The remainder on dividing 1+3+3^2+3^3++3^2021 by 50 is__ Answer: 4

Question

The remainder on dividing $1+3+3^2+3^3+\cdots+3^{2021}$ by 50 is\_\_

Answer: 4

Accepted Answer

The sum is $\dfrac{3^{2022}-1}{2}$. Using $3^{2022}=(3^4)^{505}	imes9$ and reducing modulo 100 (Euler's theorem, $\phi(50)=20$), the sum modulo 50 works out to 4.

Remainder = 4

Solutions · 1