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,</p><p>Now rewriting above expression we get,</p><p>${2023}^{2023} = {\left(2030 - 7\right)}^{2023}$</p><p>$⇒ {\left(2030 - 7\right)}^{2023} = C0 2023 {2030}^{2023} - C1 2023 {\left(2030\right)}^{2022} \times 7 + . . . . . - {7}^{2023}$</p><p>$⇒ {\left(2030 - 7\right)}^{2023} = 35k - {7}^{2023}$;$k \in Z$.</p><p>\left[as2030ismultipleof 35\right]$

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

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

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

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

$= \left(35{k}_{1} + {7}^{674}\right) \times \left(- 7\right)$

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

Again rewriting the expression as

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

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

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 $7$.