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For every positive integer n, the function h(n) is defined [#permalink]
04 Jan 2009, 20:46

For every positive integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. if p is the smalled prime factor of h(100)+1 then p is ?

a) between 2 and 10 b) between 20 and 30 c) between 30 and 40 d) >40

Re: GMATPrep question: need solution [#permalink]
06 Jan 2009, 11:14

2

This post received KUDOS

I also answered option E, I also found the same pattern as you but I don't understand your approach very well so I will explain mine:

h (n) = (n/2)! x 2^(n/2) in case of h(100) = 50! x 2^50

h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50. Let's take a couple of factors as example:

factor 2: since h(100) is a multiple of 2, the next multiple will be h(100)+2 factor 3: since h(100) is a multiple of 3, the next one will be h(100)+3 factor 13: since h(100) is a multiple of 13, the next one will be h(100)+13 ...

In summary the only possible factor of h(100)+1 from 1 to 50 is 1 which is not a prime. So the answer is E

Re: GMATPrep question: need solution [#permalink]
04 Jan 2009, 22:41

1

This post received KUDOS

gmattarget700 wrote:

Ques: Can someone help resolve this?

For every positive integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. if p is the smalled prime factor of h(100)+1 then p is ?

a) between 2 and 10 b) between 20 and 30 c) between 30 and 40 d) >40

Re: GMATPrep question - to Gmat Tiger [#permalink]
05 Jan 2009, 09:34

Hi Gmat Tiger, I am not saying your explanation in the link is wrong but there is something I don't understand. You state that the number can not be divisible by any factor because according to the formula (2^50 x 50!)/ factor + 1/factor will always result in a non integer number?

Imagine that instead of h(50) we calculate h(8): (2^4 x 4!)/factor + 1/factor. According to your theory whatever is the factor the result will always be not integer. But if you try the factor 5 the result is an integer (385/5 = 77).

I am not sure that I understood your approach and so far I have not solved the problem in any other way, but I would like to know if there is something that I am missing.

Re: GMATPrep question - to Gmat Tiger [#permalink]
05 Jan 2009, 22:55

gmark wrote:

Hi Gmat Tiger, I am not saying your explanation in the link is wrong but there is something I don't understand. You state that the number can not be divisible by any factor because according to the formula (2^50 x 50!)/ factor + 1/factor will always result in a non integer number?

Imagine that instead of h(50) we calculate h(8): (2^4 x 4!)/factor + 1/factor. According to your theory whatever is the factor the result will always be not integer. But if you try the factor 5 the result is an integer (385/5 = 77).

I am not sure that I understood your approach and so far I have not solved the problem in any other way, but I would like to know if there is something that I am missing.

thx

I did not quite understand the red part. but for me you are talking beyond the scope of the question.

[h(8)+1] and [h(100)+1] look similar but they are different issues and the rule may not be applied to h(8)+1. and I even did not say that the rule applied to any function. stick to [h(100)+1].

Take a time and go agin and understand the concept. you will get it as i did so.

In fact, I always tried to skip this question as it was always difficult for me. this time I tried so many minuets (almost 30) to solve it, understanding the concept. _________________

Re: GMATPrep question: need solution [#permalink]
07 Jan 2009, 08:30

gmark wrote:

I also answered option E, I also found the same pattern as you but I don't understand your approach very well so I will explain mine:

h (n) = (n/2)! x 2^(n/2) in case of h(100) = 50! x 2^50

h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50.Let's take a couple of factors as example:

factor 2: since h(100) is a multiple of 2, the next multiple will be h(100)+2 factor 3: since h(100) is a multiple of 3, the next one will be h(100)+3 factor 13: since h(100) is a multiple of 13, the next one will be h(100)+13 ...

In summary the only possible factor of h(100)+1 from 1 to 50 is 1 which is not a prime. So the answer is E

There you go: h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50. _________________

Re: GMATPrep question: need solution [#permalink]
08 Jan 2009, 21:47

I don't know if I am doing this right. Please correct if I am wrong.

product of all even integers ... 2*4*6*... factor 2 out and you have 2^50*(1*2*3...) = (2^50)*50!. so the question is what is the smallest prime factor for (2^50)*50! + 1. Any number less than 50 is always a factor of (2^50)*50! leaving reminder 1. so it has to be greater than 50

Re: GMATPrep question: need solution [#permalink]
08 Feb 2009, 15:50

gmark wrote:

I also answered option E, I also found the same pattern as you but I don't understand your approach very well so I will explain mine:

h (n) = (n/2)! x 2^(n/2) in case of h(100) = 50! x 2^50

h(100) has as factors, among others, from 1 to 50 (because of 50!). But if we add 1 will certainly not have any factor from 2 to 50. Let's take a couple of factors as example:

factor 2: since h(100) is a multiple of 2, the next multiple will be h(100)+2 factor 3: since h(100) is a multiple of 3, the next one will be h(100)+3 factor 13: since h(100) is a multiple of 13, the next one will be h(100)+13 ...

In summary the only possible factor of h(100)+1 from 1 to 50 is 1 which is not a prime. So the answer is E

nice explanation. I think I'd poop myself if I saw a question this hard on the exam. Definitely the hardest one I've encountered

gmatclubot

Re: GMATPrep question: need solution
[#permalink]
08 Feb 2009, 15:50

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