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by PAB2706 » Mon Mar 09, 2009 5:35 am
If n! = 40320 then n=?

I can factorize it to find n=8....but i want to know if there is any standard and faster method to obtain n for any value of n!.
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by DanaJ » Mon Mar 09, 2009 6:22 am
I don't think there is a faster way of doing it. You may notice that, since 40320 = 4032 * 10 = 4032 * 5 * 2, n is definately greater than 5 and maybe start from here:
1*2*3*4*5 = 120
6! = 720
7! = 5040
8! = 40320.
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by sureshbala » Mon Mar 09, 2009 11:31 pm
From the divisibility rule of 8, i.e the number formed by the last three digits is divisible by 8, we can conclude that 40320 is divisible by 8.

40320 = 8x5040 = 8 x 7 x 720 = 8 x 7 x 6! = 8!
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