3 × 7 22 + 2 × 10 22 - 44 when divided by 18 leaves the remainder

Given that: $3\times 7^{22}+2\times {10}^{22}-44$

$=3\times (6+1{)}^{22}+2\times (9+1{)}^{22}-44$

As we know that ${(1+x)}^n=1+x({}^{n}C_1+{}^{n}C_{2}x+.....)=1+kx ; k,x\in \mathrm{I}$

$=18\lambda +3+18\mu +2-44 ; \lambda ,\mu \in \mathrm{I}$
$=18\delta -39 ;\delta \in \mathrm{I}$
$=18\beta +15$

$\Rightarrow$remainder $=15$