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 Going maximum with factorial(s) (Posted on 2008-10-24)
Consider all possible positive integer(s) M ≥ 5, such that (M)! is expressible as the product of (M-3) consecutive positive integers. For example, we observe that: 6! = 8*9*10 .

Determine the maximum value of M.

 No Solution Yet Submitted by K Sengupta No Rating

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 algebraic exploration (proof?) Comment 2 of 2 |
(In reply to computer exploration (spoiler? -- UBASIC program) by Charlie)

If the (M-3) consecutive integers are from 5 to M+1, then M! must equal M!*(M+1)/24, and that is indeed the M=23 that was found.

If the integers were from 6 to M+2, then M! must equal M!*(M+1)*(M+2)/120, so (M+1)*(M+2) = 120, which doesn't have a positive integral solution.

From 7 to M+3: (M+1)*(M+2)*(M+3)= 720, which is satisfied by M = 7

These necessarily have smaller solutions in M as we continue this process, as the denominator of the fraction that is to equal 1 goes up by a smaller factor than the numerator when M is large.

 Posted by Charlie on 2008-10-24 13:11:27

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