To find the unit digit of $7^{95} - 3^{58}$, we can work with the properties of numbers modulo 10. This helps us identify patterns in the units digits of powers of integers. Let's break this down:

Step 1: Find the unit digit of $7^{95}$

We first observe the pattern of the units digits of powers of 7:

7^1 & = 7 \quad \text{(units digit: 7)} \\ 7^2 & = 49 \quad \text{(units digit: 9)} \\ 7^3 & = 343 \quad \text{(units digit: 3)} \\ 7^4 & = 2401 \quad \text{(units digit: 1)} \\ \end{aligned}$$ The units digits repeat every 4 powers: $$7, 9, 3, 1$$. To find $$7^{95} \mod 10$$, we divide 95 by 4 and focus on the remainder: $$95 \div 4 = 23 \text{ remainder } 3$$ So, the units digit of $$7^{95}$$ is the same as that of $$7^3$$, which is 3. ### Step 2: Find the unit digit of $$3^{58}$$ Similarly, we look for the pattern in the units digits of powers of 3: $$\begin{aligned} 3^1 & = 3 \quad \text{(units digit: 3)} \\ 3^2 & = 9 \quad \text{(units digit: 9)} \\ 3^3 & = 27 \quad \text{(units digit: 7)} \\ 3^4 & = 81 \quad \text{(units digit: 1)} \\ \end{aligned}$$ The units digits also repeat every 4 powers: $$3, 9, 7, 1$$. To find $$3^{58} \mod 10$$, we divide 58 by 4 and focus on the remainder: $$58 \div 4 = 14 \text{ remainder } 2$$ So, the units digit of $$3^{58}$$ is the same as that of $$3^2$$, which is 9. ### Step 3: Calculate the units digit of $$7^{95} - 3^{58}$$ Now, we subtract the units digits: $$7^{95} - 3^{58} \equiv 3 - 9 \mod 10 = -6 \mod 10 = 4$$ Thus, the units digit of $$7^{95} - 3^{58}$$ is **4**. Would you like further details or clarifications? ### Related Questions: 1. What is the units digit of $$9^{125}$$? 2. How do we find the units digit of $$5^{100}$$? 3. What is the cyclic pattern of units digits for powers of 6? 4. How does mod arithmetic help in simplifying power-related problems? 5. Can the units digit of $$a^b$$ ever exceed 9? ### Tip: Always look for patterns in the units digits of powers, as they often repeat in cycles, simplifying the problem.