The remainder when 19 ^ 200 + 23 ^ 200 is divided by 49 , is _____ .

Solving ${19}^{200} + {23}^{200} = {\left(21 - 2\right)}^{200} + {\left(21 + 2\right)}^{200}$

$= 2\left\{C0 200 {\left(21\right)}^{200} \times {2}^{0} + C2 200 {\left(21\right)}^{198} \times {2}^{2} . . . . C200 200 {\left(21\right)}^{0} \times {2}^{200}\right\}$

Now we know that ${21}^{2} = 441$ is divisible by $49$,

So rest of the terms will be divisible by $49$

So lets consider for $2\left\{C200 200 {\left(21\right)}^{0} \times {2}^{200}\right\} = {2}^{201}$

Now rewriting the term ${2}^{201} = {\left({2}^{3}\right)}$^{67} = {\left(7 + 1\right)}^{67}

Now ${\left(1 + 7\right)}^{67} = C0 67 \times 1 + C1 67 7 + C2 67 {7}^{2} . . . . . . . . .$

$= 1 + 67 \times 7 + 49k$

Now when dividing ${\left(1 + 7\right)}^{67}$ by $49$ we consider $1 + 67 \times 7$ as $49k$ is divisible by $49$

Now $1 + 67 \times 7 = 470$ which when divided by $49$ leaves remainder as $29$.