Question

What is the remainder when \(2^{31}\) is divided by 7?

• A. 1
• B. 2
• C. 4
• D. 5

Hint:

Use the cyclicity of \(2^n \bmod 7\), which has period 3.

Answer 1

The Correct Option is B

Write $2^{31} = (2^3)^{10}\times 2 = 8^{10}\times 2$. Since $8 \equiv 1 \pmod 7$, we get $8^{10} \equiv 1$, so $2^{31} \equiv 1 \times 2 = 2 \pmod 7$.