|
Author |
Message |
|
TAGS:
|
|
|
Intern
Joined: 02 Jan 2009
Posts: 5
Schools: HARVARD, LBS, INSEAD, CARNEGIE
Followers: 0
Kudos [?]:
0
[0], given: 0
|
For every positive integer n, the function h(n) is defined [#permalink]
 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?
|
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2528
Followers: 41
Kudos [?]:
364
[1] , given: 19
|
Re: GMATPrep question: need solution [#permalink]
04 Jan 2009, 23:41
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 discussed recently: 7-p556064?t=74417#p556064
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Intern
Joined: 02 Jan 2009
Posts: 2
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: GMATPrep question: need solution [#permalink]
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... ___________________ Great thing to know about busby seo test.
|
|
|
|
|
|
Intern
Joined: 05 Jan 2009
Posts: 10
Followers: 0
Kudos [?]:
2
[0], given: 0
|
Re: GMATPrep question - to Gmat Tiger [#permalink]
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
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2528
Followers: 41
Kudos [?]:
364
[0], given: 19
|
Re: GMATPrep question - to Gmat Tiger [#permalink]
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. 
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Intern
Joined: 05 Jan 2009
Posts: 10
Followers: 0
Kudos [?]:
2
[2] , given: 0
|
Re: GMATPrep question: need solution [#permalink]
06 Jan 2009, 12:14
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
|
|
|
|
|
|
CEO
Joined: 29 Aug 2007
Posts: 2528
Followers: 41
Kudos [?]:
364
[0], given: 19
|
Re: GMATPrep question: need solution [#permalink]
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.
_________________
Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
|
|
|
|
|
|
Intern
Joined: 22 Nov 2006
Posts: 12
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: GMATPrep question: need solution [#permalink]
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: 19
Location: Boston/Cleveland
Followers: 0
Kudos [?]:
2
[0], given: 0
|
Re: GMATPrep question: need solution [#permalink]
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
|
|
|
|
|
|
|
Re: GMATPrep question: need solution
[#permalink]
08 Feb 2009, 16:50
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
For every positive integer n, the function h(n) is defined
|
coffeeloverfreak |
3 |
06 Sep 2005, 20:18 |
|
|
|
|
For every positive integer n, the function h(n) is defined
|
chet719 |
2 |
20 Sep 2005, 11:25 |
|
|
|
|
For every positive integer n, the function h(n) is defined
|
hossainm |
1 |
25 Jun 2006, 16:06 |
|
|
|
|
For every positive integer n, the function h(n) is defined
|
uvs_mba |
3 |
15 Sep 2006, 17:14 |
|
|
|
|
For every positive integer n, the function h(n) is defined
|
lionheart187 |
1 |
31 Aug 2008, 12:11 |
|
|
|
|
|
|
close
