Problem 13 · 2025 AMC 8
Medium
Number Theory
mod-arithmeticcyclicity
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Answer: A — Histogram (A).
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Hint 1 of 2
Computing 25 remainders one by one is a trap. Adding 2 each time bumps the remainder by 2 (mod 7) — so the remainders march in a short repeating loop. Find that loop.
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Hint 2 of 2
The loop has length 7, and 25 = 3 full loops + 4 leftovers. The full loops hit every remainder equally, so it's only those 4 leftovers — the first remainders in the cycle — that decide which bars stand taller.
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Approach: find the period-7 cycle; only the leftovers break the tie
- Adding 2 each step bumps the remainder by 2 (mod 7), so 2, 4, 6, 8, 10, 12, 14, … give remainders that loop: 2, 4, 6, 1, 3, 5, 0, repeating every 7.
- There are 25 even numbers = 3 complete loops (21 numbers, each remainder appearing 3 times — so all bars start tied) plus 4 extras, which are the first four of the cycle: 2, 4, 6, 1.
- Those 4 extras push remainders 1, 2, 4, 6 up to 4 each, while 0, 3, 5 stay at 3 each. The histogram with its taller bars exactly on 1, 2, 4, 6 is choice A.
- Why this transfers: for "how often does each remainder occur" questions, find the repeating cycle, peel off the whole cycles (which tie everything), and let the short leftover tail break the tie. That beats brute-forcing every value.
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