The remainder when $(2021)^{2022}+(2022)^{2021}$ is divided by 7 is
Solution
<p>${(2021)^{2022}} + {(2022)^{2021}}$</p>
<p>$= {(7k - 2)^{2022}} + {(7{k_1} - 1)^{2021}}$</p>
<p>$$ = {\left[ {{{(7k - 2)}^3}} \right]^{674}} + {(7{k_1})^{2021}} - 2021{(7{k_1})^{2020}}\, + \,....\, - 1$$</p>
<p>$= {(7{k_2} - 1)^{674}} + (7m - 1)$</p>
<p>$= (7n + 1) + (7m - 1) = 7(m + n)$ (multiple of 7)</p>
<p>$\therefore$ Remainder = 0</p>
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
This question is part of PrepWiser's free JEE Main question bank. 193 more solved questions on Binomial Theorem are available — start with the harder ones if your accuracy is >70%.