Find the remainder when \(2^{31}\) is divided by 5.
Answer: C
We look at the pattern of remainders of powers of 2 when divided by 5.
\(2^1 \div 5 \rightarrow rem=2\)
\(2^2 \div 5 \rightarrow rem=4\)
\(2^3 \div 5 \rightarrow rem=3\)
\(2^4 \div 5 \rightarrow rem=1\)
The cycle of remainders is (2, 4, 3, 1), which has a length of 4. To find the remainder of \(2^{31}\), we find the remainder of the power when divided by the cycle length: \(31 \div 4 \rightarrow rem=3\). The required remainder is the 3rd in the cycle, which is 3.