Question

Find the remainder when $1!+2!+3!+4!+....+n!$ divided by $15$, if $n\ge 5$

Answer 1

$1!+2!+3!+4!+.........+n!$

$n\ge 5$

when n=5

$1!+2!+3+!+4!+5!=1+1\times 2+1\times 2\times 3+1\times 2\times 3\times 4+1\times 2\times 3\times 4\times 5$

$=1+2+6+24+120$

$=153$ $wherek=10$

$\frac{153}{15}=15k+3$where $k=10$

when n=6

$153+6!=153+720$

$\frac{873}{15}=15k+3$ where k=58

when $n\ge 5$, $3$ is obtained as remainder when divided by $15$