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 Factorials everywhere (Posted on 2004-08-16)
The equation a!b!=c! is trivial if we allow a or b to be 0 or 1.

However, adding that 1<a<b<c, also allows trivial solutions: which? What condition should be added to disallow such solutions?

Finally, adding that condition, can you find any solution to the problem?

 See The Solution Submitted by Federico Kereki Rating: 2.6000 (5 votes)

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 Solution | Comment 4 of 6 |

If a!b!=c!, then

a! * (1*2*3*...*b)  = 1*2*3*...*b*(b+1)*...*c

Thus a! equals the product of terms from b+1 to c. The trivial solution has been pointed out to be b=a!-1, in which case a! = c, the final single term of c!.

A non-trivial solution would be one where a! equals the product of several consecutive numbers.  For instance,

6! = 720 = 8*9*10

thus 6!*7!=10!

Edited on August 16, 2004, 12:41 pm
 Posted by Bryan on 2004-08-16 12:04:01

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