SOLUTIONS
PROBLEM OF THE WEEK
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3/9/26
Strategy: Draw some useful line segments.
The area of the shaded region is the difference between the areas of the circle and the square. Draw the diagonals of the square. The diagonals split the square into four congruent isosceles right triangles. The area of each triangle is 18 ÷ 4 = 4.5. The diagonals are also diameters of the circle, so the base and height of each of the small right triangles is a radius of the circle. The area of each triangle, ½ × r × r is 4.5, so r = 3 cm. The area of the circle is approximately 3.14 × 3² = 28.26.
The difference between the areas of the circle and square is 28.26 - 18 = 10.26.
Rounding off, the area of the shaded region is 10 sq cm.
Volume 3, Set 10, Olympiad 5, 5D. 7 Minutes
3/2/26
Strategy: Determine the number of letters in the repeating part.
In the sequence ABBCCDABBCCD..., there are 6 letters before the pattern repeats itself. The first 78 terms of the sequence contain 13 complete repetitions of the pattern and no additional letters.
The 78th letter is the same as the 6th letter, which is D.
Volume 3, Set 7, Olympiad 2, 2B. 4 Minutes
2/26/26
Strategy: Determine which digits satisfy each condition.
The single digit prime numbers are 2, 3, 5, and 7. The single digit perfect squares are 0, 1, 4, and 9, leaving 6 and 8 as neither prime nor perfect square. Choose the largest digit in each group and arrange from greatest to least to get the largest possible number, 987. But 987 is divisible by 3. To get the next greatest number, choose the next greatest digit from the group containing 7.
The greatest possible value of Janine's number is 985.
Volume 3, Set 4, Olympiad 1, 1D. 5 Minutes
2/16/26
Strategy: Work backwards.
The table below shows actions in reverse. The first column names each person, last to first. The second column shows the actions each person took. The third column states the number of coins on the table as each person approached.
56 coins were on the table to start with.
Volume 3, Set 7, Olympiad 2E, 7 Minutes
2/9/26
Strategy: Determine how many times Keri came in first.
Keri could not have finished first 3 or more times because that would give her more than 20 points. Suppose she finished first twice, a total of 16 points. Any remaining points must have come from finishing second, scoring 3 points each time. She can't have scored 4 points. The same reasoning shows she can't have finished first 0 times since 20 is not a multiple of 3. So Keri finished first once, for 8 points. The remaining 12 points came from 4 second place finishes.
The remaining 12 points came from 4 second place finishes.
Volume 3, Set 7, Olympiad 1, 1C. 5 Minutes
2/2/26
Strategy: Count horizontally, from the top layer down.
Make a table that counts cubes separately for each layer. In each case, add the number of hidden cubes to the number of visible cubes.
1 + 3 + 6 + 10 = 20 cubes are used to form the tower.
Volume 3, Set 4, Olympiad 4, 4C. 5 Minutes
1/26/26
Strategy: Combine the two cases.
Suppose 2 students are absent and Mr. Alvarez still gives each student 4 sheets. He will have the original 16 sheets left over and in addition the 4 sheets that he would have given to each of the absentees. This total of 24 sheets is enough to give each of the students who are present 1 additional sheet, with 3 left over. Then there are 24 - 3 = 21 students present.
Mr. Alvarez has 5 × 21 + 3 = 108 sheets of paper.
Volume 3, Set 13, Olympiad 2, 2C. 6 Minutes
1/19/26
Strategy: Start with the first few days; look for a pattern.
The total Lou eats for the first N days is N2. September has 30 days,
so Lou eats 900 jelly beans in September.
Volume 3, Set 10, Olympiad 5, 5C. 6 Minutes
1/12/26
Strategy: Work backwards.
Ben ended with twice as many of the 36 CDs as Ali, so Ben ended with 24 and Ali with 12.
Ali had given Ben 40%, or 2/5 of her original number, so the 12 she ended with was 3/5 of the number she started with.
Then 1/5 of her original number was 4, and Ali had 20 CDs originally.
Volume 3, Set 12, Olympiad 3, 3C. 5 Minutes
1/5/26
Strategy: Use counting principles.
Emma buys a German novel and a Spanish novel, or a German novel and a French novel, or a Spanish novel and a French novel. She has 4 choices for the German novel and 5 for the Spanish novel, so she can buy novels in those two languages in 4 x 5 = 20 ways. Likewise, she can buy a German novel and a French novel in 4 x 6 = 24 ways, and she can buy a Spanish novel and a French novel in 5 x 6 = 30 ways.
In all Emma can purchase two novels in two languages in 20 + 24 + 30 = 74 ways.
Volume 3, Set 13, Olympiad 1, 1E. 7 Minutes
12/22/25
Strategy: Minimize the number of ushers.
Group each 30 fans with 1 usher to form groups of 31. Then the 20,000 people are divided into 645 groups of 31 each, with 5 people left over. Those 5 people must contain at least 1 usher and at most 4 fans. There must be at least 645 + 1 = 646 ushers.
There are at most 20,000 - 646 = 19,354 fans that can be in attendance.
Volume 3, Set 16, Olympiad 4, 4C, 5 Minutes
12/15/25
Strategy: Draw a diagram showing time.
