Mathematics - Binomial Theorem Question with Solution | TestHub

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

$=2\left\{{}^{200}C_0{\left(21\right)}^{200}\times 2^0+{}^{200}C_2{\left(21\right)}^{198}\times 2^2....{}^{200}C_{200}{\left(21\right)}^0\times 2^{200}\right\}$

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 $2\left\{{}^{200}C_{200}{\left(21\right)}^0\times 2^{200}\right\}=2^{201}$

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

Now ${\left(1+7\right)}^{67}={}^{67}C_0\times 1+{}^{67}C_{1}7+{}^{67}C_2{7}^2.........$

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

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 $29$.