Question

The remainder, when $19^{200}+23^{200}$ is divided by $49$ , is_______

Answer 1

Answer: 29

$(21+2{)}^{200}+(21-2{)}^{200}$
$\Rightarrow 2[{}^{100}{C}_{0}21^{200}+200C_{2}21^{198}\cdot 2^2+\dots ..+{}^{200}{C}_{198}21^2{2}^{198}+2^{200}]$
$\Rightarrow 2\left[49I_1+2^{200}\right]=49I_1+2^{201}$
$\text{ Now },2^{201}=(8{)}^{67}=(1+7{)}^{67}=49I_2+{}^{67}{C}_0{}^{67}{C}_1\cdot 7=$
$49I_2+470=49I_2+49\times 9+29$
$\therefore \text{ Remainder is }29$