Find the remainder when `sum_(r=1)^(n)r!` is divided by `15`, if `n ge5`.

Explanation

Let `N=underset(r=1)overset(n)(sum)r!=1!+2!+3!+4!+5!+6!+7!+ . . .+n!`
`=(1!+2!+3!+4!)+(5!+6!+7!+ . . .+n!)`
`=33+(5!+6!+7!+ . . .+n!)`
`implies(N)/(15)=(33)/(15)+((5!+6!+7!+ . . .+n!))/(15)`
`=2+(3)/(15)+` interger [as `5!,6!`, . . . are divisible by 15]
`=(3)/(15)+`Integer
Hence, remainder is 3.