Mark the opposite shores in minutes, 0 to 35. The slower boat touches alternating shores at 7, 14, 21, 28, and 35 minutes. The faster boat touches alternating shores at 5, 10, 15, 20, 25, 30, and 35 minutes. Where the paths cross, the boats are passing.
From the diagram, the faster boat passes the slower boat 7 times.
Volume 1, Olympiad 56, #4, 6 Minutes
12/08/25
Strategy: Divide by the cost of the remaining pounds
Total Charge: $3.45
Charge for first five pounds: $1.65
Charge for remaining pounds: $1.80
Since the charge for each of the remaining pounds is 12¢ per pound, the number of remaining pounds is 1.80/12 = 15.
The total number of pounds in the package is 15 + 5 or 20.
Volume 1, Olympiad 49, #4,5 Minutes
12/01/25
Strategy: Use spatial reasoning and corner analysis
Only the 2 by 2 by 2 cubes at the eight vertices (corners) of the 3 by 3 by 3 cube have exactly three red faces.
The diagram below shows the location of one of the eight cubes.
Volume 1, Olympiad 56, #4, 6 Minutes
11/24/25
Strategy: Use the work-rate method
In one minute, the water faucet fills 1/15 of the tub and the drain empties 1/20 of the tub. Together, the faucet and drain "fill" 1/15 - 1/20 of the tub each minute. This is equivalent to 4/60 - 3/60 or 1/60 of the tub.
Therefore, in 60 minutes the faucet and drain working together will fill 60/60 or the entire tub.
Volume 1, Olympiad 56, #4, 6 Minutes
11/17/25
Strategy: Feed all the cats first.
One quart of milk feeds 6 cats so 15 cats require 2½ quarts of milk. This leaves z quart of milk.
One quart of milk feeds 10 kittens, so 5 kittens can be fed with the leftover milk.
Volume 2 Set 15, Olympiad 5, 5A. 4 Minutes
11/10/25
Strategy: Proceed one condition at a time.
The thousands digit can't be 2 or greater because then the ones digit would be more than 9. Thus, the thousands digit is 1. Also, the question reads, "Dr. Bolton WAS born..."
Condition:
The tens digit is twice the thousands digit ...... 1 _ 2 _
The ones digit is 3 times the tens digit ...... 1 _ 2 6
The hundreds digit is the sum of the other three digits ...... 1 9 2 6
Dr. Bolton was born in 1926.
Volume 2, Set 6, Olympiad 3, 3A. 3 Minutes
11/3/25
Strategy: Represent the problem visually.
The elapsed time from 8:26 to 11:26 is 3 hours, which is 180 minutes. There are 3 (not 4) passing periods between classes, for a total passing time of 12 minutes. This leaves a total of 168 minutes for the four classes.
One class period contains 42 minutes.
Volume 2, Set 11, Olympiad 1, 1C. 5 Minutes
10/27/25
Strategy: Examine the differences between scores.
The least sum is 3 x 2 = 6 and the greatest sum is 3 x 8 = 24. The 5-ring scores 3 points more
than the 2-ring and the 8-circle scores 3 points more than the 5-ring. Thus "moving" a dart from the 2-ring increases the sum by a multiple of 3. Hence, all possible sums are multiples of 3 and are between 6 and 24, inclusive;
7 sums are possible.
Volume 2, Set 11, Olympiad 1, C. 5 Minutes
10/21/25
Strategy: Choose the gumballs so as to avoid three of a color. After Anna buys 8 gumballs, she might have 2 red, 2 green, 2 yellow, and 2 purple. The next gumball she buys must be one of those same four colors, guaranteeing her three of the same color. So, the minimum number of gumballs she should expect to buy is 9.
The minimum cost that guarantees three gumballs of the same color is 5 × 9 = 45¢.
Volume 2, Set 5, Olympiad 5, 5D. 6 Minutes
10/13/25
Strategy: Draw a picture.
There are two possibilities:
This requires an extra 8m of fencing.
The least number of meters of additional fencing needed is 4 m.
Volume 2, Set 17, Olympiad 2, 2D. 56 Minutes
10/6/25
Strategy: Combine one stamp of each kind into a “super Stamp”.
Imagine one "super stamp" consisting of one of each of the stamps mentioned. This super stamp costs 50 + 20 + 10 + 5 = 85¢. There are 510 ÷ 85 = 6 of these super stamps.
Therefore Jessie has 6 fifty-cent stamps.
**This method illustrates the distributive property**
Volume 2, Set 10, Olympiad 5, 5B. 5 Minutes
9/29/25
Strategy: Draw a diagram
As shown, the front of house ABCD would be 70 - 10 - 10 = 50 feet. The side of house ABCD would be 100 - 20 - 30 = 50 feet.
The largest area the house can have is 50 x 50 = 2500 sq. ft.
Volume 2, Set 5, Olympiad 5, 5D. 6 Minutes
9/22/25
Strategy: Find the total number of points earned.
In each race 5 + 3 + 1 = 9 points are earned. The three racers earn a total of 3 × 9 = 27 points. If each racer has the same score, each would have 9 points. Thus, Beth's total score is greater than
9. Each runner's total is the sum of three odd numbers and therefore is odd. The least odd number greater than 9 is 11, which is a possible winning score. For example, the scores could be 5, 5, 1; 3, 3, 3; 1, 1, 5.
Beth's least possible total score is 11.
Volume 2, Set 10, Olympiad 2, 2C. 6 Minutes