Problem 15 · 2012 AMC 8
Medium
Number Theory
lcm
The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?
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Answer: D — Between 61 and 65.
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Hint 1 of 2
"Leaves 2 left over" every time is the clue: if you remove those 2 leftover, the number divides evenly by 3, 4, 5, and 6. So x − 2 is a common multiple of all four.
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Hint 2 of 2
This is the subtract-the-leftover shift: turn "remainder 2" into "exact multiple", then the number you need is the LCM of 3, 4, 5, 6. (Notice 3 and 6 are covered once you handle 4 and... well, 6 needs both 2 and 3.)
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Approach: shift off the leftover, then use the LCM
- Strip away the 2 leftover: x − 2 must divide evenly by 3, 4, 5, and 6 at once — so it's a common multiple of all four.
- The smallest common multiple is LCM(3, 4, 5, 6). Build it from prime needs: 4 = 2², 5, and a 3 (which also covers 6) ⇒ LCM = 2² × 3 × 5 = 60.
- So the smallest x − 2 above 0 is 60, giving x = 62 — which lands between 61 and 65.
- The reusable move: "same remainder r for several divisors" ⇒ subtract r, find the LCM, add r back. Don't list multiples one by one.
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