Smallest prime factor of product of even numbers

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gander123: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Tue Sep 25, 2012 12:47 amFollowed by: 1 member; Thanked: 3 times

by gander123 » Mon Dec 31, 2012 5:05 am

Hey guys,

Here's the problem:

"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

Correct answer: E

Believe it or not, it had to guess on this one... Would really appreciate your time-saving approaches to this problem.

Cheers,

Tobi

Source: Beat The GMAT — Problem Solving | Mon Dec 31, 2012 5:05 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Dec 31, 2012 5:53 am

gander123 wrote: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:

This question has been discussed many times in the forum.

h(100) + 1 = (2*4*6*8*10*...*100) + 1 = (2^50)*(1*2*3*4*5*...*50) + 1

Thus when [h(100) + 1] is divided by any integer (including all the primes) less than or equal to 50, it'll leave a remainder of 1. Thus p must be greater than 50 in turn p > 40.

The correct answer is E.

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gander123: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Tue Sep 25, 2012 12:47 amFollowed by: 1 member; Thanked: 3 times

by gander123 » Wed Jan 02, 2013 3:44 am

(2^50)*(1*2*3*4*5*...*50) + 1

How do you get there?! Whats the intermediate step?

But understand your reasoning ...

Thanks

Tobi

GMAT/MBA Expert

Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Wed Jan 02, 2013 5:20 am

gander123 wrote:
(2^50)*(1*2*3*4*5*...*50) + 1
How do you get there?! Whats the intermediate step?

Each term of (2*4*6*8*10*...*100) is multiple of 2. I'm just taking one 2 from each term out of the bracket. Hence, I'm taking fifty 2s out of the bracket.

For example, (2*4*6*8) can be written as [(2*1)*(2*2)*(2*3)*(2*4)] = (2*2*2*2)*(1*2*3*4) = (2^4)*(1*2*3*4)

Hope that helps.

gander123: Senior | Next Rank: 100 Posts; Posts: 50; Joined: Tue Sep 25, 2012 12:47 amFollowed by: 1 member; Thanked: 3 times

by gander123 » Wed Jan 02, 2013 5:36 am

Dear Anurag,

Thank you, great help!

cheers,

Tobi

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