amer siddiqui wrote:For every positive integer n, the function h(n) is defined as 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...

GMAT say it is "greater than 40". How was that answer arrived at? anyone?

My thoughts:

The product of the first 100 even integers from 2 to 100 will be:

2x4x6x8x10........x100

which can be written as [2x1]x[2x2]x[2x3]x[2x4]......x[2x50]

this can be further simplified to 2 to the power 50 multiplied by 50! (fifty factorial).

We then need to add one to that.
I can't go beyond that...please can others attempt?

thanks

Hi!

This is THE most commonly asked question on the forums; do a search of h(100) and you'll find many many threads and solutions.

As you rightly pointed out:
h(100) = 2^50 * 50!

This indicates that all numbers from 1 to 50 (including 50) are factors of h(100), including the prime numbers from 2 to 47. If this is true then h(100) + 1 will not have any prime factors below 50. Because when h(100)+1 is divided by any of these prime factors the remainder will always be 1, since h(100) is a multiple of each of these numbers.

To illustrate this consider the number 5040. It can be written as 5040 = 1*2*3*4*5*6*7.
If you consider 5040 + 1, using the above logic none of the numbers from 2 - 7 can be a factor of it. Because they are factors of 5040, dividing 5041 by them will give a reminder of 1. The next prime number that is a factor of 5041 has to be greater than 7, in fact the next factor is 71.

Hence the smallest prime number that is a factor of h(100) +1 has to be greater than 50. I am not sure why the answer says 'more than 40'. It could be so because the other choices are all below 40.

Hope this helps.
Arun Anand

This is a killer of a question, but as was pointed out, there are lots of discussions about it. You can see a detailed explanation and step-by-step video to solve here.

Hope that helps,
-Patrick