For how many integers $n$, with $2 \leq n \leq 80$, is $\frac{(n-1)(n)(n+1)}{8}$ equal to an integer?
(A) 10
(B) 20
(C) 59
(D) 39
(E) 49
If $n$ is an odd integer, then each of $n-1$ and $n+1$ is even.

In fact, $n-1$ and $n+1$ are consecutive even integers, so one is a multiple of 4 and the other is divisible is 2 (since it is even).

Thus, $(n-1)(n+1)$ contains at least 3 factors of 2, which tells us that $(n-1)(n)(n+1)$ does as well, ie. is divisible by 8 .

So if $n$ is an odd integer, then $\frac{(n-1)(n)(n+1)}{8}$ is an integer.

(There are 39 odd integers between 2 and 80, inclusive.)

If $n$ is an even integer, then each of $n-1$ and $n+1$ is odd.

Thus, $(n-1)(n)(n+1)$ is divisible by 8 only when $n$ is divisible by 8 .

(There are 10 multiples of 8 between 2 and 80 , inclusive.)

Therefore, there are $39+10=49$ integers $n$, with $2 \leq n \leq 80$, such that $\frac{(n-1)(n)(n+1)}{8}$ is an integer.

ANSWER: (E)