forgmat Posted November 21, 2004 Share Posted November 21, 2004 Anyone knows more about it. Please explain. :) Quote Link to comment Share on other sites More sharing options...
reactor Posted February 20, 2005 Share Posted February 20, 2005 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. :) Quote Link to comment Share on other sites More sharing options...
charanya Posted February 20, 2005 Share Posted February 20, 2005 252^5 divisible by 6^n .what is n? Need stepwise explanation please. Quote Link to comment Share on other sites More sharing options...
kathy Posted February 22, 2005 Share Posted February 22, 2005 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 Quote Link to comment Share on other sites More sharing options...
fmku Posted February 22, 2005 Share Posted February 22, 2005 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 Quote Link to comment Share on other sites More sharing options...
charanya Posted February 22, 2005 Share Posted February 22, 2005 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. Quote Link to comment Share on other sites More sharing options...
kathy Posted February 22, 2005 Share Posted February 22, 2005 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 Quote Link to comment Share on other sites More sharing options...
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