Step 1: Understand the problem.
We are given a two-digit number where the unit digit multiplied by 3 is equal to three more than the sum of the digits. Additionally, when the digits are reversed, the resulting number is 18 less than the original number.
Let the two-digit number be represented as \( 10a + b \), where \( a \) is the tens digit and \( b \) is the units digit.
Step 2: Set up the equations based on the conditions.
1. The unit digit multiplied by 3 is equal to three more than the sum of the digits:
\( 3b = a + b + 3 \)
Simplifying the equation:
\( 3b - b = a + 3 \)
\( 2b = a + 3 \)
\( a = 2b - 3 \) (Equation 1)
2. When the digits are reversed, the resulting number is 18 less than the original number:
\( 10b + a = 10a + b - 18 \)
Simplifying the equation:
\( 10b - b = 10a - a - 18 \)
\( 9b = 9a - 18 \)
\( b = a - 2 \) (Equation 2)
Step 3: Solve the system of equations.
Substitute \( a = 2b - 3 \) from Equation 1 into Equation 2:
\( b = (2b - 3) - 2 \)
Simplifying:
\( b = 2b - 5 \)
\( b - 2b = -5 \)
\( -b = -5 \)
\( b = 5 \)
Step 4: Conclusion.
The units digit of the original number is 5.
Final Answer:
The correct option is (B): 5.