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Could anyone tell me how to solve this math problem ?

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by Megumi » Fri Dec 21, 2007 7:19 pm
Hi, I just took the GMAT Practice test 1 and there is one question I don't know how to answer and hopefully someone can help me out.

Thank you for your time.

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 ?

the answer is greater than 40.
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by moneyman » Sat Dec 22, 2007 6:03 am
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 ?

According to the question, H(100)=2*4*6..100 or 2*(2*2)*(2*3)..(2*50) which equals to 2*50*(2*3*4*5..50).Therefore the smallest prime factor would be greater than 40.
Maxx
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by preciousrain7 » Sat Dec 22, 2007 3:03 pm
moneyman wrote: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 ?

According to the question, H(100)=2*4*6..100 or 2*(2*2)*(2*3)..(2*50) which equals to 2*50*(2*3*4*5..50).Therefore the smallest prime factor would be greater than 40.
Hi Max,

why is equal to 2*50*(2*3...*50). why isn't it just 2*(2*3...*50)?

And secondly I'm still not sure why the answe is 40. THANKS a bunch for your explanation!

Sonia
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by beingAndNothing » Sat Dec 22, 2007 10:09 pm
H(100) = (2.1) (2.2) (2.3).....(2.50) = (2^50) * 50!

H(100) + 1 = (2^50)*50! + 1.

I think there's a rule which says that smallest prime factor of n! + 1 should
be greater than 50. I guess the choices did not have an option >50. Do let me know if there was an option >50.
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