Binomial Theorem Part 1

Given, when x is divided by 4 , the remainder is 3 .
Let x=4p+3, then
(2020+x)2022=‌‌(2020+4p+3)2022
=‌‌(2024+4p−1)2022
=‌‌(4k−1)2022(∵2024‌ is divisible by 4) ‌
=‌‌‌2022C0(4K)2022(−1)0+‌2022C1(4K)2021
‌(−1)1+....+‌2022C2022(4A)0(−1)2022
On expansion (2020+x)2022, we get the form of 8λ+1. Since, each terms have 2022 and 4k, so if we take 2 common from 2022 we get 8. Thus, each term have 8 in common.
Hence, remainder is 1.