OG prep test question

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OG prep test question

by val1e » Sun Dec 12, 2010 5:18 pm
The function h(n) is defined as the product of all the even integers from 2 to n. If p is the smallest prime factor in h(100) + 1 then p is:

a. Between 2 and 20
b. Between 10 and 20
c. 20....30
d. 30.... 40
e. >40
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by GMATGuruNY » Sun Dec 12, 2010 5:48 pm
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
1) 2 and 10
2) 10 and 20
3) 20 and 30
4) 30 and 40
5) > 40
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. (For future reference, integers that share no factors other than 1 are called coprimes.)

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.
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by val1e » Mon Dec 13, 2010 2:13 pm
Thanks, this will help out with other problems as well.