GMAT Prob. w/ primes: For every positive integer n, h(n) is

A

artshep

New member
Joined
Sep 11, 2006
Messages
8
I'm having a little trouble with the following problem. I'm looking for the best method to approach it, but have been looking at it for a while and can not come up with anything. Any help would be appreciated.

Thanks,
Artis

*******************

For every positive integer n, 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:

- between 2 and 10
- between 10 and 20
- between 20 and 30
- between 30 and 40
- greater than 40
 
seems to me that ok... well anyhow, here is an idea:

Let p be a prime. then p divides h(n) iff 2*p is less than or equal to n. think about why this is true.

let P = {p : p is prime and 2*p<=n}.

By construction, we have included all primes which divide h(n).

Now, for every element p in P, we have that p divides h(n). But then certainly p does not divide h(n)+1 since h(n)=0 (mod p), therefore h(n)+1=1 (mod p).

The thing to notice now, is than, for all primes p where 2*p is less than n. p does not divide h(n)+1. Therefore, any prime which does divide h(n)+1 must be bigger than the biggest prime contained in the set P.

This idea will be useful for your problem
 
If you need additional help consider the following:

"The number of letters in this sentence has 1 useful property."
 
Your reply

Your first response was somewhat helpful...I'm still working it out.

Your second, however, was completely useless.


Artis
 
extra hint: the number of letters in that sentence is 41.


extra extra hint: 41 is prime


extra extra extra hint: 41*2 is less than 100
 
Primes

Look, I know you're trying to be smart and clever with your answers here...but it's probably a better idea just to be straightforward.

Even so, the response has absolutely no applicability in helping me solve the problem. The fact that it has 41 letters is totally irrelevant.

The best way to think about this problem is that any number plus 1 does not share factors (other than 1) with that original number. It also cannot share factors of those factors.

Since h(100) has all even number less than 100 as factors, then h(100) + 1 cannot have any even numbers less than 100 as factors. And since any prime less than 40 multiplied by 2 gives you an even number less than 100, the answer must be E.
 
You must log in or register to reply here.
Top