How to prove that the sum $1!+2!+...+n! \equiv 0 \pmod 9$

Question

I understand the divisibility rule, if the sum of a numbers digits are divisible by $9$, the number is divisible by $9$. I'm not sure how to integrate that, or if I even should. Could I assume that for any $n>8$, $n!$ is divisible by $9$. And thus the sum of all $n!$ where $n>8$ would too be divisible by $9$? Thanks in advance for any help.

Answer 1

This statement isn't true for $n=1$ since $1!=1$, nor for $n=2$ since $1!+2!=3$.

But for $n=3$, $1!+2!+3!=9\equiv 0$ (mod 9).

This is again not true $n=4$, since $1!+2!+3!+4!=33\equiv 6$ (mod 9).

But for $n=5$, $1!+2!+3!+4!+5!=153\equiv 0$ (mod 9).

Then $m!=0$ (mod 9) for all $m\geq 6$. Therefore, for $n\geq 6$, $1!+\dots+n!\equiv 1!+2!+3!+4!+5!\equiv 0$ (mod 9).

Answer 2

Hint 1

If $n\ge 6$ then

\begin{align} n! &=1\cdot 2\cdot \color{red}{3}\cdot 4\cdot 5\cdot \color{red}6\cdot \;\cdots\;\cdot n = \color{red}9\cdot (1\cdot 2\cdot 4\cdot 5\cdot 2\cdot\;\cdots\; \cdot n). \end{align}

Hint 2

$$1!+2!+3!+4!+5!=153=\color{red}9\cdot 17.$$

Answer 3

But it's not true for $n= 1,2$

It is for $n=3$ by examination. $1! +2! +3! = 1+2 +6=9$.

And $4! = 24$ is not divisible by $9$ but $1!+2!+3!$ is, then it can't be true for $n=4$ (and by examination for $n=4$ we git $(1!+2! +3!) + 4! = 9 + 24 = 33$).

As $4! \equiv 24 \equiv 6\pmod 9$ then $5!\equiv 6*5\equiv 30\equiv 3\pmod 9$ so $(1!+2!+3!)+4!+5! \equiv 0 + 6 + 3\equiv 0 \pmod 9$.

And for any $n\ge 6$ we have $3\times 6=18|n!$ so for all $n\ge 6$ we have $9|n!$ so as $(1!+2! +3!) + (4! + 5!) + 6! +......... + n! \equiv (0) + (6 + 3) + 0 + ....... + 0\equiv 0 \pmod 9$.

but it's not true for $n=1,2,4$.