Find the unit's digit in $(264)^{102} + (264)^{103}$.

### Concept & Logic The unit digit of a number raised to a power depends entirely on the unit digit of its base and the cyclicity of its powers. ### Step-by-Step Solution **Given:** The expression is $(264)^{102} + (264)^{103}$. **Calculation / Deduction:** * We only need to consider the unit digit of the base, which is $4$. Thus, we evaluate the unit digit of $4^{102} + 4^{103}$. * The powers of $4$ follow a simple even/odd pattern: $4^1 = 4$, $4^2 = 16$ (ends in $6$), $4^3 = 64$ (ends in $4$). * Rule: $4^{\text{odd}}$ ends in $4$, and $4^{\text{even}}$ ends in $6$. * Since $102$ is an even power, $4^{102}$ yields a unit digit of $6$. * Since $103$ is an odd power, $4^{103}$ yields a unit digit of $4$. * Adding these together: $6 + 4 = 10$. * The unit digit of $10$ is $0$. ### Exam Strategy & Shortcut Memorize the cyclicity of $4$ and $9$. Instead of dividing the exponent by $4$ (the standard cyclicity method), just look at whether the exponent is even or odd. You can instantly deduce the unit digits as $6$ and $4$, giving a sum ending in $0$. ### Common Pitfall Students often try to calculate standard cyclicity by dividing the large exponents by $4$ to find the remainder. While this works, it wastes valuable time compared to the direct even/odd shortcut for numbers ending in $4$ or $9$. ### Final Answer Therefore, the correct answer is 0.