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Senior Manager
Joined: 13 Mar 2007
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Location: Russia, Moscow

Thanks !!!



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GMATPrep, PS  prime number [#permalink]
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10 Sep 2007, 05:38
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



Manager
Joined: 29 Aug 2007
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I think the OA should be Option 1
Between 2 and 10
for H(100) the smalles prime divisor is 2
of H (100) + 1 , the Smallest prime Divisor could be 3 or 5



Manager
Joined: 02 Jul 2007
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I thought the same thing, but the OA is D!!
can't figure out how though.



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Joined: 16 Feb 2007
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gnr646 wrote: I thought the same thing, but the OA is D!!
can't figure out how though.
well that is a really tricky question...
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?



Senior Manager
Joined: 27 Aug 2007
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Can you tell us, if it is not a secret, what is the source of question?
Thnx in advance
Oh, I see GmatPrep.



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Joined: 10 Jun 2007
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I am quite sure that the OA for this is E.
This question has been discussed many times.



Senior Manager
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bkk145 wrote: I am quite sure that the OA for this is E. This question has been discussed many times.
Cann't you provide us with the link to one of those discussions??



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Ferihere wrote: bkk145 wrote: I am quite sure that the OA for this is E. This question has been discussed many times. Cann't you provide us with the link to one of those discussions??
http://gmatclub.com/forum/t46645



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my appologies, OA is E.



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FUNCTIONS [#permalink]
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13 Jan 2008, 07:59
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



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Re: FUNCTIONS [#permalink]
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13 Jan 2008, 08:48
marcodonzelli 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
A. between 2 and 10 B. between 10 and 20 C. between 20 and 30 D. between 30 and 40 E. greater than 40 http://www.gmatclub.com/forum/t46645



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GMATprep: h(n)+1 [#permalink]
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21 Apr 2008, 09:05
n is a integer, n>0 and h(n)= the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(n)+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
OA to follow pls help!



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Re: GMATprep: h(n)+1 [#permalink]
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21 Apr 2008, 11:04
Well, I can definitely get get a couple answers that coincide with choice A.
If n = 4, then h(n) = 2x4 = 8
8 + 1 = 9
Smallest (and only) prime factor = 3
If n = 6, then h(n) = 2x4x6 = 48
48 + 1 = 49
Smallest (and only) prime factor = 7
So, I'd go with A
However, the question looks very familiar, are you sure you wrote it correctly?



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Re: GMATprep: h(n)+1 [#permalink]
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21 Apr 2008, 11:38
Brindusa wrote: n is a integer, n>0 and h(n)= the product of all the even integers from 2 to n inclusive. If p is the smallest prime factor of h(n)+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
OA to follow pls help! Must be 100, not n, I guess. 7p446677



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Re: GMATprep: h(n)+1 [#permalink]
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21 Apr 2008, 14:14
A if we consider n =1,4 and 6
the close factors for the function I could get are1, 3 and 7



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GMAT Prep Question on Integers [#permalink]
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25 May 2008, 08:39
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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, 22:04, edited 1 time in total.



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Re: GMAT Prep Question on Integers [#permalink]
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25 May 2008, 17:24
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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!
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Last edited by TheGMATDoctor on 13 Jul 2010, 13:01, edited 4 times in total.



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Re: GMAT Prep Question on Integers [#permalink]
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25 May 2008, 18:38



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Re: GMAT Prep Question on Integers [#permalink]
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28 May 2008, 14:05
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.
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Re: GMAT Prep Question on Integers
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