In a two-digit natural number, the digit at th etens place is equal to the square of the digit at units place. If 54 is subtracted from the number, the digits get interchanged . Find the number.

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Define the Variables
Let:
- \( x \) = digit at the unit's place
- \( y \) = digit at the tens place

### Step 2: Formulate the Equations
From the problem statement, we have two key pieces of information:
1. The digit at the tens place is equal to the square of the digit at the units place:
\[ y = x^2 \]
2. If 54 is subtracted from the number, the digits get interchanged. The original number can be expressed as \( 10y + x \). After subtracting 54, the number becomes \( 10x + y \):
\[ 10y + x - 54 = 10x + y \]

### Step 3: Rearranging the Second Equation
Rearranging the second equation gives:
\[ 10y + x - 54 = 10x + y \]
\[ 10y - y + x - 10x = 54 \]
\[ 9y - 9x = 54 \]
Dividing the entire equation by 9:
\[ y - x = 6 \]
So, we have:
\[ y = x + 6 \]

### Step 4: Substitute the Value of \( y \)
Now, we can substitute the value of \( y \) from the first equation into the second:
\[ x^2 = x + 6 \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ x^2 - x - 6 = 0 \]

### Step 6: Factor the Quadratic Equation
We can factor this quadratic equation:
\[ (x - 3)(x + 2) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 3 \quad \text{or} \quad x = -2 \]

### Step 7: Determine Valid Values
Since \( x \) represents a digit, it must be a non-negative integer. Therefore, we discard \( x = -2 \) and keep:
\[ x = 3 \]

### Step 8: Find \( y \)
Now, substitute \( x = 3 \) back into the equation for \( y \):
\[ y = x^2 = 3^2 = 9 \]

### Step 9: Form the Original Number
Now we can find the original two-digit number:
\[ \text{Number} = 10y + x = 10(9) + 3 = 90 + 3 = 93 \]

### Conclusion
The two-digit natural number is:
\[ \boxed{93} \]