Help with AP/GP Problem

[ajdude]

1. For every positive even integer n, the function f(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 f(100) + 1, then p is

a) between 2 and 10
b) between 10 and 20
c) between 30 and 30
d) between 30 and 40
e) greater than 40

[sdasar]

I think the answer is E.

f(100) + 1 = 2*4*6*8....*100 + 1 = 2(1*2*3*4*5...50) + 1


The first term is divisible by every number from 2 to 50. On the converse, by adding 1 to this term, no number from 2 to 50 can divide it, and all of them give a reminder of 1.

Therefore the smallest factor (including prime) has to be > 50.

[jmbamba]

I think the answer is E.

f(100) + 1 = 2*4*6*8....*100 + 1 = 2(1*2*3*4*5...50) + 1

A small mistake, but has no impact on the result. f(100) + 1 = 2*4*6*8....*100 + 1 = (2^50)(1*2*3*4*5...50) + 1

And as you said it, if x=kn, there's no way x+1 equals k'n. So if n is a factor of x, n can never be a factor of x+1. So none of the factors of (2^50)(1*2*3*4*5...50) can be a factor of f(100)+1. It then follows that the smallest prime factor of f(100)+1 is greater than 50. Answer E.

[sdasar]

Good catch.

[ajdude]

Thanks very much!