The remainder when (2023)2023 is divided by 35 is \_\_\_\_\_\_\_\_.
Answer: 7
Solutions · 1
Official solution
\[(2023)^{2023}=(2030-7)^{2023}=(35K-7)^{2023}=\sum_{i=0}^{2023}{}^{2023}C_i(35K)^{2023-i}(-7)^i\]
All terms except the last are multiples of 35: (2023)2023=35N+(−7)2023=35N−72023.
\[-7^{2023}=-7\cdot7^{2022}=-7(7^2)^{1011}=-7(49)^{1011}=-7(50-1)^{1011}\]
\[=-7\left[{}^{1011}C_0\,50^{1011}-{}^{1011}C_1\,50^{1010}+\cdots-{}^{1011}C_{1011}\right]=-7(5\lambda-1)=-35\lambda+7\]
So (2023)2023=35N−35λ+7=35M+7. Remainder = 7