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H(100) + 1

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H(100) + 1

by mdoganay » Tue Mar 06, 2012 4:57 pm
Every positive even integer n, the function h(n) is defined to be product of all 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

Can somebody explain this question, I can not even figure out how to attack this question.
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by Anurag@Gurome » Tue Mar 06, 2012 7:32 pm
mdoganay wrote:Every positive even integer n, the function h(n) is defined to be product of all 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

Can somebody explain this question, I can not even figure out how to attack this question.

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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by GMATGuruNY » Wed Mar 07, 2012 4:15 am
I posted an explanation here:

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by krusta80 » Wed Mar 07, 2012 7:23 am
product(evens from 2 through n) = 2^(n/2) * (n/2)!

for n = 100, product = 2^50 * 50!

So, we need to find the smallest prime factor of 2^50 * 50! + 1

Now the key to this question starts to present itself: 50! is a multiple of all prime numbers less than 50, since 50! is the product of ALL integers from 1 through 50. Therefore, since we're adding 1 to a multiple of 50!, we know that none of the factors of 50! can be a factor of h(100)+1. This, in turn, means that none of the prime numbers less than 50 can be a factor of h(100)+1.

The smallest prime of h(100)+1 is certainly greater than 40.

E
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