Problem 19 · 2004 AMC 8
Medium
Number Theory
lcm
A whole number larger than 2 leaves a remainder of 2 when divided by each of the numbers 3, 4, 5, and 6. The smallest such number lies between which two numbers?
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Answer: B — Between 60 and 79.
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Hint 1 of 2
A remainder of 2 every time is suspicious — it means if you set aside that constant 2, the leftover divides evenly by all four numbers. So look at x − 2 instead of x.
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Hint 2 of 2
The move is shift away the common remainder: x − 2 must be a common multiple of 3, 4, 5, 6, so the smallest is their LCM. (And note 6 = 2×3 is already covered by 4 and 3, so really LCM(3,4,5).)
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Approach: subtract the remainder, then LCM
- Same remainder 2 for all divisors means x − 2 is divisible by 3, 4, 5, and 6 at once — so x − 2 is a common multiple.
- The smallest positive common multiple is LCM(3, 4, 5, 6) = 60 (the 4 supplies 2×2, the 3 and 5 the rest; 6 adds nothing new).
- So the smallest x − 2 = 60, giving x = 62, which lies between 60 and 79.
- This 'remove the remainder' trick turns any 'leaves remainder r for several divisors' question into a clean LCM problem — a staple of number-theory contests.
- Sanity check: 62 ÷ 5 = 12 r2, 62 ÷ 4 = 15 r2, 62 ÷ 3 = 20 r2, 62 ÷ 6 = 10 r2. All four give remainder 2.
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