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

Step 1: Find the cyclicity of powers of 2 modulo 7.
\(2^1 = 2\), \(2^2 = 4\), \(2^3 = 8 \equiv 1 \pmod 7\). The pattern repeats every 3 powers.
Step 2: Reduce the exponent.
\(31 = 3 \times 10 + 1\), so \(2^{31} \equiv 2^{1} \pmod 7\).
Step 3: Compute.
\[2^{31} \equiv 2 \pmod 7\] The remainder is \(\boxed{2}\).