Find the number of positive integers n such that n divides (n+1)(n+2)(n+3).
Correct: C
We are given that n divides (n+1)(n+2)(n+3).
We can expand the expression: (n+1)(n+2)(n+3) = (n^2 + 3n + 2)(n+3) = n^3 + 3n^2 + 2n + 3n^2 + 9n + 6 = n^3 + 6n^2 + 11n + 6.
For n to divide this expression, n must divide the constant term, which is 6 (since n divides n^3, 6n^2, and 11n).
So, n must be a divisor of 6. The positive divisors of 6 are 1, 2, 3, and 6.
Thus, there are 4 such positive integers.