The last digit of `2137^(753)` is

Options

A. 9
B. 7
C. 3
D. 1

Answer

To find the last digit of \( 2137^{753} \), we can simplify our problem by focusing on the last digit of the base, which is \( 7 \). Therefore, we need to find the last digit of \( 7^{753} \).

### Step 1: Identify the pattern of the last digits of powers of 7

Let's calculate the last digits of the first few powers of \( 7 \):

- \( 7^1 = 7 \) (last digit is 7)
- \( 7^2 = 49 \) (last digit is 9)
- \( 7^3 = 343 \) (last digit is 3)
- \( 7^4 = 2401 \) (last digit is 1)

Now, we can see that the last digits repeat every four powers:

- \( 7^1 \) → 7
- \( 7^2 \) → 9
- \( 7^3 \) → 3
- \( 7^4 \) → 1

### Step 2: Determine the position in the cycle for \( 7^{753} \)

Since the last digits repeat every 4 numbers, we can find the position of \( 7^{753} \) in this cycle by calculating \( 753 \mod 4 \):

\[ 753 \div 4 = 188 \quad \text{(remainder 1)} \]

This means:

\[ 753 \mod 4 = 1 \]

### Step 3: Find the last digit based on the remainder

From our earlier calculations, we know:

- If the remainder is 1, the last digit is 7.
- If the remainder is 2, the last digit is 9.
- If the remainder is 3, the last digit is 3.
- If the remainder is 0, the last digit is 1.

Since \( 753 \mod 4 = 1 \), the last digit of \( 7^{753} \) is the same as the last digit of \( 7^1 \), which is 7.

### Conclusion

Thus, the last digit of \( 2137^{753} \) is **7**.

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