A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Explanation

To solve the problem step by step, we will follow the information provided in the question and use algebraic equations.

### Step 1: Define the digits
Let the two-digit number be represented as \( 10X + Y \), where \( X \) is the tens digit and \( Y \) is the units digit.

### Step 2: Set up the equations
From the problem, we know:
1. The product of the digits is 35:
\[ X \cdot Y = 35 \]
2. If 18 is added to the number, the digits interchange their places:
\[ 10X + Y + 18 = 10Y + X \]

### Step 3: Rearrange the second equation
Rearranging the second equation gives:
\[ 10X + Y + 18 = 10Y + X \]
Subtract \( X \) and \( Y \) from both sides:
\[ 10X - X + Y - Y + 18 = 10Y - Y \]
This simplifies to:
\[ 9X - 9Y + 18 = 0 \]
Dividing the entire equation by 9:
\[ X - Y + 2 = 0 \]
Thus, we can express \( X \) in terms of \( Y \):
\[ X = Y - 2 \]

### Step 4: Substitute \( X \) in the product equation
Now substitute \( X \) in the product equation:
\[ (Y - 2) \cdot Y = 35 \]
Expanding this gives:
\[ Y^2 - 2Y = 35 \]
Rearranging leads to:
\[ Y^2 - 2Y - 35 = 0 \]

### Step 5: Factor the quadratic equation
To factor \( Y^2 - 2Y - 35 \), we look for two numbers that multiply to -35 and add to -2. The numbers are -7 and 5:
\[ (Y - 7)(Y + 5) = 0 \]
Setting each factor to zero gives:
1. \( Y - 7 = 0 \) → \( Y = 7 \)
2. \( Y + 5 = 0 \) → \( Y = -5 \) (not valid since \( Y \) must be a digit)

### Step 6: Find \( X \)
Now that we have \( Y = 7 \), we can find \( X \):
\[ X = Y - 2 = 7 - 2 = 5 \]

### Step 7: Form the two-digit number
The two-digit number is:
\[ 10X + Y = 10 \cdot 5 + 7 = 50 + 7 = 57 \]

### Step 8: Verification
1. The product of the digits \( 5 \cdot 7 = 35 \) (correct).
2. Adding 18 to the number: \( 57 + 18 = 75 \), which has digits interchanged (correct).

Thus, the two-digit number is **57**.

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