When digits a two-digit number are interchanged, then it becomes 54 more than the original number. The product of the two digits is 16. What is the original number?

Options

A. 19
B. 44
C. 82
D. 28

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Define the digits
Let the two-digit number be represented as \(10y + x\), where \(y\) is the tens digit and \(x\) is the units digit.

### Step 2: Set up the equation for the interchanged number
When the digits are interchanged, the number becomes \(10x + y\). According to the problem, this new number is 54 more than the original number. Therefore, we can write the equation:
\[ 10x + y = (10y + x) + 54 \]

### Step 3: Simplify the equation
Rearranging the equation gives:
\[ 10x + y - 10y - x = 54 \]
This simplifies to:
\[ 9x - 9y = 54 \]
Dividing the entire equation by 9, we get:
\[ x - y = 6 \quad \text{(Equation 1)} \]

### Step 4: Set up the equation for the product of the digits
We are also given that the product of the two digits is 16:
\[ xy = 16 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations simultaneously
From Equation 1, we can express \(x\) in terms of \(y\):
\[ x = y + 6 \]
Now, substitute this expression for \(x\) into Equation 2:
\[ (y + 6)y = 16 \]
This expands to:
\[ y^2 + 6y - 16 = 0 \]

### Step 6: Factor the quadratic equation
To solve the quadratic equation \(y^2 + 6y - 16 = 0\), we can factor it:
\[ (y + 8)(y - 2) = 0 \]
Thus, the solutions for \(y\) are:
\[ y = -8 \quad \text{(not valid as a digit)} \quad \text{or} \quad y = 2 \]

### Step 7: Find the value of \(x\)
Substituting \(y = 2\) back into the expression for \(x\):
\[ x = 2 + 6 = 8 \]

### Step 8: Form the original number
Now we can form the original two-digit number:
\[ 10y + x = 10(2) + 8 = 20 + 8 = 28 \]

### Conclusion
The original number is **28**.