JEE Main 2023 (Online) 10th April Evening ShiftQuestion 1
Let the number $(22)^{2022}+(2022)^{22}$ leave the remainder $\alpha$ when divided by $3$ and $\beta$ when divided by $7$. Then $\left(\alpha^{2}+\beta^{2}\right)$ is equal to
Binomial Theorem · Remainder and Divisibility
Solution
Concept. Reduce each base modulo $3$ and modulo $7$; the bases become $0$ or $\pm1$. Formulas used. (1) $a\equiv b\pmod m\Rightarrow a^k\equiv b^k\pmod m$Step 1 - Modulo $3$. $22=21+1\equiv1$ and $2022\equiv0$ (digit sum $6$). So the number $\equiv1+0=1$: $\alpha=1$. Step 2 - Modulo $7$. $22\equiv1$ and $2022=7\cdot288+6\equiv-1$. So the number $\equiv1+(-1)^{22}=2$: $\beta=2$. Step 3 - Value. $\alpha^2+\beta^2=1+4=5$. Answer: (A)