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gmat prep question - integers and functions

Expert replies
by gxj007 » Sun Aug 30, 2009 1:12 pm
Can anyone explain how to come up with the solution fo this one. thanks


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 between?

2 and 10
10 and 20
30 and 40
Greater than 40



Answer:
E - greater than 40
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Source: — Problem Solving |

by deagez » Mon Aug 31, 2009 8:30 pm
This is a high level problem that is actually quite simple if you have a good command over number properties.

The first premise is 2*4*6*....*100+1

Lets simplify this by pulling out a 2,

now we have 2(1*2*3*4*...*50)+1

So, not using the +1, the product of all those numbers is something humongous. Lets make up a number 454,413

If 454,413 is divisible by 1,2,3,4...,50 then 454,414 can not be divisible by any of them (besides 1, but it does not matter as 1 is not prime). So the smallest possible prime that it is divisible by has to be larger than 50. Hence the answer is E.

Lets look at it in a simpler way. 18 is divisible by 6,3,2,9,18. But if you add 1 making the number 19, it is not divisible by any of those. So its factors (19,1) are larger than 18 (not counting 1 for original problem because it is not a prime #)
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