Problem 23 · 2019 AMC 8
Hard
Algebra & Patterns
divisibilitycasework
After Euclid High School's last basketball game, it was determined that 14 of the team's points were scored by Alexa and 27 were scored by Brittany. Chelsea scored 15 points. None of the other 7 team members scored more than 2 points. What was the total number of points scored by the other 7 team members?
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Answer: B — 11 points.
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Hint 1 of 2
Scores are whole numbers, so T/4 and 2T/7 must be integers. That forces T to be a multiple of both 4 and 7 — i.e. a multiple of 28. Suddenly only a few totals are possible.
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Hint 2 of 2
The 7 others score at most 2 each, so their total is between 0 and 14. Write "others" in terms of T, then test T = 28, 56, … until it lands in that window.
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Approach: divisibility narrows T to multiples of 28, then bound it
- Let T be the team total. Alexa's T/4 and Brittany's 2T/7 must both be whole numbers, so T is a multiple of lcm(4, 7) = 28.
- Others' points = T − T4 − 2T7 − 15 = 13T28 − 15, and this must sit in [0, 14] since 7 players score ≤ 2 each.
- Test multiples of 28: T = 28 gives 13 − 15 = −2 (impossible); T = 56 gives 26 − 15 = 11 ✓ (in range); T = 84 gives 39 − 15 = 24 (too big). Only T = 56 works.
- Why this transfers: when unknowns are split by fractions, the denominators force the total to be a multiple of their lcm — combine that with a realistic bound (here, ≤14) and the candidate list is tiny.
Mark:
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