The product of digits of a 2-digit number is 24. If we add 45 to the number, the new number obtained is a number formed by interchanging the digits. What is the original number?

Options

A. 54
B. 83
C. 38
D. 45

Answer

To solve the problem step by step, we will denote the two-digit number as \(10x + y\), where \(x\) is the tens digit and \(y\) is the units digit.

### Step 1: Set up the equations based on the problem statement.
1. The product of the digits is given as:
\[ x \cdot y = 24 \quad \text{(Equation 1)} \]
2. Adding 45 to the original number results in the number formed by interchanging the digits:
\[ 10x + y + 45 = 10y + x \quad \text{(Equation 2)} \]

### Step 2: Simplify Equation 2.
Rearranging Equation 2:
\[ 10x + y + 45 = 10y + x \]
Subtract \(x\) and \(y\) from both sides:
\[ 10x - x + y - y + 45 = 10y - y \]
This simplifies to:
\[ 9x + 45 = 9y \]
Dividing the entire equation by 9:
\[ x + 5 = y \quad \text{(Equation 3)} \]

### Step 3: Substitute Equation 3 into Equation 1.
From Equation 3, we can express \(y\) in terms of \(x\):
\[ y = x + 5 \]
Now substitute \(y\) in Equation 1:
\[ x(x + 5) = 24 \]
Expanding this gives:
\[ x^2 + 5x - 24 = 0 \]

### Step 4: Solve the quadratic equation.
To solve the quadratic equation \(x^2 + 5x - 24 = 0\), we can factor it:
\[ (x + 8)(x - 3) = 0 \]
Setting each factor to zero gives us:
\[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \quad \text{(not valid since } x \text{ must be a digit)} \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Step 5: Find the value of \(y\).
Substituting \(x = 3\) back into Equation 3:
\[ y = 3 + 5 = 8 \]

### Step 6: Form the original number.
The original two-digit number is:
\[ 10x + y = 10(3) + 8 = 30 + 8 = 38 \]

### Conclusion:
The original number is **38**.