The remainder when (2021)2023 is divided by 7 is :
Solution
(2021)<sup>2023</sup></p>
<p>= (2016 + 5)<sup>2023</sup> [here 2016 is divisible by 7]</p>
<p>= <sup>2023</sup>C<sub>0</sub> (2016)<sup>2023</sup> + .......... + <sup>2023</sup>C<sub>2022</sub> (2016) (5)<sup>2022</sup> + <sup>2023</sup>C<sub>2023</sub> (5)<sup>2023</sup></p>
<p>= 2016 [<sup>2023</sup>C<sub>0</sub> . (2016)<sup>2022</sup> + ....... + <sup>2023</sup>C<sub>2022</sub> . (5)<sup>2022</sup>] + (5)<sup>2023</sup></p>
<p>= 2016$\lambda$ + (5)<sup>2023</sup></p>
<p>= 7 $\times$ 288$\lambda$ + (5)<sup>2023</sup></p>
<p>= 7K + (5)<sup>2023</sup> ...... (1)</p>
<p>Now, (5)<sup>2023</sup></p>
<p>= (5)<sup>2022</sup> . 5</p>
<p>= (5<sup>3</sup>)<sup>674</sup> . 5</p>
<p>= (125)<sup>674</sup> . 5</p>
<p>= (126 $-$ 1)<sup>674</sup> . 5</p>
<p>= 5[<sup>674</sup>C<sub>0</sub> (126)<sup>674</sup> + ......... $-$ <sup>674</sup>C<sub>673</sub> (126) + <sup>674</sup>C<sub>674</sub>]</p>
<p>= 5 $\times$ 126 [<sup>674</sup>C<sub>0</sub>(126)<sup>673</sup> + ....... $-$ <sup>674</sup>C<sub>673</sub>] + 5</p>
<p>= 5 . 7 . 18 [<sup>674</sup>C<sub>0</sub>(126)<sup>673</sup> + ....... $-$ <sup>674</sup>C<sub>673</sub>] + 5</p>
<p>= 7$\lambda$ + 5</p>
<p>Replacing (5)<sup>2023</sup> in equation (1) with 7$\lambda$ + 5, we get,</p>
<p>(2021)<sup>2023</sup> = 7K + 7$\lambda$ + 5</p>
<p>= 7(K + $\lambda$) + 5</p>
<p>$\therefore$ Remainer = 5
About this question
Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion
This question is part of PrepWiser's free JEE Main question bank. 193 more solved questions on Binomial Theorem are available — start with the harder ones if your accuracy is >70%.