs of numbers each adding to
1 + 100 = · · · = 50 + 51 = 101, so the original sum is (7)(50)(101) = 7(5050) = 35350.
Question 15. For which of the following values of a does the polynomial x3 − 6x2 + (5 + a)x − a have only two
distinct roots? [Hint: x = 1 is a root.]
(A) 3
(B) 1/3
(C) 5
(D) 25
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Question 21. For the function f (x) = x2 + 8x − 9, the range of values output from f is:
(A) all y ≥ −1
(B) all y ≥ −9
2
D all y ≥ −25
(E) all y ≥ −36
Answer. Completing the square, we have f (x) = (x + 8x + 16) − 25 = (x + 4) − 25. Since (x + 4)2 can take any
value y ≥ 0, the range of f is y ≥ −25.
√
√
1+ 7
1− 7
and β =
, then α3 + β 3 is equal to
Question 22. If α =
2
2
√
√
√
11 + 5 7
11
(E)
(C) 4 + 7
D
(A) 4
(B) 5 7
2
2
√
√
√
√
√
√
1
1
Answer. α3 = [1 − 3( 7) + 3( 7)2 − ( 7)3 ], while β 3 = [1 + 3( 7) + 3( 7)2 + ( 7)3 ]. Adding, we get
8
8
2
√
44
1
11
(2 + 6( 7)2 ) =
=
.
8
8
2
E 25/4
2
Answer. Factoring the polynomial, we have (x − 1)(x − 5x + a). To have only two values for a root, either
x2 − 5x + a must have root 1, and so x2 − 5x + a = (x − 1)(x − 4) and a = 4, or it must have a double root, and so
5
25
25
x2 − 5x + a = (x − )2 = x2 − 5x + . So a =
.
2
4
4
(C) all y ≥ −16
Question 23. A circle is inscribed inside of a quarter circle, touching
both sides and the circle. The ratio of the radius of the
quarter circle to the radius of the circle is:
Question 16. A certain positive whole number, when multiplied by 192, is a perfect square. What is the smallest
that the number can be?
(A) 1
(B) 2
C 3
(D) 6
(E) 12
Answer. 192 = 2 · 96 = 2 · 6 · 16 = 22 · 3 · 24 = 26 · 31 . To be a perfect square we want the exponents to all be even,
so multiplying by 3 is needed (and enough), yielding 192 · 3 = 576 = (24)2 .
(A)
√
1+ 3
2
(B)
√
2
(C)
√
3
(D) 2
E 1+
√
2
Answer. Draw radius lines (of length r) from the center of the circle to the three points where it touches. The
diagonal line lies on a√radius of the quarter circle, while the other two are the sides of a square. The diagonal of the
square,
√ with length
√ r 2, is the remainder of
√ the radius of the quarter circle. The quarter circle therefore has radius
r + r 2 = (1 + 2)r, giving a ratio of 1 + 2.
Question 17. A cube with side length 4 has a square hole with side length
1 drilled all of the way through the center of each face.
What is the volume remaining from the cube?
Question 24. A student has 10 pairs of socks: 5 black and 5 blue. If after each wash-dry cycle a sock disappears
without a trace, what is the probability that after two cycles there will be an even number of black socks?
(A) 32
(B) 36
(C) 48
(D) 51
E 54
Answer. The cube has volume 43 = 64, and each hole individually has volume 4. But the very center cube with side
length 1 would be counted 3 times if we just subtracted off the volumes of the three holes, and so the true volume
remaining is 64 − 3(4) + 2 = 54.
Question 18. What is the largest n so that 6n evenly divides 18! ?
(A) 5
(B) 6
(C) 7
D 8
(E) 9
Answer. 18! has 9 + 4 + 2 + 1 = 16 factors of 2 (counting the number of multiples of 2, then 4, then 8, then 16),
and 6 + 2 = 8 factors of 3. This means that we can assemble 8 factors of 6 in 18!.
Question 19. If the length of a rectangle is increased by 20% and the width is decreased by 25%, then the area is:
(A) increased by 10%
(B) increased by 5%
(C) unchanged
(D) decreased by 5%
E decreased by 10%
Answer. If we let be the original length and w the width, then the new rectangle has area (1.2)(0.75w) = (0.9)w,
a decrease of 10% from the original area w.
(A)
4
9
B
9
19
(C)
B 3
(C) 5
(D) 7
Question 25. If
5
9
(E)
3
5
A
7
6
x + 2y
x + 3y
= 2, then what is
?
2x + y
3x + y
(B)
5
4
(C)
7
5
(D)
8
5
(E)
5
3
Answer. Clearing the denominator, we have x + 3y = 2(2x + y) = 4x + 2y, so y = 3y − 2y = 4x − x = 3x. Plugging
7
y = 3x into the second expression and eliminating x, we end up with .
6
(E) 9
Answer. 34 = 81 ends in 1, and so 20134 , and any power of 20134 , ends in 1. Since 20132013 = 20134·503+1
= (20134 )501 · 2013 = (10N + 1) · 2013, 20132013 ends in a 3.
3
(D)
Answer. We start with 20 socks! Having started with an even number (10) of black socks, we must lose either zero
10 9
9
or two black socks during the two wash cycles. The probability of losing two blue socks in two cycles is
·
=
,
20 19
38
9
10 9
·
=
. Since the two possibilities are distinct, the total probability
while that of losing two black socks is (also)
20 19
38
9
9
9
+
=
.
is
38 38
19
Question 20. What digit does 20132013 end in?
(A) 1
1
2
4