Question

What is the remainder when1!+2!+3!+....1000 is divided by 12

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{1!+2!+3!+\;.\;.\;.\;.+1000!}$

$\underline{\textbf{To find:}}$

$\textsf{The remainder when}$

$\mathsf{1!+2!+3!+\;.\;.\;.\;.+1000!\;\;is\;divided\;by\;12}$

$\underline{\textbf{Solution:}}$

$\textsf{We know that,}$

$\textbf{From 4! onwards, each n! is a multiple of 12}$

$\mathsf{Consider,}$

$\mathsf{1!+2!+3!+\;.\;.\;.\;.+1000!}$

$\mathsf{=1!+2!+3!+(Multiple\;of\;12)}$

$\mathsf{=1+2+6+(Multiple\;of\;12)}$

$\mathsf{=9+(Multiple\;of\;12)}$

$\therefore\textsf{The remainder is 9 when}$

$\mathsf{1!+2!+3!+\;.\;.\;.\;.+1000!\;is\;divided\;by\;12}$