Mathematical Olympiads
November 18, 2014
for Elementary & Middle Schools
SOLUTIONS AND ANSWERS
1A METHOD 1: Strategy: Use grouping.
Observe that the six numbers being added are consecutive and can be grouped into three pairs that will have the same sum. Starting with the two middle numbers: 3 + 4 = 7; and working outwards: 2 + 5 = 7 and 1 + 6 = 7. This makes three groups of 7. The sum is equal to 3 × N, therefore N = 7. METHOD 2: Strategy: Use the commutative property and grouping.
Observe that 3 × N means 3 groups of N or equivalently N groups of 3. Count the number of groups of 3. One group of 3 is 1 + 2, 3 alone is a second group of 3, 4 + 5 is three more groups of 3 and 6 is two more groups of 3. A total of 7 groups of 3 so N = 7. =
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METHOD 3: Strategy: Use arithmetic and algebra.
1A
N=7
1B
64
1C
Add the numbers on the left side of the equation and then divide both sides of the equation by 3. Thus 21 = 3 × N, so N = 7. 3
FOLLOW-UP: The average of a set of consecutive odd integers is 36. If the sum of all the
integers in this set is 288, what is the greatest integer in the set? [43]
1B METHOD 1: Strategy: Find a pattern for the sums of the numbers in each row.
The sums of the numbers in each row starting at the top of the triangle are: 1, 2, 4, 8, and 16. Each subsequent row is double the previous row so the sixth row adds to 32 and the seventh row has a sum of 64. 1D
36
METHOD 2: Strategy: Continue the diagram and then add.
The sixth row is 1 5 10 10 5 1 and the seventh row is 1 6 15 20 15 6 1. The sum of the numbers in the seventh row is 64.
sq cm
METHOD 3: Strategy: Use the diagram to determine the pattern.
1E
Note that from the top to the bottom of the diagram every row has two arrows from each number, representing the fact that in the row below, each of the numbers in the row above will be counted twice. Thus the sum in each succeeding row is twice the sum of the previous row. The sixth and seventh rows add to 32 and 64, respectively. FOLLOW-UP: The sum of the numbers in the thirteenth row is 4,096. If the sum of the
numbers in row P is subtracted from the sum of the numbers in the thirteenth row, the
difference is 3,840. What row does P represent? [9]
Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
19
Olympiad 1, Continued
1C METHOD 1: Strategy: Create a Venn diagram.
Start the diagram by placing the 12 in the region that displays students who like both video games and cartoons. Since 16 students like video games, 16 – 12 = 4 students like only video games. Since 20 students like cartoons, 20 – 12 = 8 students like only cartoons. Adding the number of students in these 3 categories accounts for 4 + 12 + 8 = 24 students. Since the class has 27 students, the number of students who do not like either is 27 – 24 = 3. METHOD 2: Strategy: Make a table to represent each student.
Let V represent the students who like video games and let C represent those who like cartoons. V V V V V V V V V V V V V V V V C C C C C C C C C C C C C C C C C C C C The three empty spaces at the end of the table represent the students who like neither. FOLLOW-UP: A survey of 91 fifth grade students found that 44 students like only dogs and 29
like only cats. If there are twice as many students who like both dogs and cats, as there are
students who like neither, how many students like dogs? [56]
1D Strategy: Consider the rectangles without any overlap.
If the two rectangles did not overlap, the total length would be 13 + 13 = 26 cm. If we slide one of the rectangles over the other until the total length is 22 cm, there would be 26 – 22 = 4 cm of overlap. Since the height is still 9 cm, the area of the overlap is 4 × 9 = 36 sq cm. 1E METHOD 1: Strategy: Use the shapes as items placed on a balance scale.
Place 2 Δs and 1 ◊ on one side of the scale and an 18 unit weight on the other side. Then add the remaining shapes to the first side and the corresponding weights to the other side of the scale. On one side we have (Δ + Δ + ◊) + (◊ + ◊ + ⌂) + (⌂ + ⌂ + Δ) and on the other side 18 + 22 + 17. Therefore three of each shape equals 57, so one of each shape equals 57/3 = 19. METHOD 2: Strategy: Make a table.
Δ ◊ ⌂ 0 18 X 1 16 X 2 14 X 3 12 X 4 10 2 5 8 6 6 6 10 7 4 14 8 2 18 9 0 22 Fill in the table with all possible values of Δ and compute the other values based upon the first and second equations. There is only one set of values that satisfy the third equation. These are shown in bold and 5 + 8 + 6 = 19. FOLLOW-UP: Given the equation 2 × Δ – 2 × ◊ + ⌂ = Δ – 2 × ◊ + 2 × ⌂ – 6, if Δ = 15, what
is the value of ⌂? [21]
NOTE: Other FOLLOW-UP problems related to some of the above can be found in our three
contest problem books and in “Creative Problem Solving in School Mathematics.”
Visit www.moems.org for details and to order.