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by chufus » Fri Nov 25, 2011 12:43 am
So here is the question:

For every positive 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
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by chufus » Fri Nov 25, 2011 1:10 am
See the solution below. I know its not perfect but still serves as a reference point for something you can remember.

https://www.manhattangmat.com/forums/post5477.html

Answer is E.. Greater than 40
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by neelgandham » Fri Nov 25, 2011 2:24 am
chufus wrote:So here is the question:

For every positive 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
Chufus - Thanks for pasting the content !

Solution:
h(100)= (2*4*6*.....*100)
h(100)= 2^50(1*2*3*.....*50)
So, we know that h(100) is divisible by any prime number between 2 and 50. If h(100) is divisible by any prime number between 2 and 50, then h(100)+1 should leave a remainder 1 when divided by any prime number between 2 and 50. For p to be a factor of h(100)+1, it should be > 50.

Option E

p.s: I am not sure if you can post content from a different forum in here !
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by GMATGuruNY » Fri Nov 25, 2011 6:13 pm
I posted an explanation here:

https://www.beatthegmat.com/tough-prime- ... 25475.html
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by Anurag@Gurome » Fri Nov 25, 2011 6:14 pm
girish3535 wrote:.....
This question has been discussed many times on this forum.


h(100) = 2 * 4 * 6 * ... * 100
= (2 * 1) * (2 * 2) * (2 * 3) * ... * (2 * 50)
= 2^(50) * (1 * 2 * 3 ... * 50)
Then h(100) + 1 = 2^(50) * (1 * 2 * 3 ... * 50) + 1
Now, h(100) + 1 cannot have any prime factors 50 or below, because dividing this value by any of these prime numbers will give a remainder of 1.

The correct answer is E.
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