The sum of the digits of a 3-digit number is subtracted from the number. The resulting number is always

### Concept & Formula Any generic number can be expanded algebraically into its base-10 place values. This is the key to proving universal divisibility rules for digits. $$ \text{A 3-digit number} = 100x + 10y + z $$ $$ \text{Where } x, y, \text{ and } z \text{ are the hundreds, tens, and units digits respectively.} $$ ### Step-by-Step Solution * **Step 1: Set up the original number.** Let our 3-digit number be represented as $100x + 10y + z$. * **Step 2: Set up the sum of the digits.** The literal sum of its digits is simply $(x + y + z)$. * **Step 3: Perform the subtraction.** The problem states to subtract the digit sum from the original number: * Result $= (100x + 10y + z) - (x + y + z)$ * Combine like terms: Result $= 100x - x + 10y - y + z - z$ * Result $= 99x + 9y$ * **Step 4: Factor the result.** * Result $= 9(11x + y)$ * **Conclusion:** Because the final algebraic expression has a global factor of 9, the resulting number will always be a perfect multiple of 9, regardless of what the original digits were. ### Exam Strategy & Shortcut **Pick a real number:** Don't waste time setting up algebra if you can do quick mental math. Pick any random 3-digit number, for example, 123. Sum of its digits $= 1 + 2 + 3 = 6$. Subtract the sum from the number: $123 - 6 = 117$. Check the options: Is 117 divisible by 9? Check the digit sum of 117: $1 + 1 + 7 = 9$. Yes, it is perfectly divisible by 9. ### Common Pitfall Students sometimes randomly guess divisibility by 6 (option d) because they pick a test number like 111 (Sum = 3. $111 - 3 = 108$. 108 is divisible by both 9 and 6). You must pick a number that breaks the false patterns. 123 yields 117, which is NOT divisible by 6, proving option (d) false and leaving option (b) as the universal truth. ### Final Answer Therefore, the correct answer is **divisible by 9**.