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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]

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21 Mar 2008, 03:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

For every positive even integer n, the function h(n) is defined to be product of all 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 then 40
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21 Mar 2008, 16:28
h(100) + 1 can be reduced to $$2^{50}*50! + 1$$. Any prime factor has to be greater than 50. Hence E.
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21 Mar 2008, 16:37
neelesh wrote:
h(100) + 1 can be reduced to $$2^{50}*50! + 1$$. Any prime factor has to be greater than 50. Hence E.

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21 Mar 2008, 16:49
For any n! + 1 the smallest factor has to be greater than any of the numbers within the the factorial

2! + 1 = 3: smallest factor = 3
3! + 1 = 7: smallest factor = 7
4! + 1 = 25: smallest factor = 5

50! + 1 some number whose smallest factor has to be greater thnn 50
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21 Mar 2008, 19:39
neelesh wrote:
For any n! + 1 the smallest factor has to be greater than any of the numbers within the the factorial

2! + 1 = 3: smallest factor = 3
3! + 1 = 7: smallest factor = 7
4! + 1 = 25: smallest factor = 5

50! + 1 some number whose smallest factor has to be greater thnn 50

Thanks much! Out of curiosity - is this mentioned as a trick somewhere or did you try with different numbers to solve this problem? I just want to make sure I cover any reading material if I need to.
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21 Mar 2008, 21:12
neelesh wrote:
For any n! + 1 the smallest factor has to be greater than any of the numbers within the the factorial

2! + 1 = 3: smallest factor = 3
3! + 1 = 7: smallest factor = 7
4! + 1 = 25: smallest factor = 5

50! + 1 some number whose smallest factor has to be greater thnn 50

still unclear for me.

but there is 2^50 in 50!+1. how do you include that and find the smallest prime factor?
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21 Mar 2008, 21:26
Hi:

Let's see:

We know that: If n is diviable by K, then (n+1) can NOT be diviable by K

h(n)=2*4*..*100= 2^50 * (1*2*3*..*50)

--> h(n) is diviable by 2, 3, 4,...,50 --> h(n)+1 can not be diviable by any of intergers among {2,3,..,50}.

Therefore, if x is a factor of h(n)+1, x must NOT be {1,2,..,50}--> x> 50--> E

(for any x, not neccessary a prime factor)

Hope this helps.

--NotaHug
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24 Mar 2008, 07:38
notahug, what about 5*7 +1 = 35 + 1 = 36, 36 = 9*4 = (2^2)*(3^2), 2<5<7
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24 Mar 2008, 07:50
az780 wrote:
notahug, what about 5*7 +1 = 35 + 1 = 36, 36 = 9*4 = (2^2)*(3^2), 2<5<7

I got what Notahug means
for 5*7 +1=35+1=36
I think he did not mean that for 36, factors must be more than 5 or 7. Rather, it must not be equal to 5 or 7.

back to the question, since factors of h(n) are all ranging from 2 to 50, then it means that factors of h(n)+1 cannot be 2,3,4,5,....50. That's why he answered E

Re: Function, evens   [#permalink] 24 Mar 2008, 07:50
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