H(100) + 1

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mdoganay: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Fri Jan 20, 2012 4:38 pm

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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Source: Beat The GMAT — Problem Solving | Tue Mar 06, 2012 4:57 pm

krusta80: Master | Next Rank: 500 Posts; Posts: 143; Joined: Mon Mar 14, 2011 3:13 amFollowed by: 5 members; Thanked: 34 times

by krusta80 » Tue Mar 06, 2012 5:41 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.

Me neither, but I never let that stop me from finding more interesting ways of organizing the givens.

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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killer1387: Legendary Member; Posts: 581; Joined: Sun Apr 03, 2011 7:53 amFollowed by: 5 members; Thanked: 52 times

by killer1387 » Tue Mar 06, 2012 7:05 pm

E +1

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