The sum of the digits of a two-digit number is 10. The number formed by reversing the digits is 18 less than the original number. Find the original number.

Answer

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

### Step 1: Set up the equations based on the problem statement.
1. The sum of the digits is 10:
\[ x + y = 10 \quad \text{(Equation 1)} \]

2. The number formed by reversing the digits is 18 less than the original number:
\[ 10y + x = (10x + y) - 18 \quad \text{(Equation 2)} \]

### Step 2: Simplify Equation 2.
Rearranging Equation 2 gives:
\[ 10y + x + 18 = 10x + y \]
\[ 10y - y + x + 18 = 10x \]
\[ 9y + 18 = 9x \]
Dividing the entire equation by 9:
\[ y + 2 = x \quad \text{(Equation 3)} \]

### Step 3: Substitute Equation 3 into Equation 1.
Now, substitute \(x\) from Equation 3 into Equation 1:
\[ y + (y + 2) = 10 \]
\[ 2y + 2 = 10 \]
Subtracting 2 from both sides:
\[ 2y = 8 \]
Dividing by 2:
\[ y = 4 \]

### Step 4: Find \(x\) using the value of \(y\).
Now substitute \(y = 4\) back into Equation 3:
\[ x = y + 2 = 4 + 2 = 6 \]

### Step 5: Form the original number.
The original two-digit number can be formed as:
\[ 10y + x = 10(4) + 6 = 40 + 6 = 46 \]

### Conclusion:
The original number is **46**.

---