Question

What is the remainder when $1!+2!+3!+\cdots+100!$ is divided by $7$? 

• A. 0
• B. 5
• C. 3
• D. 6

Hint:

For factorial sums mod a prime $p$, everything from $p!$ onward is $0 \pmod p$.

Answer 1

The Correct Option is B

For $n\ge 7$, $n!$ is a multiple of $7$, so it contributes $0\pmod 7$. Hence \[ 1!+2!+\cdots+100!\equiv 1!+2!+3!+4!+5!+6!\pmod 7. \] Compute mod $7$: $1!\equiv1$,$2!\equiv2$, $3!\equiv6$,$4!=24\equiv3$, $5!=120\equiv1$, $6!=720\equiv6$. Sum $=1+2+6+3+1+6=19\equiv \boxed{5}\ (\bmod 7)$.