If the six digit number `15 x 1y 2` is divisible by 44 then (x + y) is equal to :

Options

A. A) 7
B. B) 8
C. C) 6
D. D) 9

Answer

To find the values of \( x \) and \( y \) in the six-digit number \( 15x1y2 \) such that it is divisible by 44, we need to check for divisibility by both 4 and 11.

### Step 1: Check divisibility by 4
A number is divisible by 4 if the number formed by its last two digits is divisible by 4. Here, the last two digits are \( y2 \).

- The possible values for \( y \) are 0 to 9. We will check which values make \( y2 \) divisible by 4:
- \( 02 \div 4 = 0.5 \) (not divisible)
- \( 12 \div 4 = 3 \) (divisible)
- \( 22 \div 4 = 5.5 \) (not divisible)
- \( 32 \div 4 = 8 \) (divisible)
- \( 42 \div 4 = 10.5 \) (not divisible)
- \( 52 \div 4 = 13 \) (divisible)
- \( 62 \div 4 = 15.5 \) (not divisible)
- \( 72 \div 4 = 18 \) (divisible)
- \( 82 \div 4 = 20.5 \) (not divisible)
- \( 92 \div 4 = 23 \) (divisible)

Thus, the possible values for \( y \) that make \( y2 \) divisible by 4 are \( y = 1, 3, 5, 7, 9 \).

### Step 2: Check divisibility by 11
A number is divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is either 0 or a multiple of 11.

For the number \( 15x1y2 \):
- Odd positions: \( 1 + x + y \)
- Even positions: \( 5 + 1 + 2 = 8 \)

We need to set up the equation:
\[ (1 + x + y) - 8 = 0 \quad \text{or} \quad (1 + x + y) - 8 = 11 \]
This simplifies to:
\[ x + y - 7 = 0 \quad \text{or} \quad x + y - 7 = 11 \]
Thus:
1. \( x + y = 7 \)
2. \( x + y = 18 \)

### Step 3: Solve for \( x \) and \( y \)
Now we will check the possible values of \( y \) from step 1 against the equations from step 2.

1. If \( y = 1 \):
- \( x + 1 = 7 \) → \( x = 6 \)
2. If \( y = 3 \):
- \( x + 3 = 7 \) → \( x = 4 \)
3. If \( y = 5 \):
- \( x + 5 = 7 \) → \( x = 2 \)
4. If \( y = 7 \):
- \( x + 7 = 7 \) → \( x = 0 \)
5. If \( y = 9 \):
- \( x + 9 = 7 \) → \( x = -2 \) (not valid)

Now we have the valid pairs:
- \( (x, y) = (6, 1) \) → \( x + y = 7 \)
- \( (x, y) = (4, 3) \) → \( x + y = 7 \)
- \( (x, y) = (2, 5) \) → \( x + y = 7 \)
- \( (x, y) = (0, 7) \) → \( x + y = 7 \)

### Conclusion
In all valid cases, \( x + y = 7 \).

Thus, the final answer is:
\[ \boxed{7} \]