21+...Cn23762 Now, (2021)3762 will be divisible by 17 all the terms except the last one for last one. ∴‌‌(2021)3762=17µ−23762 =17µ−22(23760)=17µ−4(16)235 =17µ−4⋅(17−1)235 (17−1)235=(−1)(1−17)235 =−(C0−C117+C2172−...) =−C0+17γ=−1+17γ 17µ−4(17−1)235=17µ−4[−1+17γ] =17(µ−4γ)+4 ∴‌‌(2021)3762=17k+4 Hence, 4 is the remainder.