Let a number of three digits have for its middle digit the sum of the other two digits. Then it is a multiple of

### Concept & Logic A standard three-digit number with digits $a$ (hundreds), $b$ (tens), and $c$ (units) can be expressed algebraically as $100a + 10b + c$. Substituting given digit constraints into this algebraic form reveals its divisibility properties. ### Step-by-Step Solution * Let the three-digit number be represented as $100a + 10b + c$, where $a, b,$ and $c$ are the digits. * **Given constraint:** The middle digit is the sum of the other two digits. $$b = a + c$$ * Substitute $b$ into the original algebraic expansion of the number: $$100a + 10(a + c) + c$$ * Expand and simplify the expression: $$100a + 10a + 10c + c$$ $$110a + 11c$$ * Factor out the common term: $$11(10a + c)$$ * Since the entire number can be expressed as $11$ multiplied by an integer $(10a + c)$, the number must be perfectly divisible by $11$. ### Exam Strategy & Shortcut **Option Elimination via Examples:** Pick any three-digit number satisfying the condition. Example: Let first digit = 1, last digit = 2. Middle digit = $1+2 = 3$. The number is $132$. Test $132$ against options: Not a multiple of $10$ or $50$. $132 / 18$ is not an integer. $132 / 11 = 12$. Option (b) is the only valid choice. ### Common Pitfall Trying to memorize divisibility rules without understanding algebraic expansion. While you might know the rule for $11$ (alternating digit sum), mapping it inversely from the algebraic equation is more robust for abstract questions. ### Final Answer Therefore, the correct answer is **11**.