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JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Binomial Theorem

The remainder when

(2021)^{2023}

is divided by 7 is :

(A)1
(B)2
(C)5✓ Correct
(D)6

Answer: (C)

Solution

Note that

2021 \equiv - 2 (mod 7)

.

So

(2021)^{2023} \equiv (- 2)^{2023} (mod 7)

.

By Fermat's Little Theorem,

2^{6} \equiv 1 (mod 7)

.

Write

2023 = 6 \cdot 337 + 1

, so

2^{2023} = 2^{6 \cdot 337 + 1} = (2^{6})^{337} \cdot 2 \equiv 1^{337} \cdot 2 = 2 (mod 7)

.

Therefore,

(- 2)^{2023} = - 2^{2023} \equiv - 2 \equiv 5 (mod 7)

.

The remainder is 5.

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option

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