To solve the problem, we need to analyze the two-digit number \( x \) whose digits differ by 4. We will find the positive difference between the squares of the digits of \( x \) and compare it to 15.
### Step-by-Step Solution:
1. **Define the digits of the two-digit number \( x \)**:
Let the two digits be \( a \) and \( b \). Since \( x \) is a two-digit number, we can express it as \( x = 10a + b \), where \( a \) is the tens digit and \( b \) is the units digit.
2. **Set up the condition for the digits**:
The problem states that the digits differ by 4. This gives us two possible equations:
\[
a - b = 4 \quad \text{(1)}
\]
or
\[
b - a = 4 \quad \text{(2)}
\]
3. **Find possible pairs of digits**:
From equation (1):
- If \( a - b = 4 \), then \( a = b + 4 \).
- The possible values for \( b \) (which must be a digit from 0 to 9) are \( 0, 1, 2, 3, 4, 5 \). Thus, the pairs are:
- \( (4, 0) \) → \( 40 \)
- \( (5, 1) \) → \( 51 \)
- \( (6, 2) \) → \( 62 \)
- \( (7, 3) \) → \( 73 \)
- \( (8, 4) \) → \( 84 \)
- \( (9, 5) \) → \( 95 \)
From equation (2):
- If \( b - a = 4 \), then \( b = a + 4 \).
- The possible values for \( a \) (which must be a digit from 1 to 9) are \( 1, 2, 3, 4, 5 \). Thus, the pairs are:
- \( (1, 5) \) → \( 15 \)
- \( (2, 6) \) → \( 26 \)
- \( (3, 7) \) → \( 37 \)
- \( (4, 8) \) → \( 48 \)
- \( (5, 9) \) → \( 59 \)
4. **Calculate the positive difference between the squares of the digits**:
For each pair of digits \( (a, b) \), we calculate \( |a^2 - b^2| \):
- For \( (4, 0) \): \( |4^2 - 0^2| = |16 - 0| = 16 \)
- For \( (5, 1) \): \( |5^2 - 1^2| = |25 - 1| = 24 \)
- For \( (6, 2) \): \( |6^2 - 2^2| = |36 - 4| = 32 \)
- For \( (7, 3) \): \( |7^2 - 3^2| = |49 - 9| = 40 \)
- For \( (8, 4) \): \( |8^2 - 4^2| = |64 - 16| = 48 \)
- For \( (9, 5) \): \( |9^2 - 5^2| = |81 - 25| = 56 \)
- For \( (1, 5) \): \( |1^2 - 5^2| = |1 - 25| = 24 \)
- For \( (2, 6) \): \( |2^2 - 6^2| = |4 - 36| = 32 \)
- For \( (3, 7) \): \( |3^2 - 7^2| = |9 - 49| = 40 \)
- For \( (4, 8) \): \( |4^2 - 8^2| = |16 - 64| = 48 \)
- For \( (5, 9) \): \( |5^2 - 9^2| = |25 - 81| = 56 \)
5. **Compare the results with 15**:
- The positive difference between the squares of the digits for the pairs we calculated is always greater than 15. For example, the smallest value we found is 16 (from the pair (4, 0)).
### Conclusion:
The positive difference between the squares of the digits of \( x \) is always greater than 15. Therefore, Column A is greater than Column B.
**Final Answer**: Column A > Column B.
