rapsadoodle_doo Posted June 20, 2006 Share Posted June 20, 2006 For every positive even integer n, the function is defined to be the product of all 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 Quote Link to comment Share on other sites More sharing options...
manish8109 Posted June 20, 2006 Share Posted June 20, 2006 Already been discussed before several times refer to http://www.www.urch.com/forums/gmat-math/50366-gmatprep-quant-questions-another-3-a.html Quote Link to comment Share on other sites More sharing options...
800Bob Posted June 20, 2006 Share Posted June 20, 2006 h(100) = (2^50)(50!) and is therefore a multiple of every integer from 1 through 50. h(100) + 1, then, will be a multiple of no integer from 2 through 50. Therefore the least prime factor of h(100) + 1 has to be greater than 50. The answer is E. Quote Link to comment Share on other sites More sharing options...
stucco005 Posted June 20, 2006 Share Posted June 20, 2006 h(100) + 1, then, will be a multiple of no integer from 2 through 50. I have seen this question before. I have always wanted clarification on exactly why adding 1 to the product of 2^50(50!) makes all possible prime factors of this sum greater than 50. Can anyone elaborate with a detailed, broken down analysis? For example, when I create the following mini-case, it seems that primes can exist within that sum: consider the product of 2, 5, 7 = 70. Adding 1 to this = 71. 71 has primes of 5 and 13. So, 5 is still in the orginal number and does not neccessarily have to be greater than 71. What am I missing? Thanks! Quote Link to comment Share on other sites More sharing options...
800Bob Posted June 20, 2006 Share Posted June 20, 2006 71 has primes of 5 and 13.What does that mean ? ? ? ? ? 71 is a prime number and has no prime factors. I don't know what it means to say that 71 has primes of 5 and 13. Consecutive integers can have no common factor greater than 1. If n is a multiple of 2, n+1 is not a multiple of 2. The next multiple of 2 is n+2. If n is a multiple of 3, n+1 is not a multiple of 3. The next multiple of 3 is n+3. If n is a multiple of 4, n+1 is not a multiple of 4. The next multiple of 4 is n+4. If n is a multiple of 5, n+1 is not a multiple of 5. The next multiple of 5 is n+5. etc. Quote Link to comment Share on other sites More sharing options...
stucco005 Posted June 20, 2006 Share Posted June 20, 2006 What does that mean ? ? ? ? ? 71 is a prime number and has no prime factors. I don't know what it means to say that 71 has primes of 5 and 13. :smack: Bob! Sorry, I was thinking of a different number and got carried away! You are totally correct, and thank you for finally clarifing this problem! Quote Link to comment Share on other sites More sharing options...
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