Problem 12 · 2017 AMC 8
Easy
Number Theory
divisibility
The smallest positive integer greater than 1 that leaves a remainder of 1 when divided by 4, 5, and 6 lies between which of the following pairs of numbers?
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Answer: D — Between 60 and 79.
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Hint 1 of 2
'Remainder 1' means the number is just one past a clean multiple. So subtract that 1: n − 1 leaves no remainder under 4, 5, or 6 — it's a common multiple of all three.
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Hint 2 of 2
Smallest common multiple = lcm. Find lcm(4, 5, 6), then add the 1 back. Careful: lcm(4,6) = 12, not 24.
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Approach: shift by the remainder, then take the lcm
- If n leaves remainder 1 under each of 4, 5, 6, then n − 1 is exactly divisible by all three — a common multiple. Peeling off the remainder turns three conditions into one lcm.
- lcm(4, 5, 6) = 60 (note 4 and 6 share a factor of 2, so it's 60, not 120). The smallest such n > 1 is 60 + 1 = 61.
- 61 lies between 60 and 79.
- Why this transfers: 'same remainder r for several divisors' always means n = lcm(…)·k + r; subtract r first to expose the lcm.
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