The remainder when 2023 2023 is divided by 35 is

To find the remainder when ${2023}^{2023}$ is divided by $35$ we will use binomial expansion,

Now rewriting above expression we get,

${2023}^{2023}={\left(2030-7\right)}^{2023}$ 

$\Rightarrow {\left(2030-7\right)}^{2023}={}^{2023}C_0{2030}^{2023}-{}^{2023}C_1{\left(2030\right)}^{2022}\times 7+.....-7^{2023}$

$\Rightarrow {\left(2030-7\right)}^{2023}=35k-7^{2023}$; $k\in Z$.

$\left[\mathrm{as} 2030 \mathrm{is} \mathrm{multiple} \mathrm{of} 35\right]$

Now remainder will be $-7^{2023}$. Rewriting $-7^{2023}$, we get

$-7^{2023}=-{\left(7^3\right)}^{674}\times 7$

$=-{\left(7^3\right)}^{674}\times 7=-{\left(343\right)}^{674}\times 7$

$=-{\left(343\right)}^{674}\times 7=-{\left(350-7\right)}^{674}\times 7$

$=\left(35k_1+7^{674}\right)\times \left(-7\right)$

So, again using binomial we get remainder as $-7\times 7^{674}$

Again rewriting the expression as 

$-7\times 7^{674}=-7^{675}$

$\Rightarrow -7^{675}=-{\left(7^3\right)}^{225}={\left(350-7\right)}^{75}$

Again using binomial we get remainder as 

${\left(350-7\right)}^{75}={\left(-7\right)}^{75}=-{\left(350-7\right)}^{25}$

Again using binomial we get remainder as $7^{25}$.

Now again rewriting ${\left(7^3\right)}^8\times 7={\left(350-7\right)}^8\times 7$

Using binomial remainder will be $7^9$ which can be written as ${\left(7^3\right)}^3={\left(350-7\right)}^3$,

Using binomial remainder will be $-7^3$ which can be written as $-343$, so remainder will be $350-343=7$,

Hence, the remainder when ${2023}^{2023}$ when divided by $35$ is $7$.