Discussion
What is the remainder when \(3^{21}\) is divided by 5?
Answer: C
We find the cyclicity of remainders of powers of 3 when divided by 5.
\(3^1 \div 5 \rightarrow rem=3\)
\(3^2 \div 5 \rightarrow rem=4\) (from 9)
\(3^3 \div 5 \rightarrow rem=2\) (from 27)
\(3^4 \div 5 \rightarrow rem=1\) (from 81)
The cycle length is 4. We find the remainder of the power \(21\) when divided by 4, which is \(21 \div 4 \rightarrow rem=1\). The required remainder is the 1st in the cycle, which is 3.
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