Discussion
What is the remainder if \(8^{145}\) is divided by 5?
Answer: C
The remainder is 3.
\(8 \equiv 3 \pmod 5\), and the powers of 3 modulo 5 repeat in a cycle of four: 3, 4, 2, 1.
\[ \begin{aligned} 145 &= 4 \times 36 + 1 \\ 8^{145} &\equiv 3^{1} \equiv 3 \pmod 5 \end{aligned} \]
Equivalently, the units digits of powers of 8 cycle 8, 4, 2, 6, and \(8^{145}\) ends in 8, which leaves 3 on division by 5. 6 cannot be a remainder on division by 5, 0 would need a factor 5, and 4 is the next term of the cycle, an off-by-one.
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