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

Step 1: Since \( 6!=6\times5! \), the terms \( 5! \) and \( 6! \) combine as \[ 5!+6!=5!(1+6)=7\times5!, \] which is exactly divisible by \( 7 \) and so contributes nothing to the remainder.

Step 2: So only \( 1!+2!+3!+4!=1+2+6+24=33 \) affects the remainder, since every term from \( 7! \) onward is also a multiple of \( 7 \).

Step 3: Dividing, \( 33=7\times4+5 \), so \( 33\equiv5\pmod7 \).
\[ \boxed{5} \]