Division
E
Mathematical Olympiads
December 16, 2014
for Elementary & Middle Schools
SOLUTIONS AND ANSWERS
2A METHOD 1: Strategy: Pair numbers that add to 100.
9 + 26 + 83 + 55 + 45 + 17 + 74 + 91 = (9 + 91) + (26 + 74) + (83 + 17) + (55 + 45)
= 4(100) = 400.
Contest
2
2A
400
METHOD 2: Strategy: Add the numbers in the order written.
9 + 26 + 83 + 55 + 45 + 17 + 74 + 91 = 400.
FOLLOW-UP: Find the value of
. [20]
2B
2B METHOD 1: Strategy: Consider which numbers have an odd number of factors.
The number 6 factors into 6 × 1 or 3 × 2. Therefore 6 has four factors, 1, 2, 3, and 6. The
number 9 factors into 9 × 1 and 3 × 3. The number 9 has only three factors, 1, 3, and 9.
To have an odd number of factors, the number must factor into two factors that are the
same. Therefore it must be a perfect square. Count the number of perfect squares less
than 20. These numbers are: 1, 4, 9, and 16. Therefore there are 4 numbers less than 20 2C
that have an odd number of factors.
METHOD 2: Strategy: List the factors of all non-prime numbers less than 20. Prime
numbers only have 2 factors.
1: 1
9: 1, 3, 9
15: 1, 3, 5, 15
4: 1, 2, 4
10: 1, 2, 5, 10
16: 1, 2, 4, 8, 16
6: 1, 2, 3, 6
12: 1, 2, 3, 4, 6, 12
18: 1, 2, 3, 6, 9, 18
8: 1, 2, 4, 8
14: 1, 2, 7, 14
The four bolded results each have an odd number of factors.
FOLLOW-UP: How many positive integers less than 50 have exactly two factors? [15]
2C Strategy: Determine the least common denominator for the two fractions.
The least common denominator for the fractions 2/3 and 3/4 is 12. Therefore the
fewest possible number of games played in each half is 12. Team Alpha won
games in the first half and
games in the second half. Thus the fewest number
of games won for the season was 8 + 9 = 17 games.
Copyright © 2014 by Mathematical Olympiads for Elementary and Middle Schools, Inc. All rights reserved.
4
17
2D
66
sq cm
2E
22
Olympiad 2, Continued
2D METHOD 1: Strategy: Label the cubes.
There are 4 cubes with 5 red faces, 10 cubes with 4 red faces, and 2 cubes with 3 red faces. The
total area of all the red faces is 4 × 5 + 10 × 4 + 2 × 3 = 20 + 40 + 6 = 66 sq cm.
METHOD 2: Strategy: Divide the object into similar shaped pieces.
The 4 prongs of three cubes each have 4 × 13 = 52 painted faces. The cubes that join these sets
of three cubes each have 3 painted faces and 2 × 3 = 6. The two cubes joining the two groups of
seven cubes each have 4 faces painted red and 2 × 4 = 8. The total area of all the painted cubes is
52 + 6 + 8 = 66 sq cm.
METHOD 3: Strategy: Subtract the number of unpainted faces from the total number of faces.
Sixteen cubes have a total of 16 × 6 = 96 sq cm. There are 15 places where one cube shares a face
with another cube. Since these faces are common to two cubes, we subtract 2 × 15 = 30 from 96
to get 96 – 30 = 66 sq cm of painted cubes.
2E Strategy: Start with cases that factor uniquely.
The only way to get 14 with three different factors is 14 = 1 × 2 × 7. The only way to get 15 with
three different factors is 15 = 1 × 3 × 5. Therefore the upper left corner must be 1. Since 5 must
be in the top row, it must be in the third column because 5 is a factor of 180 but not of 144. In
column 1, either the 2 or the 7 must be in the third row. The 7 must be in
the lower left corner since 7 is not a factor of 64. Note that 378/7 = 54
so the remaining numbers in the third row are 6 and 9. If 6 were in the
lower right corner the number above it would be 180/(5 × 6) = 6, which is
not possible. The fourth corner is 9 and the number above it will be
180/(5 × 9) = 4. The sum of the numbers in the 4 corners is 1 + 5 + 7 + 9 = 22.
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.