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 $=1+2+6+3+1+6=19\equiv \boxed{5}\ (\bmod 7)$. 

Answer 2

Since every factorial from \( 7! \) onward is divisible by \( 7 \), only the first six terms affect the remainder. Track the cumulative sum modulo \( 7 \) after adding each factorial in turn, then match the final result against the given options.

1. Option A (0): Building the running total step by step, \[ 1,\ 1+2=3,\ 3+6=9\equiv2,\ 2+24=26\equiv5,\ 5+120=125\equiv6,\ 6+720=726\equiv5\ (\text{all mod }7), \] the final cumulative value is \( 5 \), not \( 0 \), so this option is rejected.
2. Option B (5): The running total above ends at \( 726\equiv5\pmod7 \) after all six terms, and every later term \( 7!,8!,\ldots,100! \) adds nothing modulo \( 7 \). This matches exactly.
3. Option C (3): The running total touches \( 3 \) only midway (after \( 2! \)), not at the end, so this option is rejected.
4. Option D (6): The value \( 6 \) appears only as an intermediate result (after \( 5! \)), not as the final cumulative remainder, so this option is rejected too.

Tracking the cumulative sum term by term confirms the final remainder is \( 5 \).

Hence, the correct answer is 5.