Discussion

What is the remainder when \(2^{31}\) is divided by 5?

Answer: C

We can use the concept of cyclicity of remainders.

\(2^1 \div 5\) gives remainder 2.

\(2^2 \div 5\) gives remainder 4.

\(2^3 \div 5\) gives remainder 3.

\(2^4 \div 5\) gives remainder 1.

\(2^5 \div 5\) gives remainder 2.

The cycle of remainders is (2, 4, 3, 1), which has a length of 4.

To find the remainder of \(2^{31} \div 5\), we divide the power 31 by the cyclicity 4.

\(31 \div 4\) gives a remainder of 3.

The remainder will be the 3rd term in the cycle, which is 3.

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