If a number of two digits is $k$ times the sum of its digits, then the number formed by interchanging the digits is the sum of the digits multiplied by

### Concept & Logic The fundamental property of two-digit numbers and their reversals. The sum of any two-digit number and the number formed by reversing its digits is always a perfect multiple of 11, specifically: $$ \text{Original} + \text{Reversed} = 11 \times (\text{Sum of Digits}) $$ ### Step-by-Step Solution Let the original two-digit number be $10x + y$, where $x$ and $y$ are its digits. The problem states the number is $k$ times the sum of its digits: $10x + y = k(x + y)$ --- (Equation 1) The number formed by interchanging the digits is $10y + x$. Let this reversed number be $m$ times the sum of its digits: $10y + x = m(x + y)$ --- (Equation 2) Add Equation 1 and Equation 2 together: $(10x + y) + (10y + x) = k(x + y) + m(x + y)$ Group the terms on the left side: $11x + 11y = (k + m)(x + y)$ Factor out 11 on the left: $11(x + y) = (k + m)(x + y)$ Since $(x + y)$ is not zero, we can divide both sides by $(x + y)$: $11 = k + m$ We need to find the multiplier for the interchanged number, which is $m$: $m = 11 - k$ ### Exam Strategy & Shortcut Use the Assumption Method to bypass the abstract algebra entirely. Pick an easy, real two-digit number, say 24. * Sum of digits = 6. * The number 24 is $4 \times 6$, so $k = 4$. * The interchanged number is 42. * $42 = 7 \times 6$, so the new multiplier is 7. Now check the options using $k = 4$ to see which one yields 7: (a) $4 - 1 = 3$ (b) $11 - 4 = 7$ (Matches!) (c) $9 + 4 = 13$ (d) $10 - 4 = 6$ Option (b) is the only correct choice. ### Common Pitfall Getting hopelessly bogged down trying to isolate individual variables $x$ and $y$ algebraically. This leads to a dead end. The trick relies on recognizing the structural relationship of adding the original and reversed numbers. ### Final Answer Therefore, the correct answer is **$11 - k$**.