When a certain number is multiplied by 18, the product consists entirely of 2's. What is the minimum number of 2's in the product?

### Concept & Logic The question asks for the smallest number of the form $2222\dots$ that is exactly divisible by 18. Since 18 is composed of co-prime factors 2 and 9 ($18 = 2 \times 9$), the number must satisfy the divisibility rules for both 2 and 9. ### Step-by-Step Solution - **Calculation / Deduction:** A number consisting entirely of 2s is naturally even, so it is always divisible by 2. We only need to focus on the rule for 9. - For a number to be divisible by 9, the sum of its digits must be a multiple of 9. - Let $n$ be the number of 2s. The sum of the digits is $2 \times n$, or $2n$. - We need the smallest integer $n$ such that $2n$ is a multiple of 9. - Testing multiples: 2, 4, 6, 8, 10, 12, 14, 16, 18... - The smallest multiple of 9 in this sequence is 18. - So, $2n = 18 \implies n = 9$. - The product requires exactly nine 2s (which is 222222222). ### Exam Strategy & Shortcut Do not perform long division of $22222\dots$ by 18! Use divisibility rules. Since 18 requires divisibility by 9, just ask yourself: "How many 2s do I need to add together to reach 9, 18, or 27?" Nine 2s equal 18, instantly giving you the answer. ### Common Pitfall Wasting precious exam time attempting to set up a long division of 22222... by 18. This brute-force method is prone to counting errors and takes significantly longer than applying the rule of 9. ### Final Answer Therefore, the correct answer is **9**.