The last digit in `(25"_")^(32)and(25"_")^(33)` both is 6. The missing digit is :

Options

A. 2
B. 8
C. 6
D. 4

Answer

To solve the problem of finding the missing digit in \( 25\_^{32} \) and \( 25\_^{33} \) such that both have a last digit of 6, we can follow these steps:

### Step 1: Define the missing digit
Let the missing digit be represented by \( x \). Thus, we can express the numbers as \( 25x \).

### Step 2: Analyze the last digit of \( 25x \)
The last digit of \( 25x \) is determined by the last digit of \( x \) because the last digit of 25 is 5. Therefore, we need to focus on the last digit of \( x \).

### Step 3: Consider the powers
We need to evaluate \( (25x)^{32} \) and \( (25x)^{33} \). The last digit of a number raised to a power depends on the last digit of the base.

### Step 4: Find the last digit for different values of \( x \)
We will analyze the last digits of \( x \) from 0 to 9:

- If \( x = 0 \): Last digit of \( 25x = 0 \) → Last digit of \( 0^{32} = 0 \) and \( 0^{33} = 0 \)
- If \( x = 1 \): Last digit of \( 25x = 5 \) → Last digit of \( 5^{32} = 5 \) and \( 5^{33} = 5 \)
- If \( x = 2 \): Last digit of \( 25x = 0 \) → Last digit of \( 0^{32} = 0 \) and \( 0^{33} = 0 \)
- If \( x = 3 \): Last digit of \( 25x = 5 \) → Last digit of \( 5^{32} = 5 \) and \( 5^{33} = 5 \)
- If \( x = 4 \): Last digit of \( 25x = 0 \) → Last digit of \( 0^{32} = 0 \) and \( 0^{33} = 0 \)
- If \( x = 5 \): Last digit of \( 25x = 5 \) → Last digit of \( 5^{32} = 5 \) and \( 5^{33} = 5 \)
- If \( x = 6 \): Last digit of \( 25x = 6 \) → Last digit of \( 6^{32} = 6 \) and \( 6^{33} = 6 \)
- If \( x = 7 \): Last digit of \( 25x = 5 \) → Last digit of \( 5^{32} = 5 \) and \( 5^{33} = 5 \)
- If \( x = 8 \): Last digit of \( 25x = 0 \) → Last digit of \( 0^{32} = 0 \) and \( 0^{33} = 0 \)
- If \( x = 9 \): Last digit of \( 25x = 5 \) → Last digit of \( 5^{32} = 5 \) and \( 5^{33} = 5 \)

### Step 5: Identify valid values of \( x \)
From the analysis, the only value of \( x \) that gives a last digit of 6 for both \( (25x)^{32} \) and \( (25x)^{33} \) is \( x = 6 \).

### Conclusion
Thus, the missing digit \( x \) is **6**.