The remainder when (2021)2023 is divided by 7 is :

Question

The remainder when (2021)2023 is divided by 7 is : (A) 1 (B) 2 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

(2021)2023=(7 λ-2)2023 = 2023 C 0(7 A )2023- ldots . 2023 C 2023 22023 =7 t -22023 ∴-22023=-2 × 22022 =-2 ×(23)674 =-2(1+7 μ)674 =-(7 α+2) ⇒ remainder =-2 or +5

Q. The remainder when $(2021)^{2023}$ is divided by $7$ is :

1

2

5

6

$(2021)^{2023} = (7 λ - 2)^{2023}$
$=^{2023} C_{0} (7 A)^{2023} - \dots .^{2023} C_{2023} 2^{2023}$
$= 7 t - 2^{2023}$
$∴ - 2^{2023} = - 2 \times 2^{2022}$
$= - 2 \times (2^{3})^{674}$
$= - 2 (1 + 7 μ)^{674}$
$= - (7 α + 2)$
$\Rightarrow$ remainder $= - 2$ or $+ 5$

Q. The remainder when (2021)2023 is divided by 7 is :

1

2

5

6

(2021)2023=(7λ−2)2023 =2023C0​(7A)2023−….2023C2023​22023 =7t−22023 ∴−22023=−2×22022 =−2×(23)674 =−2(1+7μ)674 =−(7α+2) ⇒ remainder =−2 or +5

Q. The remainder when $(2021)^{2023}$ is divided by $7$ is :

$(2021)^{2023} = (7 λ - 2)^{2023}$
$=^{2023} C_{0} (7 A)^{2023} - \dots .^{2023} C_{2023} 2^{2023}$
$= 7 t - 2^{2023}$
$∴ - 2^{2023} = - 2 \times 2^{2022}$
$= - 2 \times (2^{3})^{674}$
$= - 2 (1 + 7 μ)^{674}$
$= - (7 α + 2)$
$\Rightarrow$ remainder $= - 2$ or $+ 5$