JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 18)
If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)2022 is divided by 8 is __________.
Answer
1
Explanation
Let x = 4k + 3
(2020 + x)2022
= (2020 + 4k + 3)2022
= (4(505) + 4k + 3)2022
= (4P + 3)2022
= (4P + 4 $$-$$ 1)2022
= (4A $$-$$ 1)2022
2022C0(4A)0($$-$$1)2022 + 2022C1(4A)1($$-$$1)2021 + ......
= 1 + 2022(4A)(-1) + .....
= 1 + 8$$\lambda$$
$$ \therefore $$ Reminder is 1.
(2020 + x)2022
= (2020 + 4k + 3)2022
= (4(505) + 4k + 3)2022
= (4P + 3)2022
= (4P + 4 $$-$$ 1)2022
= (4A $$-$$ 1)2022
2022C0(4A)0($$-$$1)2022 + 2022C1(4A)1($$-$$1)2021 + ......
= 1 + 2022(4A)(-1) + .....
= 1 + 8$$\lambda$$
$$ \therefore $$ Reminder is 1.
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