by RonPurewal Sat Sep 29, 2007 5:21 am
Well, g(80) is divisible by all of 2, 4, 6, 8, ..., 80, which means it's divisible by all of 1, 2, 3, 4, 5, ..., 40 (and, specifically, by all of the prime numbers in that list).
Which means that g(80) + 1 is divisible by NONE of the prime numbers in that list.
So it's E.
This problem reminds me of OG #153 (the one in which one of the statement reads 13! <= k <= 13! + 13), a problem that we occasionally use in lectures. The underlying principle is the same.