A number consists of two digits. The product of these digits is 14. If 45 is subtracated from the number, the digits interchange their places. Find the number.

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Define the digits
Let the two-digit number be represented as \( AB \), where \( A \) is the digit in the tens place and \( B \) is the digit in the units place. Thus, we can express the number as:
\[ \text{Number} = 10A + B \]

### Step 2: Set up the equations
According to the problem, we have two conditions:
1. The product of the digits is 14:
\[ A \times B = 14 \quad \text{(Equation 1)} \]
2. If 45 is subtracted from the number, the digits interchange their places:
\[ 10A + B - 45 = 10B + A \quad \text{(Equation 2)} \]

### Step 3: Simplify Equation 2
Rearranging Equation 2 gives:
\[ 10A + B - A - 10B = 45 \]
This simplifies to:
\[ 9A - 9B = 45 \]
Dividing through by 9:
\[ A - B = 5 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \( A \times B = 14 \) (Equation 1)
2. \( A - B = 5 \) (Equation 3)

From Equation 3, we can express \( A \) in terms of \( B \):
\[ A = B + 5 \]

### Step 5: Substitute into Equation 1
Substituting \( A \) into Equation 1:
\[ (B + 5) \times B = 14 \]
Expanding this gives:
\[ B^2 + 5B = 14 \]
Rearranging it leads to:
\[ B^2 + 5B - 14 = 0 \]

### Step 6: Factor the quadratic equation
Now we need to factor the quadratic equation:
\[ B^2 + 7B - 2B - 14 = 0 \]
Factoring gives:
\[ (B - 2)(B + 7) = 0 \]
Thus, the possible values for \( B \) are:
\[ B = 2 \quad \text{or} \quad B = -7 \]
Since \( B \) must be a digit, we take \( B = 2 \).

### Step 7: Find \( A \)
Substituting \( B = 2 \) back into Equation 3:
\[ A - 2 = 5 \implies A = 7 \]

### Step 8: Form the number
Now we have:
- \( A = 7 \)
- \( B = 2 \)

Thus, the two-digit number is:
\[ \text{Number} = 10A + B = 10 \times 7 + 2 = 72 \]

### Final Answer
The number is \( 72 \).

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