Medium INTEGER +4 / -1 PYQ · JEE Mains 2021

If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)2022 is divided by 8 is __________.

Answer (integer) 1

Solution

Let x = 4k + 3<br><br>(2020 + x)<sup>2022</sup><br><br>= (2020 + 4k + 3)<sup>2022</sup><br><br>= (4(505) + 4k + 3)<sup>2022</sup> <br><br>= (4P + 3)<sup>2022</sup><br><br>= (4P + 4 $-$ 1)<sup>2022</sup><br><br>= (4A $-$ 1)<sup>2022</sup><br><br><sup>2022</sup>C<sub>0</sub>(4A)<sup>0</sup>($-$1)<sup>2022</sup> + <sup>2022</sup>C<sub>1</sub>(4A)<sup>1</sup>($-$1)<sup>2021</sup> + ...... <br><br>= 1 + 2022(4A)(-1) + ..... <br><br>= 1 + 8$\lambda$<br><br>$\therefore$ Reminder is 1.

About this question

Subject: Mathematics · Chapter: Binomial Theorem · Topic: Binomial Expansion

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