If in a three-digit number the last two digits places are interchanged, a new number is formed which is greater than the original number by 45. What is the difference between the last two digits of that number ?

Options

A. 5
B. 8
C. 6
D. 9

Answer

To solve the problem step by step, we will define the three-digit number and analyze the situation based on the information given.

### Step 1: Define the three-digit number
Let the three-digit number be represented as \( 100x + 10y + z \), where:
- \( x \) is the hundreds place digit,
- \( y \) is the tens place digit,
- \( z \) is the units place digit.

### Step 2: Interchange the last two digits
When the last two digits are interchanged, the new number becomes \( 100x + 10z + y \).

### Step 3: Set up the equation based on the problem statement
According to the problem, the new number is greater than the original number by 45. Therefore, we can set up the equation:
\[ (100x + 10z + y) - (100x + 10y + z) = 45 \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 100x + 10z + y - 100x - 10y - z = 45 \]
This simplifies to:
\[ 10z + y - 10y - z = 45 \]
Combining like terms gives:
\[ 9z - 9y = 45 \]

### Step 5: Factor out the common term
We can factor out 9 from the left side:
\[ 9(z - y) = 45 \]

### Step 6: Solve for the difference between the last two digits
Now, divide both sides by 9:
\[ z - y = 5 \]

### Conclusion
The difference between the last two digits \( z \) and \( y \) of the original three-digit number is 5.