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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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 computer exploration (spoiler? -- UBASIC program) | Comment 1 of 2
10   for M=6 to 1006
20    F=!(M)
30    R=M-3
40    Rt=int(F^(1/R))
50    St=Rt+int(-R/2)-1
60    Fin=St+R-1
65    if St<1 then St=1:Fin=St+R-1
68    Trial=0
70    while Trial<F
72      Trial=!(Fin)/!(St-1)
80      if Trial=F then print M;F;St;Fin
82      St=St+1:Fin=Fin+1
85    wend
90   next

produces only

6  720  8  10
7  5040  7  10
23  25852016738884976640000  5  24

indicating that from M=6 through M=1006 (the most that UBASIC can handle with no overflow), M can be only 6, 7 or 23 to produce the desired result. It shows 7! as being the product of the numbers 7 through 10, or 5040, and 23! as being the product of the numbers 5 through 24, or 25,852,016,738,884,976,640,000.

 Posted by Charlie on 2008-10-24 12:34:10

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