wlvoh wrote:The following is a doozy of a PS problem from a practice test on the Official GMAT Prep Software from GMAC. If anyone can help explain how to solve this one, I'd greatly appreciate it.

"For every positive even integer n, the function 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:

A.) between 2 and 10

B.) between 10 and 20

C.) between 20 and 30

D.) between 30 and 40

E.) greater than 40

We are given that h(n) is defined to be the product of all the even integers from 2 to n, inclusive. For example, h(8) = 2 x 4 x 6 x 8.

We need to determine the smallest prime factor of h(100) + 1. Before determining the smallest prime factor of h(100) + 1, we must recognize that h(100) and h(100) + 1 are consecutive integers, and consecutive integers will never share the same prime factors.

Thus, h(100) and h(100) + 1 must have different prime factors. However, rather than determining all the prime factors of h(100), let's determine the largest prime factor of h(100). Since h(100) is the product of the even integers from 2 to 100 inclusive, let's find the largest even integer less than 100 that is the product of 2 and a prime number.

That even integer is 94 since since 2 x 47 = 94, which is less than 100. That means the largest prime factor of h(100) is 47. Notice that the next prime after 47 is 53, and 2 x 53 = 106, which is greater than 100.

Since 47 is the largest prime number that is a factor of h(100), all prime numbers from 2 to 47 are included in the prime factorization of h(100). Since we have mentioned that h(100) + 1 will not have any of the prime factors of h(100), all the prime factors in h(100) + 1, including the smallest one, must be greater than 47. Looking at the answer choices, only choice E can be the correct answer.

Answer: E

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, the p is

A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40

Important Concept: If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313

Now let's examine h(100)
h(100) = (2)(4)(6)(8)....(96)(98)(100)
= (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50)
Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)....(48)(49)(50)]

Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100)+1 (based on the above rule)

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Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100)+1 (based on the above rule)

So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47.

Answer = E

Cheers,
Brent