Question

If a, b, c be last three digits of the number (6! + 1)${}^{6!}$, then (a + b + c) is

Answer 1

$(720+1{)}^{720}$
${\Rightarrow }^{720}{C}_0{720}^{720}+....{.}^{720}{C}_{719}.720.1{+}^{720}{C}_{720}{.1}^{720}$
$\Rightarrow \underset{1000K}{\underset{}{{}^{720}{C}_0.{720}^{720}+.......}}\underset{....400}{\underset{}{720.720}+1}$
So we have to check last two terms ... 401 hence last 3 digits are 401 $\Rightarrow$ a + b + c = 5