When the digits of a two -digit number are reversed , then the original number is 63 more than the new number formed and square of units digit of the original number is 63 more than square of tens digit of the original number . Find the original number .

Options

A. 59
B. 94
C. 49
D. 81

Answer

To solve the problem step by step, let's 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. When the digits are reversed, the new number becomes \( 10y + x \).
2. According to the problem, the original number is 63 more than the new number:
\[ (10x + y) = (10y + x) + 63 \]

### Step 2: Simplify the first equation

Rearranging the equation gives:
\[ 10x + y - 10y - x = 63 \]
\[ 9x - 9y = 63 \]
Dividing the entire equation by 9:
\[ x - y = 7 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation

The problem also states that the square of the units digit is 63 more than the square of the tens digit:
\[ y^2 = x^2 - 63 \]
Rearranging gives:
\[ x^2 - y^2 = 63 \]

### Step 4: Factor the second equation

Using the difference of squares:
\[ (x - y)(x + y) = 63 \]

### Step 5: Substitute Equation 1 into the second equation

From Equation 1, we know \( x - y = 7 \). Substitute this into the factored equation:
\[ 7(x + y) = 63 \]
Dividing by 7:
\[ x + y = 9 \quad \text{(Equation 2)} \]

### Step 6: Solve the system of equations

Now we have two equations:
1. \( x - y = 7 \)
2. \( x + y = 9 \)

Adding these two equations:
\[ (x - y) + (x + y) = 7 + 9 \]
\[ 2x = 16 \]
\[ x = 8 \]

### Step 7: Find \( y \)

Substituting \( x = 8 \) back into Equation 2:
\[ 8 + y = 9 \]
\[ y = 1 \]

### Step 8: Form the original number

The original number is:
\[ 10x + y = 10(8) + 1 = 81 \]

### Final Answer

The original number is **81**.

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