JEE MAIN - Mathematics (2022 - 27th July Morning Shift - No. 5)
The remainder when $$(2021)^{2022}+(2022)^{2021}$$ is divided by 7 is
0
1
2
6
Explanation
$${(2021)^{2022}} + {(2022)^{2021}}$$
$$ = {(7k - 2)^{2022}} + {(7{k_1} - 1)^{2021}}$$
$$ = {\left[ {{{(7k - 2)}^3}} \right]^{674}} + {(7{k_1})^{2021}} - 2021{(7{k_1})^{2020}}\, + \,....\, - 1$$
$$ = {(7{k_2} - 1)^{674}} + (7m - 1)$$
$$ = (7n + 1) + (7m - 1) = 7(m + n)$$ (multiple of 7)
$$\therefore$$ Remainder = 0
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