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 4k1 so if we take 2 common from 2022 we get 8 . Thus, each term have 8 in common. Hence, remainder is 1.