The digit in the unit's place of the number $123^{99}$ is

### Concept & Logic The unit digit of an exponentiation $a^b$ depends solely on the unit digit of the base $a$ and the cyclicity of its powers. ### Step-by-Step Solution **Given:** Find the unit digit of $123^{99}$. **Calculation:** * We only care about the unit digit of the base, which is $3$. The problem simplifies to finding the unit digit of $3^{99}$. * The unit digits of powers of $3$ cycle every $4$ steps: $3, 9, 7, 1$. * Find the remainder when the exponent $99$ is divided by the cyclicity $4$: $$99 \pmod 4 = 3$$ * Since the remainder is $3$, we map this to the $3$rd step of our cycle, which is equivalent to $3^3$. * $3^3 = 27$, so the unit digit is $7$. ### Exam Strategy & Shortcut Quickly divide the exponent by $4$: $100$ is perfectly divisible by $4$, so $99$ is $1$ less, meaning the remainder is $3$. The $3$rd power of $3$ ends in $7$. ### Common Pitfall Students sometimes mistake a remainder of $0$ to mean the $0$th power (which would equal $1$). While correct for base $3$, the rule is that a remainder of $0$ corresponds to the $4$th item in the cycle. Always be careful to map remainders correctly ($1 \rightarrow 1$st, $2 \rightarrow 2$nd, $3 \rightarrow 3$rd, $0 \rightarrow 4$th). ### Final Answer **Therefore, the correct answer is 7.**