The digit in the unit place of the number represented by $(7^{95} - 3^{58})$ is

### Concept & Formula When finding the unit digit of a subtraction operation between two exponentiated terms, calculate the unit digits individually using cyclicity. If the first unit digit is smaller than the second, you must "borrow" a $10$, exactly as you do in standard column subtraction. ### Step-by-Step Solution **Given:** Evaluate the unit digit of $(7^{95} - 3^{58})$. **Calculation:** * **Step 1: Find unit digit of $7^{95}$** * Cyclicity of $7$ is $4$. * $95 \pmod 4 = 3$. * Maps to $7^3 = 343$, so the unit digit is $3$. * **Step 2: Find unit digit of $3^{58}$** * Cyclicity of $3$ is $4$. * $58 \pmod 4 = 2$. * Maps to $3^2 = 9$, so the unit digit is $9$. * **Step 3: Perform Subtraction** * We need the unit digit of $(\dots 3) - (\dots 9)$. * Since $3 < 9$, we cannot simply write $-6$. We must borrow from the tens place of the larger overall number. * Borrowing $10$ makes it $13 - 9 = 4$. ### Exam Strategy & Shortcut Memorize the cyclicity charts for $2, 3, 7,$ and $8$ (they all have cycles of $4$). Determine $95 \pmod 4 = 3 \rightarrow 7^3 \rightarrow 3$. Determine $58 \pmod 4 = 2 \rightarrow 3^2 \rightarrow 9$. If the first number is smaller, mentally add $10$ before subtracting: $13 - 9 = 4$. ### Common Pitfall The most dangerous pitfall here is getting $3 - 9$ and just subtracting the smaller from the larger to get $6$, or getting a negative number and not knowing how to handle it. You are subtracting the *entire* large number $3^{58}$ from $7^{95}$, so normal borrowing rules of subtraction apply! ### Final Answer **Therefore, the correct answer is 4.**