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GMATPrep, PS - prime number [#permalink]
10 Sep 2007, 04:38

For every positive 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 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 would have voted for E since in the product of the even integers, every prime til 47 (2*47=96) is included and thus the smalles prime factor of h(100)+1 should be greater than 47... ?? and thus E?? are you sure with OA D?

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

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

GMAT Prep Question on Integers [#permalink]
25 May 2008, 07:39

1

This post received KUDOS

For every positive even integern, 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 1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40

Last edited by ritula on 25 May 2008, 21:04, edited 1 time in total.

Re: GMAT Prep Question on Integers [#permalink]
25 May 2008, 16:24

12

This post received KUDOS

This question is testing the concept of coprimes. 2 positive integers are coprime when their greatest common factor (their only common factor) is 1. Now note that two different prime numbers are always coprime. For example, 3 and 7 are coprime. So are 13 and 19. But the two integers need not be prime numbers in order to be coprime. For example, 4 and 9 are coprime (1 is their only common factor). Also, Important! Two consecutive integers are always coprime. The question is testing you on this concept. Let's solve it now: h(n) = 2*4*6*....................*n. ----n is even. h(100) = 2*4*6*........ 94*96*98*100. h(100) = (2^50)*(1*2*3*.......47*48*49*50). Note:I have pooled together all the 2s from all the even integers from 2 to 100; that's how I got 2^50. Now, the largest prime number involved in the above factorization is 47. All the prime from 2 to 47 are also involved in the above factorization. Actually, 47 is the greatest prime factor of h(100). Since h(100) and h(100) + 1 are consecutive integers, they are necessarily coprime (see above). h(100) and h(100) + 1 have no common factor except 1, so they have no common prime factor either. The smallest prime factor of h(100) +1 must then be greater than 47. Clearly, this prime factor is greater than 40.

That's all folks! _________________

Dakar Azu is The GMAT Doctor. Dakar is an experienced GMAT teacher who can be reached at http://700gmatclub.com. He prepares aspiring business students thoroughly to get them well over the GMAT 700-score hurdle through his online GMAT courses.

Last edited by TheGMATDoctor on 13 Jul 2010, 12:01, edited 4 times in total.

Re: GMAT Prep Question on Integers [#permalink]
28 May 2008, 13:05

TheGMATDoctor wrote:

This question is testing the concept of coprimes. 2 positive integers are coprime when their greatest common factor (their only common factor) is 1. Now note that two different prime numbers are always coprime. For example, 3 and 7 are coprime. So are 13 and 19. But the two integers need not be prime numbers in order to be coprime. For example, 4 and 9 are coprime (1 is their only common factor). Also, Important! Two consecutive integers are always coprime. The question is testing you on this concept. Let's solve it now: h(n) = 2*4*6*....................*n. ----n is even. h(100) = 2*4*6*........ 94*96*98*100. h(100) = (2^50)*(1*2*3*.......47*48*49*50). Note:I have pooled together all the 2s from all the even integers from 2 to 100; that's how I got 2^50. Now, the largest prime number involved in the above factorization is 47. All the prime from 2 to 47 are also involved in the above factorization. Actually, 47 is the greatest prime factor of h(100). Since h(100) and h(100) + 1 are consecutive integers, they are necessarily coprime (see above). h(100) and h(100) + 1 have no common factor except 1, so they have no common prime factor either. The smallest prime factor of h(100) +1 must then be greater than 47. Clearly, this prime factor is greater than 40.

That's all folks! Asan Azu, The GMAT Doctor.

Wow, lot of info. I understood everything except for this part: h(100) = (2^50)*(1*2*3*.......47*48*49*50) Could you explain this part a little more clearly? Thanks. _________________

Factorials were someone's attempt to make math look exciting!!!

gmatclubot

Re: GMAT Prep Question on Integers
[#permalink]
28 May 2008, 13:05

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...