what is the remainder when 1!+2!+3!+4!+⋯+49! is divided by 7?
MyApproch:
By taking individual numbers as ($1$!+$2$!+$3$!+$4$!+⋯+$49$!)/$7$
I get $1$+$2$+$6$+$3$+$1$+$6$+$0$+$0$.....=$19$/$7$=$5$
But the approach will be too long.I am not able to figure out any other way
$\endgroup$ 5Can anyone guide me how to approach the problem?
2 Answers
$\begingroup$For $k \geq 7$, we have $$7\; \vert\; k!.$$ Hence $$\sum_{k=1}^{49} k! = 1!+2!+3!+4!+5!+6!\equiv1+2+(-1)+3+1+(-1)\equiv5 \pmod{7}.$$
$\endgroup$ $\begingroup$If you're concerned that evaluating the higher factorials will be time consuming, note that $5! + 6! = 5! \times (1 + 6) \equiv 0 \pmod 7$
$\endgroup$
