The sum of a number consisting of two digits and the number formed by interchanging the digits is always divisible by

### Concept & Formula To solve digit-manipulation problems, represent the two-digit number algebraically using its base-10 place values. $$ \text{A two-digit number} = 10x + y $$ ### Step-by-Step Solution * **Step 1: Define the primary number.** Let the original two-digit number have a tens digit $x$ and units digit $y$. Its value is $10x + y$. * **Step 2: Define the interchanged number.** When the digits swap places, $y$ becomes the tens digit and $x$ becomes the units digit. The new value is $10y + x$. * **Step 3: Calculate their sum.** * Sum = $(10x + y) + (10y + x)$ * Combine like terms: $= 11x + 11y$ * **Step 4: Factor the expression.** * $= 11(x + y)$ * **Conclusion:** The algebraic sum is $11$ multiplied by the sum of the digits. Therefore, this total will always be a multiple of 11. ### Exam Strategy & Shortcut **Pick a real number and test it:** Let's choose a simple two-digit number, like 23. Reverse the digits to get 32. Add them together: $23 + 32 = 55$. Check the options: 55 is clearly divisible by 11. It is not divisible by 7, 9, or 10. You can verify the theoretical rule with practical numbers instantly. ### Common Pitfall The main danger here is mixing up this rule with the subtraction counterpart. * Sum of reversed numbers = divisible by 11. * Difference of reversed numbers = divisible by 9. Keeping these two properties distinctly memorized saves massive amounts of time on standard aptitude tests. ### Final Answer Therefore, the correct answer is **11**.