Mathematics - Binomial Theorem Question with Solution | TestHub

Given that $\alpha$ be the remainder when ${\left(22\right)}^{2022}+{\left(2022\right)}^{22}$ is divided by $3$ and $\beta$ be the remainder when the same is divided by $7$.

$\Rightarrow {\left(22\right)}^{2022}+{\left(2022\right)}^{22}={\left(21+1\right)}^{2022}+{\left(2022\right)}^{22}$.

Here ${\left(2022\right)}^{22}$ is divisible by $3$ as $2022$ is divisible by $3$.

So on expanding ${\left(21+1\right)}^{2022}$, we get

$\Rightarrow {\left(21+1\right)}^{2022}={}^{2022}C_0{\left(21\right)}^{2022}+{}^{2022}C_1{\left(21\right)}^{2021}+......+{}^{2022}C_{2022}{\left(1\right)}^{2022}$

$=3\left(3^{2021}\times 7^{2022}+{}^{2022}C_1\times 3^{2020}\times 7^{2021}+......\right)+1$

$=3k_1+1$

In this case the remainder is $1$

Hence, $\alpha =1$

Now, $\Rightarrow {\left(22\right)}^{2022}+{\left(2022\right)}^{22}={\left(21+1\right)}^{2022}+{\left(2023-1\right)}^{22}$

Take ${\left(2023-1\right)}^{22}$

$\Rightarrow {\left(2023-1\right)}^{22}={}^{22}C_0{\left(2023\right)}^{22}-{}^{22}C_1{\left(2023\right)}^{21}+......+{}^{22}C_{22}{\left(-1\right)}^{22}$

$=7\left({}^{22}C_0{\left(7\right)}^{21}{\left(289\right)}^{22}-{}^{22}C_1{\left(7\right)}^{20}{\left(289\right)}^{21}+......\right)+1$

$=7k_2+1$

$\Rightarrow {\left(21+1\right)}^{2022}+{\left(2023-1\right)}^{22}=7k_1+1+7k_2+1$

$=7\mu +2$

$\Rightarrow \beta =2$.

Hence, ${\alpha }^2+{\beta }^2=1^2+2^2=5$.

Therefore, the required answer is $5$.