The digit in the units place of the number represented by `(7^(95) - 3^(58))` is:

Options

A. 6
B. 7
C. 0
D. 4

Answer

To find the digit in the units place of the expression \( (7^{95} - 3^{58}) \), we can follow these steps:

### Step 1: Determine the units digit of \( 7^{95} \)

The units digits of powers of 7 follow a repeating cycle:
- \( 7^1 \) has a units digit of 7
- \( 7^2 \) has a units digit of 9
- \( 7^3 \) has a units digit of 3
- \( 7^4 \) has a units digit of 1

This cycle (7, 9, 3, 1) repeats every 4 terms. To find the units digit of \( 7^{95} \), we can calculate \( 95 \mod 4 \):

\[ 95 \div 4 = 23 \quad \text{(remainder 3)} \]

Thus, \( 95 \mod 4 = 3 \). According to our cycle, the units digit of \( 7^{95} \) corresponds to the units digit of \( 7^3 \), which is 3.

### Step 2: Determine the units digit of \( 3^{58} \)

The units digits of powers of 3 also follow a repeating cycle:
- \( 3^1 \) has a units digit of 3
- \( 3^2 \) has a units digit of 9
- \( 3^3 \) has a units digit of 7
- \( 3^4 \) has a units digit of 1

This cycle (3, 9, 7, 1) also repeats every 4 terms. To find the units digit of \( 3^{58} \), we calculate \( 58 \mod 4 \):

\[ 58 \div 4 = 14 \quad \text{(remainder 2)} \]

Thus, \( 58 \mod 4 = 2 \). According to our cycle, the units digit of \( 3^{58} \) corresponds to the units digit of \( 3^2 \), which is 9.

### Step 3: Calculate the units digit of \( (7^{95} - 3^{58}) \)

Now that we have the units digits:
- Units digit of \( 7^{95} \) is 3
- Units digit of \( 3^{58} \) is 9

We can now perform the subtraction:

\[ 3 - 9 = -6 \]

Since we need the units digit, we can convert this negative result into a positive one by adding 10:

\[ -6 + 10 = 4 \]

### Final Answer

The digit in the units place of \( (7^{95} - 3^{58}) \) is **4**.

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