Mathematical Olympiads
February 10, 2015
for Elementary & Middle Schools
SOLUTIONS AND ANSWERS
4A
4A METHOD 1: Strategy: Group nickels and dimes together.
6
Pair up every nickel with a dime. There are five dimes left over, and those are worth $.50. Therefore the pairs of nickels and dimes make up a total of $1.40 – $.50 = $.90. Each nickel-­‐dime pair is worth $.15, so there are 90/15 = 6 pairs of nickels and dimes. Pablo has 6 nickels. METHOD 2: Strategy: Use an organized guess and check.
Since the sum of all the coins does not end in a five, there must be an even number of nickels. If there are 2 nickels and 7 dimes the total amount is $.80. With 4 nickels and 9 dimes, the total is $1.10. With 6 nickels and 11 dimes the total amount is $1.40. FOLLOW-UP: On a plane are 150 passengers. There are 20 more men than women. How
many women are on the plane? [65]
4B
2000
4B METHOD 1: Strategy: Apply the distributive property.
Recognize that (18 × 40) + (12 × 40) = (18 + 12) × 40 = 30 × 40. Then recognize that (20 × 28) + (20 × 12) = 20 × (28 + 12) = 20 × 40. Add 30 × 40 + 20 × 40 = (30 + 20) × 40 = 50 × 40 = 2000. METHOD 2: Strategy: Create a common factor.
(18 × 40) + (12 × 40) + (20 × 28) + (20 × 12) = (18 × 40) + (12 × 40) + (40 × 14) + (40 × 6) = 40(18 + 12 + 14 + 6) = 40(50) = 2000. FOLLOW-UP: What is the value of (15 × 20) + (47 × 20) + (38 × 20) + (100 × 12) +
(100 × 5) + (100 × 3)? [4000]
4C
210
4D
4C METHOD 1: Strategy: Use the least common multiple (LCM).
If a number is divisible by 3, 5, and 7, it is divisible by their least common multiple. The LCM of 3, 5, and 7 is their product, since the three numbers have no common factors other than 1. Since 3 × 5 × 7 = 105, which is not between 200 and 300, we must multiply this result by 2. This gives us 105 × 2 = 210 which is divisible by 3, 5, and 7 and is also between 200 and 300. The number is 210. METHOD 2: Strategy: Consider the multiples of 7 between 200 and 300.
We first consider multiples of 7 to lessen the number of possibilities. The multiples of 7 in the given interval are 203, 210, 217, …, 294. Since the number must be divisible by 5 as well it must end in either a 5 or a 0. A number that is divisible by 3 has the sum of its digits divisible by 3. The number 210 satisfies all three conditions so 210 is the answer.
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10
4E
181
sq cm
Olympiad 4, Continued
4D Strategy: Count in an organized fashion.
If the first digit is 6, the other two digits must add to 18. There is only one such 3-­‐digit number whose digit sum is 24. That number is 699. If the first digit is 7, the other two digits must add to 17. There are two such numbers: 789 and 798. If the first digit is 8, the remaining two digits must add to 16. We can use 7 and 9 or 8 and 8. There are three such numbers: 879, 897, and 888. If the first digit is 9, the other two digits add to 15. There are four such numbers: 969, 996, 987, and 978. The total number of counting numbers for which the digit sum is 24 is 1 + 2 + 3 + 4 = 10. 4E METHOD 1: Strategy: Count the painted areas for corner and edge cubes separately.
The top cube has 5 painted faces, a total of 5 sq cm on that layer. For each layer below the top, there are two kinds of cubes: corner cubes that have 3 painted faces and edge cubes which have only 2 painted faces. Each layer will have 4 corner cubes so there are 4 × 4 = 16 corner cubes for a total of 16 × 3 = 48 sq cm. On a layer with n cubes on a side, there will be n – 2 edge cubes on each of four sides. Therefore in the 3 × 3 layer there are 4 × (3 – 2) = 4 edge cubes for a total of 4 × 2 = 8 sq cm. In the 5 × 5 layer there is a total of 4 × (5 – 2) = 12 edge cubes with an area of 12 × 2 = 24 sq cm. The remaining two layers have areas of 4 × (7 – 2) × 2 = 40 sq cm and 4 × (9 – 2) × 2 = 56 sq cm. The total painted area is 5 + 48 + 8 + 24 + 40 + 56 = 181 sq cm. METHOD 2: Strategy: Look at the picture from different angles.
Looking down from the top, the exposed area is just the area of a 9 × 9 rectangle or 81 sq cm. Look at the shape from the side and the exposed area is 1 + 3 + 5 + 7 + 9 = 25 sq cm. Since there are 4 sides the total area that would be painted is 81 + 4 × 25 = 181 sq cm. FOLLOW-UP: Continue the pyramid by adding rows beneath the 9 by 9 row. Follow the same
pattern for 5 more rows. Find the volume of the new pyramid. [1330 cu cm]
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.