What is the remainder when 1! + 2! + 3! + … + 100! is divided by 15?

From 5! onwards every factorial contains both a 3 and a 5, so each one is a multiple of 15 and adds nothing to the remainder — for instance 5! $=120=15\times 8$. Only the first four terms matter: 1 + 2 + 6 + 24 = 33, and $33=15\times 2+3$. The remainder is 3. Cutting the tail off at the wrong place is the usual error; the safe rule is to stop at the first factorial that already contains every prime factor of the divisor, which for $15=3\times 5$ means 5!.