Question

Take a break from integrals: Find the remainder when \(\sum\limits_{n=1}^{2015}n!\) is divided by \(15\).

Answer 1

as

\(\large\tt \begin{align} \color{black}{5!+6!+7!+............2015!\hspace{1.5em}\\
}\end{align}\)

all are divisible by

\(\large\tt \begin{align} \color{black}{15\hspace{1.5em}\\}\end{align}\)

so

\(\large\tt \begin{align} \color{black}{\dfrac{1!+2!+3!+4!+5!.......2015!}{15}\hspace{1.5em}\\~\\
\implies =\dfrac{1!+2!+3!+4!}{15}\hspace{1.5em}\hspace{1.5em}\\~\\
\implies =\dfrac{33}{15}\hspace{1.5em}\hspace{1.5em}\\~\\~\\
\Large \implies \equiv3\hspace{1.5em}\hspace{1.5em}\\~\\
}\end{align}\)

Answer 2

That's correct!