JEE Main 2023 — Mathematics Algebra

To find the remainder when ${2023}^{2023}$20232023 is divided by $35$35 we will use binomial expansion,

Now rewriting above expression we get,

$${2023}^{2023}={(2030-7)}^{2023}$$20232023=(2030−7)2023

$$\Rightarrow {(2030-7)}^{2023}=C02023{2030}^{2023}-C12023{(2030)}^{2022}\times 7+.....-{7}^{2023}$$⇒(2030−7)2023=C0202320302023−C12023(2030)2022×7+.....−72023

$\Rightarrow {(2030-7)}^{2023}=35k-{7}^{2023}$⇒(2030−7)2023=35k−72023; $k\in Z$k∈Z.

$$[\mathrm{as}2030\mathrm{is}\mathrm{multiple}\mathrm{of}35]$$[as2030ismultipleof35]

Now remainder will be $-{7}^{2023}$−72023. Rewriting $-{7}^{2023}$−72023, we get

$$-{7}^{2023}=-{({7}^{3})}^{674}\times 7$$−72023=−(73)674×7

$$=-{({7}^{3})}^{674}\times 7=-{(343)}^{674}\times 7$$=−(73)674×7=−(343)674×7

$$=-{(343)}^{674}\times 7=-{(350-7)}^{674}\times 7$$=−(343)674×7=−(350−7)674×7

$$=(35{k}_{1}+{7}^{674})\times (-7)$$=(35k1​+7674)×(−7)

So, again using binomial we get remainder as $-7\times {7}^{674}$−7×7674

Again rewriting the expression as

$$-7\times {7}^{674}=-{7}^{675}$$−7×7674=−7675

$$\Rightarrow -{7}^{675}=-{({7}^{3})}^{225}={(350-7)}^{75}$$⇒−7675=−(73)225=(350−7)75

Again using binomial we get remainder as

$${(350-7)}^{75}={(-7)}^{75}=-{(350-7)}^{25}$$(350−7)75=(−7)75=−(350−7)25

Again using binomial we get remainder as ${7}^{25}$725.

Now again rewriting ${({7}^{3})}^{8}\times 7={(350-7)}^{8}\times 7$(73)8×7=(350−7)8×7

Using binomial remainder will be ${7}^{9}$79 which can be written as ${({7}^{3})}^{3}={(350-7)}^{3}$(73)3=(350−7)3,

Using binomial remainder will be $-{7}^{3}$−73 which can be written as $-343$−343, so remainder will be $350-343=7$350−343=7,

Hence, the remainder when ${2023}^{2023}$20232023 when divided by $35$35 is $7$7.