JEE Main 2022 (Online) 27th July Morning ShiftQuestion 1
The remainder when $(2021)^{2022}+(2022)^{2021}$ is divided by $7$ is
Binomial Theorem · Remainder and Divisibility
Solution
Concept. Reduce each base modulo $7$; then $8=7+1$ finishes it. Formulas used. (1) $8^N=(1+7)^N\equiv1\pmod7$Step 1 - Reduce the bases. $2021=7\cdot288+5\equiv-2$ and $2022\equiv-1\pmod7$. Step 2 - First term. $(-2)^{2022}=2^{2022}=\left(2^3\right)^{674}=8^{674}\equiv1$ by (1). Step 3 - Second term. $(-1)^{2021}=-1$. Step 4 - Add. $1-1=0$, so the remainder is $0$. Answer: (A)