So H(n) = n!, then H(100) = 100!, then the smallest prime factor of H(100) or 100! must be 2. IMO: A
How 'bout OA?

It says "EVEN NUMBERS" so 100! wont be correct.

OA is E

This question was discussed many times in this forum..Please search this question.you ll get tons of explanations.

pushkin1982 wrote:Q)For Every Positive Integer n, The Function H(n) is defined to be the product of all the even integers from 2 to n (both inclusive). If p is the smallest prime factor of H(100), Then p is between

a) 2 & 10
b) 10 & 20
c) 20 & 30
d) 30 & 40
e) Greater than 40

Please explain

Here is the rule that is being tested with this problem:

If x is a positive integer, the only factor common both to x and to x+1 is 1. They share no other factors.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1. So 1 is the only factor common both to x and to x+1. They share no other factors.

Thus, in the problem above, we know that 1 is the only factor common both to h(100) and to h(100) + 1. They share no other factors.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100

Factoring out 2, we get:

h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100). This means that NONE of the prime numbers between 1 and 50 is a factor of h(100) + 1, because h(100) and h(100) + 1 share no factors other than 1.

So the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.

Thanks a Ton!

tons ton

Here is how I did the problem:

I thought for a quick second to myself that this is similar recurring question that you may get on the GMAT or practice materials. I felt as if this will be fairly easy to solve if I can get a prime number that is above 40 to work for this instance. I am asked: What is the greatest prime factor of 2x4x6...x100. I think to myself for a minute.. well ... 11x 2 =22 .. 13 x 2= 26.. I should be able to go over 40.. Quick and easy... I pick 47 x 2 =94. This is sufficient to conclude that the answer is E; since 94 is a factor of this equation, it can be split into 2x47 and thus 47 is prime and larger than 40.

The answer is E

Arcane,

You have an interesting approach, but unfortunately it doesn't work for two reasons. The easiest reason is that this problem is searching for the SMALLEST prime factor, whereas you found the GREATEST prime factor, 47.

The more complicated reason gets to the heart of the problem. You are finding factors for the value of h(100), whereas the question wants to know about the factors for h(100) + 1. Basically, all of the prime factors of h(100) are offset by 1 because we have added 1 to the value.

Think of it this way. You could find all the prime factors for 33, but if I add 1 to 33, none of those factors are the same. Therefore, finding the factors of 33 is useless for finding the factors of 33 + 1.

As a side note, I love this problem. It reminds me of my favorite question from OG 10 that did not make it to OG 12. It is a data sufficiency question about a very similar subject.

you're right about that I don't know what I was thinking..