A 2-digit number is such that the product of the digit is 20. If 9 is subtracted from the number, the digits interchange their places. Find the number.

Answer

To solve the problem step by step, let's define the variables and set up the equations based on the information provided.

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

### Step 2: Set up the equations
From the problem, we have two conditions:
1. The product of the digits is 20:
\[ xy = 20 \]
2. If 9 is subtracted from the number, the digits interchange their places:
\[ 10x + y - 9 = 10y + x \]

### Step 3: Rearranging the second equation
Rearranging the second equation gives:
\[ 10x + y - 9 = 10y + x \]
\[ 10x - x + y - 10y = 9 \]
\[ 9x - 9y = 9 \]
Dividing the entire equation by 9:
\[ x - y = 1 \]
This simplifies to:
\[ x = y + 1 \]

### Step 4: Substitute \(x\) in the product equation
Now substitute \(x\) from the equation \(x = y + 1\) into the product equation \(xy = 20\):
\[ (y + 1)y = 20 \]
Expanding this gives:
\[ y^2 + y - 20 = 0 \]

### Step 5: Factor the quadratic equation
Now we need to factor the quadratic equation:
\[ y^2 + y - 20 = 0 \]
To factor, we look for two numbers that multiply to \(-20\) and add to \(1\). The numbers \(5\) and \(-4\) work:
\[ (y + 5)(y - 4) = 0 \]

### Step 6: Solve for \(y\)
Setting each factor to zero gives us:
\[ y + 5 = 0 \quad \Rightarrow \quad y = -5 \quad (\text{not valid since } y \text{ must be a digit}) \]
\[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \]

### Step 7: Find \(x\)
Now substitute \(y = 4\) back into the equation \(x = y + 1\):
\[ x = 4 + 1 = 5 \]

### Step 8: Find the two-digit number
Now we can find the two-digit number:
\[ 10x + y = 10(5) + 4 = 50 + 4 = 54 \]

### Final Answer
The two-digit number is \(54\).

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