Question

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

Step-1: Use the division algorithm in the statement.

Dividend $=$ Quotient $\times$ Divisor $+$ remainder

Here Dividend$=x$, Quotient $=$ let $k$, Divisor$=4$, Remainder $=3$

$\therefore x=4k+3$

Put $x=4k+3$ in ${(2020+x)}^{2022}$.

${(2020+x)}^{2022}={(2020+4k+3)}^{2022}$

Write this in the form ${(x-a)}^n$ we add and subtract $1$ in above.

$$\begin{gathered} {(2020+4k+3+1-1)}^{2022}=(2024+4k-1{)}^{2022} \\ \left[(2024+4k)=4(506+k)=4\left(A\right)[let(500+k)=A] \\ (2024+4k)=4A\right] \\ \therefore {(2024+4k-1)}^{2022}={(4A-1)}^{2022} \end{gathered}$$

Step-2: Expand ${(4A-1)}^{2022}$ by the binomial theorem.

$$\begin{gathered} {(4A-1)}^{2022}={}^{2022}c_0{\left(4A\right)}^{2022}{(-1)}^0+{}^{2022}c_1{\left(4A\right)}^{2021}{(-1)}^1+...{}^{2022}c_{2021{\left(4A\right)}^1{(-1)}^{2021}+{}^{2022}c_{2022{(-1)}^{2022}}} \\ ={\left(4A\right)}^{2022}-2022{\left(4A\right)}^{2021}+...-2022\left(4A\right)+1 \\ \because {}^{2022}c_0=1,{}^{2022}c_1=2022,{}^{2022}c_{2021}=2022,and{}^{2022}c_{2022}=1 \\ ={\left(4A\right)}^{2022}-2022{\left(4A\right)}^{2021}+...-2022\left(4A\right)+1 \\ Thisisoftheform \\ =4\times 2\mu +1 \\ =8\mu +1 \end{gathered}$$

This is of the form $8\mu +1$. By division algorithm, the remainder is 1.

Therefore, the remainder is 1.