🇺🇸 AMC 8 ⇄ switch contest
2014 AMC 8

Problem 8

Problem 8 · 2014 AMC 8 Easy
Number Theory divisibility-by-11

Eleven members of the Middle School Math Club each paid the same integer amount for a guest speaker to talk about problem solving at their math club meeting. In all, they paid their guest speaker $1A2. What is the missing digit A of this 3-digit number?

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Answer: D — A = 3.
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Hint 1 of 2
"11 people each paid the same whole amount" is the whole clue: the total is 11 × something, so 1A2 must be a multiple of 11.
Still stuck? Show hint 2 →
Hint 2 of 2
Test divisibility by 11 with the alternating digit sum: add every other digit, subtract the rest, and check for a multiple of 11.
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Approach: translate the word clue into a divisibility-by-11 test
  1. Equal split among 11 people ⇒ total is a multiple of 11, so 1A2 must be divisible by 11.
  2. Alternating sum (test for 11): 1 − A + 2 = 3 − A must be a multiple of 11.
  3. A is one digit (0–9), so the only way 3 − A hits a multiple of 11 is 3 − A = 0, giving A = 3.
  4. Check: 132 = 11 × 12. ✓
  5. You'll reuse this: "n equal shares" always means "total divisible by n" — the fastest way to turn a money/sharing sentence into a number-theory condition.
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