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by Khalid.B » Tue Apr 17, 2012 3:49 pm
For every 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

Answer A
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by Anurag@Gurome » Tue Apr 17, 2012 7:56 pm
Khalid.B wrote:For every 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

Answer A
h(100) = 2 * 4 * 6 * ... * 100
= (2 * 1) * (2 * 2) * (2 * 3) * ... * (2 * 50)
= 2^(50) * (1 * 2 * 3 ... * 50)
Then h(100) + 1 = 2^(50) * (1 * 2 * 3 ... * 50) + 1
Now, h(100) + 1 cannot have any prime factors 50 or below, because dividing this value by any of these prime numbers will give a remainder of 1.

The correct answer is E. Please check the OA.
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by aneesh.kg » Tue Apr 17, 2012 9:27 pm
h(100) + 1 = 2*4*6*...100 + 1
let one of the even numbers multiplied in '2*4*6*...100' be 2x where, 1 =< x =< 50

2*4*6*...100 is a multiple of 2x, so it will also be a multiple of x.
If 2*4*6*...100 is a multiple of x, '2*4*6*...100 + 1' cannot be a multiple of x. Infact, h(100) will leave a remainder of 1 when divided by any number from 2 to 50.

Therefore, the Prime factor p > 50.

Option (E) is correct
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by Khalid.B » Wed Apr 18, 2012 9:48 am
You are both correct, the answer is (E), Thanks.
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