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#1 (permalink) |
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I JUST got here.
Join Date: Jul 2006
Posts: 18
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How to solve this question(short cut)..I have no clue.
Problem solving For every positive even integer n, the function H(n) is defined to be the product of all the even integers from 2 to n, inclusive. If P is the smallest prime factor of H(100)+1 then P is ? a) between 2 and 10 b) between 10 and 20 c) between 20 and 30 d) between 30 and 40 e) greater than 40 answer (highlight): E |
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#2 (permalink) |
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TestMagic Guru
![]() ![]() ![]() ![]() Join Date: Oct 2004
Posts: 1,149
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h(n)=2*4*6*8*----*n
h(100)=2*4*6*8*----*100 =2^50 (1*2*3*------*50) so, h(100) will be divisible by all the prime numbers b/w 2 & 50, h(100)+1 will not be divisible by any of the prime numbers b/w 2 & 50 since it will leave remainder 1 after dividing h(100)+1 by any of the prime number b/w 2&50. so, the next prime number by which it could be divisible should be greater than 50, the smallest prime factor of h(100)+1 should be greater than 40 i.e. answer (E) Or por favor, El Seņor refer to http://www.urch.com/forums/gmat-prob...ction-h-n.html (Function h(n)) Last edited by manish8109 : 2006 August 5th at 09:58 AM. Reason: Automerged post |
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