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 LCM and Equality Puzzle (Posted on 2015-09-07)
Each of M and N is a positive integer such that:
LCM (M, M+5) = LCM(N, N+5)

 No Solution Yet Submitted by K Sengupta No Rating

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 Analytical Solution (spoiler) Comment 1 of 1
Yes, it would seem that M always = N

a) If M is not a multiple of 5, then no factor of M can be a factor of M + 5, so the LCM of (M, M+5) = M*(M+5).
This is not divisible by 5, so N cannot be divisible by 5, so the LCM(N,N+5) = N*(N+5).
Therefore M*(M+5) = N*(N+5)
Therefore, M = N.

b) If M is a multiple of 5, the LCM of (M, M+5)
= 5*LCM(M/5,(M+5)/5)
= 5*LCM(M/5,M/5+1)
= 5*(M/5)*(M/5+1)
This is divisible by 5, so N is 5, so the LCM(N,N+5) = 5*(N/5)*(N/5+1)
Therefore 5*(M/5)*(M/5+1) = 5*(N/5)*(N/5+1)
Therefore M*(M+5) = N*(N+5)
Therefore, M = N.

There is nothing unique about 5.  For any prime P, with M and N positive integers,
M = N if and only if LCM(M,M+P) = LCM(N,N+P)

 Posted by Steve Herman on 2015-09-08 17:14:41

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