All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Going maximum with factorial(s) (Posted on 2008-10-24) Difficulty: 3 of 5
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

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts 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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information