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

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For every positive even integer n, the function h(n) is [#permalink] New post 17 Apr 2006, 00:48
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For every positive even 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 smallest prime factor of h(100)+1, then p is

a. between 2 and 10

b. betwen 10 and 20

c. between 20 and 30

d. between 30 and 40

e. greater than 40
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 [#permalink] New post 17 Apr 2006, 01:09
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E...


For h(100) the smallest prime factor is 53... So h(100) +1 should be more than 53.

Hence more than 40 works!
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 [#permalink] New post 17 Apr 2006, 06:44
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I was wrong :-D

Last edited by bqminh1983 on 17 Apr 2006, 13:24, edited 1 time in total.
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 [#permalink] New post 17 Apr 2006, 10:05
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bqminh1983 wrote:
h(100) = 2 + 4 + 6+ ... + 100 = 2( 1+ 2 + 3+ ... +50) = 2*50*51/2 = 2550

h(100) + 1 = 2551

2551 is not divisible by any number smaller than 40 so the answer is E


Actually, the question says product of the numbers.
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 [#permalink] New post 17 Apr 2006, 10:09
sm176811 wrote:
E...


For h(100) the smallest prime factor is 53... So h(100) +1 should be more than 53.

Hence more than 40 works!


Just wondering, the smallest prime for number 9 is 3, but for 10, 2. Don't know if that principle works.
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 [#permalink] New post 17 Apr 2006, 11:15
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:oops: I still don't understand how to solve this problem. Can someone explain it step by step? Thanks
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 [#permalink] New post 17 Apr 2006, 13:05
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Here's what I think-
h(100) = 2*4*6*8*...100
so you can factor out 2^50 and get 2^50(1*2*3*...50) which is 2^50*50!
so h(100)+1 = 2^50*50!+1, so you know that every factor of that number up to 50 will have a remainder of 1, thus, any factor will be greater than 50.
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 [#permalink] New post 17 Apr 2006, 19:57
Could you explain how you got 2^50 (50!)?
  [#permalink] 17 Apr 2006, 19:57
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