The remainder when \(1! + 2! + 3! + ... + 95!\) is divided by \(15\) is

The remainder is $3$. Every factorial beginning with $5!$ contains factors $3$ and $5$, so the remainder comes from $1!+2!+3!+4!=33=2\cdot15+3$. A remainder of $9$ omits $4!$, even though $24$ contributes to the remainder.