A two digit number is such that the product of its digits is 12. When 9 is added to the number, the digits interchange their places, find the number :

Options

A. 62
B. 34
C. 26
D. 43

Answer

To solve the problem, we need to find a two-digit number whose digits multiply to 12, and when we add 9 to this number, the digits interchange their places. Let's denote the two-digit number as \(10X + Y\), where \(X\) is the tens digit and \(Y\) is the units digit.

### Step-by-Step Solution:

1. **Set Up the Equations**:
- The product of the digits is given as:
\[ XY = 12 \quad \text{(1)} \]
- When 9 is added to the number, the digits interchange, which gives us:
\[ 10X + Y + 9 = 10Y + X \quad \text{(2)} \]

2. **Rearranging Equation (2)**:
- Rearranging equation (2) gives:
\[ 10X + Y + 9 - X - 10Y = 0 \]
- Simplifying this leads to:
\[ 9X - 9Y + 9 = 0 \]
- Dividing through by 9:
\[ X - Y + 1 = 0 \quad \text{(3)} \]
- Rearranging gives:
\[ X = Y - 1 \quad \text{(4)} \]

3. **Substituting Equation (4) into Equation (1)**:
- Substitute \(X\) from equation (4) into equation (1):
\[ (Y - 1)Y = 12 \]
- Expanding this gives:
\[ Y^2 - Y - 12 = 0 \quad \text{(5)} \]

4. **Factoring Equation (5)**:
- We need to factor the quadratic equation:
\[ Y^2 - Y - 12 = (Y - 4)(Y + 3) = 0 \]
- Setting each factor to zero gives:
\[ Y - 4 = 0 \quad \Rightarrow \quad Y = 4 \]
\[ Y + 3 = 0 \quad \Rightarrow \quad Y = -3 \quad \text{(not valid since Y must be a digit)} \]

5. **Finding \(X\)**:
- Now substituting \(Y = 4\) back into equation (4):
\[ X = Y - 1 = 4 - 1 = 3 \]

6. **Finding the Two-Digit Number**:
- The two-digit number is:
\[ 10X + Y = 10(3) + 4 = 30 + 4 = 34 \]

### Conclusion:
The two-digit number we are looking for is **34**.