In a two digit number, the ten's digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.

Answer

To solve the problem step by step, we will define the variables and set up equations based on the information given in the question.

### Step 1: Define the Variables
Let:
- \( x \) = the digit in the tens place
- \( y \) = the digit in the ones place

### Step 2: Set Up the Equations
From the problem, we have the following information:
1. The tens digit is bigger than the ones digit: \( x > y \)
2. The product of the digits is 27: \( x \cdot y = 27 \)
3. The difference between the two digits is 6: \( x - y = 6 \)

### Step 3: Express \( y \) in Terms of \( x \)
From the equation \( x - y = 6 \), we can express \( y \) as:
\[ y = x - 6 \]

### Step 4: Substitute \( y \) in the Product Equation
Now, substitute \( y \) in the product equation \( x \cdot y = 27 \):
\[ x \cdot (x - 6) = 27 \]

### Step 5: Expand and Rearrange the Equation
Expanding the equation gives:
\[ x^2 - 6x = 27 \]
Now, rearranging it to form a standard quadratic equation:
\[ x^2 - 6x - 27 = 0 \]

### Step 6: Factor the Quadratic Equation
Next, we will factor the quadratic equation \( x^2 - 6x - 27 = 0 \). We need two numbers that multiply to \(-27\) and add to \(-6\). The numbers are \(-9\) and \(3\):
\[ (x - 9)(x + 3) = 0 \]

### Step 7: Solve for \( x \)
Setting each factor to zero gives us:
1. \( x - 9 = 0 \) → \( x = 9 \)
2. \( x + 3 = 0 \) → \( x = -3 \) (not valid since \( x \) must be a digit)

Thus, the only valid solution for \( x \) is:
\[ x = 9 \]

### Step 8: Find \( y \)
Now, substitute \( x \) back into the equation for \( y \):
\[ y = x - 6 = 9 - 6 = 3 \]

### Step 9: Form the Two-Digit Number
The two-digit number can be formed as:
\[ \text{Number} = 10x + y = 10 \cdot 9 + 3 = 90 + 3 = 93 \]

### Final Answer
The two-digit number is:
\[ \boxed{93} \]

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