Question

The sum of the digits of a two-digit number is 9. If nine times the number equals twice the number obtained by reversing its digits, the number is

• A. 27
• B. 36
• C. 18
• D. 81

Hint:

Using options is the fastest way to solve this!
Check Option (C) 18:
Sum of digits: \(1 + 8 = 9\) (Correct).
Reversed number: 81.
\(9 \times 18 = 162\).
\(2 \times 81 = 162\).
They match perfectly! This takes less than 10 seconds.

Answer 1

The Correct Option is C

Step 1: Turn the second condition into a ratio right away.
Let the tens digit be $x$ and the units digit be $y$, so the number is $10x + y$ and its reverse is $10y + x$. The condition $9(10x + y) = 2(10y + x)$ expands to \[ 90x + 9y = 20y + 2x \quad \Rightarrow \quad 88x = 11y \quad \Rightarrow \quad 8x = y \] So the digits are in the ratio $x : y = 1 : 8$.
Step 2: Split the digit sum in this ratio.
We are told $x + y = 9$. Since $x : y = 1 : 8$, the total of 9 parts splits into 1 part and 8 parts, one part being $9 \div 9 = 1$.
Step 3: Read off the digits.
\[ x = 1 \times 1 = 1, \qquad y = 8 \times 1 = 8 \]
Step 4: Form the number.
The number is $10x + y = 10 + 8 = 18$.
\[ \boxed{18} \]