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 Post subject: Smallest Prime Factor
 Post Posted: Wed Dec 30, 2009 3:03 pm 
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Course Students


Posts: 4
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) btwn 2 and 10
B) btwn 10 and 20
C) btwn 20 and 30
D) btwn 30 and 40
E) greater than 40


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 Post subject: Re: Smallest Prime Factor
 Post Posted: Tue Jan 05, 2010 9:28 pm 
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Students


Posts: 14
This seems like a real tough one. Hmmm...I could try an explanation but I'm not very sure at this point.

h(100) + 1 = (2 * 4 * 6 * ... * 98 * 100) + 1

Taking 2 as a common factor:

= 2 ^ 50 * (1 * 2 * 3 * ... * 47 * 48 * 49 * 50) = 2 ^ 50 * 50!

We could observe that the above value contains all the prime factors from 2 thru 47,

Therefore,
(h (100) + 1) % 2 = 1
(h (100) + 1) % 3 = 1
(h (100) + 1) % 5 = 1
:::::::::::::::::::::::
(h (100) + 1) % 47 = 1

% refers here to a remainder...

From that we could logically deduce that if the above value h(100) + 1 has a prime factor it should be greater then 47 and that value would be the smallest prime factor (the firs t one). It is evident that all the prime factors from 2 thru 47 DO NOT divide h(100) + 1.

I think, the answer is E.


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 Post subject: Re: Smallest Prime Factor
 Post Posted: Sat Jan 09, 2010 5:59 am 
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ManhattanGMAT Staff


Posts: 7146
look here:

for-every-positive-even-integer-n-the-function-h-n-t1152.html


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