Question

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

Answer 1

$(21+2)^{200}+(21-2)^{200} $
$\Rightarrow 2\left[{ }^{100} C _0 21^{200}+200 C _2 21^{198} \cdot 2^2+\ldots . .+{ }^{200} C _{198} 21^2\right. \left.2^{198}+2^{200}\right] $
$ \Rightarrow 2\left[49 I _1+2^{200}\right]=49 I_1+2^{201} $
$ \text { Now }, 2^{201}=(8)^{67}=(1+7)^{67}=49 I _2+{ }^{67} C _0{ }^{67} C _1 \cdot 7= $
$ 49 I _2+470=49 I _2+49 \times 9+29 $
$ \therefore \text { Remainder is } 29$