KVPY 2020-21 SA Exam Previous Papers

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Question : 11 of 80

Marks: +1, -0

Let

r

be the remainder when

2021^{2020}

is divided by

2020^{2}

. Then

r

lies between

0 and 5
10 and 15
20 and 100
107 and 120

Solution:

(2021)^{2020}=(1+2020)^{2020}

=2020 C_{0}+2020 C_{1} \.2020

+2020 C_{2}(2020)^{2}+\.......

=1+(2020)^{2}+(2020)^{2} \×\λ

, where

\λ\∈ z^{+}

when divided by

(2020)

remainder will be 1.

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