JEE Main 2023 — Mathematics Algebra

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

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

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

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

$$\Rightarrow {(21+1)}^{2022}=C02022{(21)}^{2022}+C12022{(21)}^{2021}+......+C20222022{(1)}^{2022}$$⇒(21+1)2022=C02022(21)2022+C12022(21)2021+......+C20222022(1)2022

$$=3({3}^{2021}\times {7}^{2022}+C12022\times {3}^{2020}\times {7}^{2021}+......)+1$$=3(32021×72022+C12022×32020×72021+......)+1

$$=3{k}_{1}+1$$=3k1​+1

In this case the remainder is $1$1

Hence, $\alpha =1$α=1

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

Take ${(2023-1)}^{22}$(2023−1)22

$$\Rightarrow {(2023-1)}^{22}=C022{(2023)}^{22}-C122{(2023)}^{21}+......+C2222{(-1)}^{22}$$⇒(2023−1)22=C022(2023)22−C122(2023)21+......+C2222(−1)22

$$=7(C022{(7)}^{21}{(289)}^{22}-C122{(7)}^{20}{(289)}^{21}+......)+1$$=7(C022(7)21(289)22−C122(7)20(289)21+......)+1

$$=7{k}_{2}+1$$=7k2​+1

$$\Rightarrow {(21+1)}^{2022}+{(2023-1)}^{22}=7{k}_{1}+1+7{k}_{2}+1$$⇒(21+1)2022+(2023−1)22=7k1​+1+7k2​+1

$$=7\mu +2$$=7μ+2

$\Rightarrow \beta =2$⇒β=2.

Hence, ${\alpha }^{2}+{\beta }^{2}={1}^{2}+{2}^{2}=5$α2+β2=12+22=5.

Therefore, the required answer is $5$5.