Problem 15 · 2008 AMC 8
Medium
Number Theory
divisibilityaverage-as-integer
In Theresa's first 8 basketball games, she scored 7, 4, 3, 6, 8, 3, 1 and 5 points. In her ninth game, she scored fewer than 10 points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than 10 points and her points-per-game average for the 10 games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
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Answer: B — 40.
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Hint 1 of 2
"Average is a whole number" is a divisibility clue in disguise: the running total must be a multiple of the number of games.
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Hint 2 of 2
Each new score is under 10, so the total can only land in a tiny window — usually just one multiple fits, pinning the score exactly.
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Approach: turn "integer average" into "total is a multiple"
- First 8 scores sum to 37. Adding game 9 (1–9 points) lands the 9-game total between 38 and 46, and it must be a multiple of 9. Only 45 fits — so game 9 = 45 − 37 = 8.
- Adding game 10 (under 10) lands the 10-game total between 46 and 54, and it must be a multiple of 10. Only 50 fits — so game 10 = 50 − 45 = 5.
- Product: 8 · 5 = 40.
- Why this transfers: "the average is an integer" always means "the sum is divisible by the count" — pair that with a bounded range and the value is usually forced.
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