Problem Solving

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

Brent@GMATPrepNow wrote:

rakeshd347 wrote: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 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40

OA is E

Important Concept: If k is a positive integer that's greater than 1, and if 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

Turksonaaron wrote:"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"

However, you initially stated that no factor is common to both h(100) and h(100)+1 except 1, which is not prime. Why then do you say again that the smallest prime factor COULD BE greater than 47 when we know that no such factor ever exists. The answer should be none but why did GMAC not put that option there.

What you stated logically means that such prime number could be there.

Please correct me if I'm wrong.

GMATGuruNY wrote:

Turksonaaron wrote:"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"

However, you initially stated that no factor is common to both h(100) and h(100)+1 except 1, which is not prime. Why then do you say again that the smallest prime factor COULD BE greater than 47 when we know that no such factor ever exists. The answer should be none but why did GMAC not put that option there.

What you stated logically means that such prime number could be there.

Please correct me if I'm wrong.

As you have noted, there is no prime factor common to h(100) and h(100)+1, since h(100) and h(100)+1 are COPRIMES and thus share no factors other than 1.
But the question stem does not ask for the smallest prime factor common to h(100) and h(100)+1.
It asks merely for the smallest prime factor of h(100)+1.
Since all of the prime numbers up to 47 are factors of h(100) -- and h(100) and h(100)+1 share no factors other than 1 -- NONE of the prime numbers up to 47 is a factor of h(100)+1.
Thus, the smallest prime factor of h(100)+1 must be GREATER THAN 47, with the result that the correct answer is E.