The sum of the squares of a two digit number is 10. If we add 18 to this number we get another number consisting of the same digits written in reverse order. The original number is :

Options

A. 10
B. 46
C. 13
D. none of (a), (b), ©

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
We know two things from the problem:
1. The sum of the squares of the digits is 10:
\[ x^2 + y^2 = 10 \]
2. If we add 18 to the number, we get the number with its digits reversed:
\[ (10x + y) + 18 = 10y + x \]

### Step 2: Rearranging the second equation
Rearranging the second equation gives us:
\[ 10x + y + 18 = 10y + x \]
Subtract \(x\) and \(y\) from both sides:
\[ 10x - x + y - y + 18 = 10y - y \]
This simplifies to:
\[ 9x + 18 = 9y \]
Dividing the entire equation by 9 gives:
\[ x + 2 = y \]
So we have:
\[ y = x + 2 \]

### Step 3: Substitute \(y\) in the first equation
Now, we substitute \(y\) in the first equation:
\[ x^2 + (x + 2)^2 = 10 \]
Expanding \((x + 2)^2\):
\[ x^2 + (x^2 + 4x + 4) = 10 \]
Combining like terms:
\[ 2x^2 + 4x + 4 = 10 \]
Subtracting 10 from both sides:
\[ 2x^2 + 4x - 6 = 0 \]
Dividing the entire equation by 2:
\[ x^2 + 2x - 3 = 0 \]

### Step 4: Factor the quadratic equation
To factor \(x^2 + 2x - 3\):
\[ (x + 3)(x - 1) = 0 \]
Setting each factor to zero gives:
\[ x + 3 = 0 \quad \text{or} \quad x - 1 = 0 \]
Thus:
\[ x = -3 \quad \text{or} \quad x = 1 \]
Since \(x\) must be a digit (0-9), we take \(x = 1\).

### Step 5: Find \(y\)
Using \(y = x + 2\):
\[ y = 1 + 2 = 3 \]

### Step 6: Form the original number
The original two-digit number is:
\[ 10x + y = 10(1) + 3 = 13 \]

### Step 7: Verify the conditions
1. The sum of the squares of the digits:
\[ 1^2 + 3^2 = 1 + 9 = 10 \quad \text{(True)} \]
2. Adding 18 gives:
\[ 13 + 18 = 31 \quad \text{(Reversed digits of 13)} \]

Thus, the original number is **13**.