Brainstorm
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pending
What is the smallest positive integer that leaves a remainder of 1 when divided by 2, remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and so on up to a remainder of 9 when divided by 10?
The answer is 2519.
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1 solution
N + 1 must be a multiple of 2
N + 1 must be a multiple of 4 but if it is a multiple of 2 & 4 it is necessarily a multiple of 8
N + 1 must be a multiple of 5 but if it is a multiple of 2 & 5 it is necessarily a multiple of 10
N + 1 must be a multiple of 6 but if it is a multiple of 2 & 9 (18) it is necessarily a multiple of 6
N + 1 must be a multiple of 7
N + 1 must be a multiple of 9
Hence 2 x 4 x 5 x 7 x 9 = 2520
so N+1=2520 N=2519