Problem 21 · 2014 AMC 8
Hard
Number Theory
divisibility-by-3mod-arithmetic
The 7-digit numbers 74A52B1 and 326AB4C are each multiples of 3. Which of the following could be the value of C?
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Answer: A — C = 1.
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Hint 1 of 2
You can't find A and B individually — and you don't need to. Both numbers share the same A + B, so write each divisibility-by-3 condition and subtract them to cancel A + B entirely.
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Hint 2 of 2
What's left is a single condition on C mod 3. Only one answer choice fits.
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Approach: subtract the two digit-sum conditions so the shared A+B cancels
- Divisibility by 3 = digit sum divisible by 3. First number: 7+4+5+2+1 = 19, so 19 + A + B ≡ 0 ⇒ A + B ≡ 2 (mod 3).
- Second number: 3+2+6+4 = 15, so 15 + A + B + C ≡ 0 ⇒ A + B + C ≡ 0 (mod 3).
- Subtract the first from the second: the A + B cancels, leaving C ≡ −2 ≡ 1 (mod 3).
- Among the choices {1, 2, 3, 5, 8}, only C = 1 is ≡ 1 (mod 3).
- Why this transfers: when an unknown blocks two equations the same way, subtract to eliminate it — the same move that solves systems of equations works on mod conditions too.
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