Hard quant

jsl wrote:If N is an integer, What is the units' digit of (N-1)! + N! + (N+1)! +2N*3N+1?

(1) N is greater than 1
(2) N = 4K + 2, where K is an integer equal to or greater than 1

It's not strictly necessary, but I'd start by factoring out (N-1)! from the factorials:

(N-1)! + N! + (N+1)! +2N*3N+1 = (N-1)!*(1 + N + N*(N+1)) + 2N*3N + 1

Now, if N > 5, (N-1)! will be at least 5!, so will be divisible by 5 and by 2, and so will end in 0. That means, if we know N > 5, the units digit of (N-1)! is 0, so the units digit of (N-1)!*(1 + N + N*(N+1)) will also be 0 (the units digit of a product comes from the product of the units digits). If N > 5, we then only care about the 2N*3N + 1 = 6N^2 + 1, since the units digit of the sum of the other terms must be zero. However, if N = 6, this will end in 7, while if N = 10, this will end in 1. Since even using both statements together, N could be 6 and could be 10, the statements are insufficient together, and the answer is E.

I'd add that at first, reading the question, I wondered whether it was meant to read 2N*(3N + 1) at the end, but it makes little difference to the solution in either case.