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forgmat
11-21-2004, 06:48 AM
Anyone knows more about it. Please explain. :)

reactor
02-20-2005, 03:07 AM
This is factorial and means (n-1)(n-2)(n-3).........[n-(n-2)][n-(n-1)] = (n-1)!
i.e 5! = 1 x 2 x 3 x 4 x 5 = 120 6!/5! = 6 in the same way (n-1)! can be expressed as n!/n
Factorials are widely used in Combinatorics to calculate probabilities. :)

charanya
02-20-2005, 04:43 AM
252^5 divisible by 6^n .what is n?
Need stepwise explanation please.

kathy
02-22-2005, 12:02 AM
252^5 divisible by 6^n .what is n?
Need stepwise explanation please.

Hi Charanya,

Ur question is not clear to me . I guess u r missing some part of question.

If I solve above problem I get various answers....0,1,2,3,4,5,6 etc....there is no specific answer so this can't be a GRE question

fmku
02-22-2005, 07:35 AM
252^5 = (2^2 * 3^2 * 7)^5 = 6^10 * 7^5
6^10 * 7^5 % 6^n = 0 (divisible) when n = [0,10]
Is this correct

charanya
02-22-2005, 02:58 PM
which is the greatest value of n that makes 252^5 divisable by 6^n??
I'm i clear now?

Ya, ans is 10 .but iam not able to understand u.

kathy
02-22-2005, 08:34 PM
252^5=(7x6^2)^5=(7^5)x(6^2)^5
=(7^5)x(6^10)
252^5 is divisible by 6^10 hence n=10