Problem solving integer question

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Problem solving integer question

by thedude » Sun Jan 20, 2013 10:44 am
I'm struggling to understand what the "trick" is here with this question. any help would be greatly appreciated.

Question:

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 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

I was struggling to understand how to disprove h(100) + 1 might not have 3, e.g. as its smallest prime.

thanks in advance[/i]
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by GMATGuruNY » Sun Jan 20, 2013 11:40 am
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by Anurag@Gurome » Mon Jan 21, 2013 3:58 am
thedude wrote:I'm struggling to understand what the "trick" is here with this question. any help would be greatly appreciated.

Question:

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 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

I was struggling to understand how to disprove h(100) + 1 might not have 3, e.g. as its smallest prime.

thanks in advance[/i]
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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