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
Can anyone guide me how to approach the problem?