Question

Find the remainder when (17) (9!) + 2 (18!) is divided by (9!) 17408___

Answer 1

$\frac{\left[17\right(9!)+2(18!\left)\right]}{9!.17408}=\frac{17.9!}{9!{17.2}^{10}}+\frac{2.18!}{9!.17{.2}^{10}}$
$\frac{17.9!}{9!{17.2}^{10}}=\frac{1}{2^{10}}andremainderof\frac{1}{2^{10}}=1$
Hence, the Remainder of $\left(\frac{17.9!}{9!{17.2}^{10}}\right)=17.9!$
$2.18!=\mathrm{2.18.17.16.15.14.13.12.11.10.9}!$
In 18, 16, 14, 12, 10 the number of 2s are 1, 4, 1, 2 and 1 respectively.
So $2.18!=\mathrm{2.18.17.16.15.14.13.12.11.10.9}!=2^{10}.17k.9!=$ Where k is an integer
Remainder $\left(\frac{2^{10}.17k.9!}{9!.\left(17\right)2^{10}}\right)=0$
So the required remainder is $17.9!+0=17.9!$