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

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For every positive integer n, the function h(n) is defined [#permalink] New post 06 Sep 2005, 19:18
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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 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

Kindly explain answers, thanks!
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 [#permalink] New post 07 Sep 2005, 01:44
alrady posted 1-2 days ag..belive in advanced math..check it out

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 [#permalink] New post 07 Sep 2005, 06:45
Sorry about that. I hadn't found it on an initial search, but I see now I must have misspelled a search term. My apologies for the repeat posting.

I read the explanations in the previous threads and I'm not sure I follow them, though. I wanted to reply in the original thread but it's locked:
http://www.gmatclub.com/phpbb/viewtopic.php?t=18802&highlight=h+100.

I'm not sure I understand the explanation of how if n=2m, then h(n)=2^m*m! Can anyone else shed some light?
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 [#permalink] New post 07 Sep 2005, 22:50
coffeeloverfreak wrote:
Sorry about that. I hadn't found it on an initial search, but I see now I must have misspelled a search term. My apologies for the repeat posting.

I read the explanations in the previous threads and I'm not sure I follow them, though. I wanted to reply in the original thread but it's locked:
http://www.gmatclub.com/phpbb/viewtopic.php?t=18802&highlight=h+100.

I'm not sure I understand the explanation of how if n=2m, then h(n)=2^m*m! Can anyone else shed some light?


It is written clearly that

h(2)=2=2*1
h(4)=2*4=(2*1)*(2*2)=2^2*(1*2)
h(6)=2*4*6=(2*1)*(2*2)*(2*3)=2^3*(1*2*3)
h(8)=2*4*6*8=(2*1)*(2*2)*(2*3)*(2*4)=2^4*(1*2*3*4)

so by the same token

h(2m)= 2^m * m!

For the answer AJB expalantion is quite clear
  [#permalink] 07 Sep 2005, 22:50
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