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A tough one - h(100) + 1

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by papumba2011 » Mon Apr 12, 2010 12:30 pm
For every postive 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
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Source: — Problem Solving |

by indiantiger » Mon Apr 12, 2010 12:49 pm
I think this question has been answered on the forum, just search for h(n) and you will get multiple hits.
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by Stuart@KaplanGMAT » Mon Apr 12, 2010 1:00 pm
indiantiger wrote:I think this question has been answered on the forum, just search for h(n) and you will get multiple hits.
It's been answered h(n) + 1 times at least!
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by dxgamez » Mon Apr 12, 2010 10:57 pm
lol that's a good one, Stuart...

@papumba

the number properties which is important to note in this qn is that consecutive integers do not share any primes. that would mean that h(100) and h(100) + 1 do not share any primes. you can get the solutions from the forum.

thought i should write it down, to remind myself too :)
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by akhpad » Tue Apr 13, 2010 4:03 am
This problem has been answered many times.

Please type first few words in google. You will get it at several places.
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