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, we need to find a two-digit number where the ten's digit is larger than the unit's digit, the product of the digits is 27, and the difference between the digits is 6. Let's denote the ten's digit as \( A \) and the unit's digit as \( B \).

### Step 1: Set up the equations
From the problem, we have the following equations based on the information given:
1. The product of the digits:
\[ A \times B = 27 \quad \text{(1)} \]
2. The difference between the digits:
\[ A - B = 6 \quad \text{(2)} \]

### Step 2: Express \( A \) in terms of \( B \)
From equation (2), we can express \( A \) in terms of \( B \):
\[ A = B + 6 \quad \text{(3)} \]

### Step 3: Substitute \( A \) in equation (1)
Now, we substitute equation (3) into equation (1):
\[ (B + 6) \times B = 27 \]
Expanding this gives:
\[ B^2 + 6B = 27 \]

### Step 4: Rearrange the equation
Rearranging the equation to standard quadratic form:
\[ B^2 + 6B - 27 = 0 \quad \text{(4)} \]

### Step 5: Factor the quadratic equation
Next, we need to factor the quadratic equation (4). We look for two numbers that multiply to \(-27\) and add to \(6\). The numbers \(9\) and \(-3\) work:
\[ (B + 9)(B - 3) = 0 \]

### Step 6: Solve for \( B \)
Setting each factor to zero gives us:
1. \( B + 9 = 0 \) → \( B = -9 \) (not valid since \( B \) must be a digit)
2. \( B - 3 = 0 \) → \( B = 3 \)

### Step 7: Find \( A \)
Now that we have \( B = 3 \), we can find \( A \) using equation (3):
\[ A = B + 6 = 3 + 6 = 9 \]

### Step 8: Write the two-digit number
The two-digit number is formed by \( A \) and \( B \):
\[ \text{The number is } 93 \]

### Final Answer
The two-digit number is **93**.

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