Sadowski wrote:I've seen a post about this question, but no one ever actually answered the question.

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

1) Between 2 and 10

2) Between 10 and 20

3) Between 20 and 30

4) Between 30 and 40

5) >40

3rd question on my 1st GMAT Prep

I would know how to do this if it were the summation of all even integers from 2 to n, but I'm not sure how to deal with the product of all even integers. Anyone?

hmmm .. are u sure bcoz i remeber to have answered a similar question on the forum ...

nywayz this is one of those GMAT questions where math alone wont help u solve the problem ..

first thing first .. i dont think there would be any GMAt question that actually needs you to multiply 50 numbers together to find the answer ..

.. now to the question .. the question says that h(n) = 2*4*6*8* ... *n ...

.. so for example .. h(4) = 2*4 = 2^2(1*2)

h(12) = 2*4*6*8*10*12 = 2^6( 1*2*3*4*5*6)

So h(100)= 2*4*6*8*....*100 = 2^50(1*2*3*4*5*6* ... *50) ...

Now, what are the prime factors for h(100) .. that is what are the prime numbers that divide the expression h(100)=2^50(1*2*3*4*....*50) .. the prime factors would be 2,3,5,7 and so on .. that is all the prime numbers between 1 and 50 would be a factor of the expression h(100) = 2^50(1*2*3*4 ....*50) ..

.. so now consider the expression h(100)+1 = 2^50(1*2*3*4*....*50) + 1 .. since 1 is not exactly divisible by any of the prime numbers, none of the prime factors between 1 and 50 will be able to exactly divide h(100) + 1 .. so the answer is > 40 .. that is E ..

... i hope the explanation made some sense ..