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# For every positive integer n, the function h(n) is defined

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Intern
Joined: 02 Jan 2009
Posts: 5
For every positive integer n, the function h(n) is defined [#permalink]

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04 Jan 2009, 21:46
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

Ques: Can someone help resolve this?

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Joined: 29 Aug 2007
Posts: 2452
Re: GMATPrep question: need solution [#permalink]

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04 Jan 2009, 23:41
1
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

discussed recently: 7-p556064?t=74417#p556064
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Intern
Joined: 02 Jan 2009
Posts: 2
Re: GMATPrep question: need solution [#permalink]

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04 Jan 2009, 23:56
what is this kind you tell what is this GMATPrep so i can tell my opinion.. thank you, by the way what kind of solution do you want to ask...

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Intern
Joined: 05 Jan 2009
Posts: 9
Re: GMATPrep question - to Gmat Tiger [#permalink]

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05 Jan 2009, 10:34
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.

thx
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Joined: 29 Aug 2007
Posts: 2452
Re: GMATPrep question - to Gmat Tiger [#permalink]

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05 Jan 2009, 23:55
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.
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Intern
Joined: 05 Jan 2009
Posts: 9
Re: GMATPrep question: need solution [#permalink]

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06 Jan 2009, 12:14
2
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
SVP
Joined: 29 Aug 2007
Posts: 2452
Re: GMATPrep question: need solution [#permalink]

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07 Jan 2009, 09:30
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.
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Intern
Joined: 22 Nov 2006
Posts: 12
Re: GMATPrep question: need solution [#permalink]

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08 Jan 2009, 22:47
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
Intern
Joined: 07 Jan 2009
Posts: 18
Location: Boston/Cleveland
Re: GMATPrep question: need solution [#permalink]

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08 Feb 2009, 16:50
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

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: GMATPrep question: need solution   [#permalink] 08 Feb 2009, 16:50
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# For every positive integer n, the function h(n) is defined

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