JEE Main 2022 (Online) 25th July Evening ShiftQuestion 1
The remainder when $(11)^{1011}+(1011)^{11}$ is divided by $9$ is
Binomial Theorem · Remainder and Divisibility
Solution
Concept. Reduce each base modulo $9$; $2^6\equiv1$ and $3^2\equiv0$. Formulas used. (1) $2^6=64=63+1\equiv1\pmod9$Step 1 - First term. $11\equiv2$, and $1011=6\cdot168+3$. By (1), $11^{1011}\equiv2^3=8$. Step 2 - Second term. $1011\equiv1+0+1+1=3\pmod9$, so $1011^{11}\equiv3^{11}$. Since $9\mid3^2$, $3^{11}\equiv0$. Step 3 - Add. $8+0=8$. Answer: (D)