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Factor problem

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Factor problem

by Anindya Madhudor » Mon Nov 12, 2012 11:57 am
Can anyone please help me show how to solve the following?

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) 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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Source: — Data Sufficiency |
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by eaakbari » Mon Nov 12, 2012 12:04 pm
*Please post in the right designated forum*


IMO A

Use strategy of picking numbers,

If you pick 4 as n,

then h(4) + 1 = 9 which makes 3 the only prime factor

Since the function is true for all n, the answer must be A, as its the only answer choice whose range includes 3.

Hence A
Last edited by eaakbari on Tue Nov 13, 2012 7:07 am, edited 1 time in total.
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by Anindya Madhudor » Mon Nov 12, 2012 1:44 pm
The question is specifically asking for factor of h(100)+1. It is not asking for factor of all h(n)+1.
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by Anindya Madhudor » Mon Nov 12, 2012 3:37 pm
Sorry about posting in the wrong area. I am new to this.
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by eaakbari » Tue Nov 13, 2012 8:57 am
Anindya Madhudor wrote:The question is specifically asking for factor of h(100)+1. It is not asking for factor of all h(n)+1.
Yes, I misread the question and stand corrected.

By taking 2 as common, we can rewrite h(100) as

2^50x(1x2x3x...x50).

Hence every number from 1 to 50 is a factor of 100

So every h(100) + 1 will not be a factor and will leave remainder as one (including primes).

Hence the smallest prime will be above 50.

Answer E seems to fit the description.

Hence E

P.S. Whats the source of the question.
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by Anindya Madhudor » Tue Nov 13, 2012 9:49 am
It came up in one of the two sample tests provided by official GMAC.
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