x and y are 2 different digits. If the sum of the two digit numbers formed by using both the digits is a perfect square, then find `x + y`.

Options

A. 10
B. 11
C. 12
D. 13

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) such that the sum of the two-digit numbers formed by these digits is a perfect square. Let's break down the steps:

### Step 1: Define the two-digit numbers
We can form two two-digit numbers using the digits \( x \) and \( y \):
1. The first number with \( x \) in the tens place and \( y \) in the units place is \( 10x + y \).
2. The second number with \( y \) in the tens place and \( x \) in the units place is \( 10y + x \).

### Step 2: Write the expression for the sum
Now, we can find the sum of these two numbers:
\[ (10x + y) + (10y + x) = 10x + y + 10y + x = 11x + 11y \]
This simplifies to:
\[ 11(x + y) \]

### Step 3: Determine when the sum is a perfect square
We need \( 11(x + y) \) to be a perfect square. For this to happen, \( x + y \) must be such that \( 11(x + y) \) is a perfect square. Since 11 is a prime number, \( x + y \) must be a multiple of 11 for \( 11(x + y) \) to be a perfect square.

### Step 4: Find possible values for \( x + y \)
The only two different digits \( x \) and \( y \) can take values from 0 to 9. The only possible sum \( x + y \) that is a multiple of 11 and can be formed by two different digits is:
\[ x + y = 11 \]

### Step 5: Conclusion
Thus, the value of \( x + y \) is:
\[ \boxed{11} \]