Solution: A number divisible by both 12 and 15 must be divisible by their least common multiple. Since $ \text{lcm}(12, 15) = 60 $, we are looking for the smallest three-digit number divisible by 60 and ending in 0.

["Title: Finding the Smallest Three-Digit Number Divisible by Both 12 and 15 — Using LCM", "When solving problems involving divisibility by multiple numbers, one powerful mathematical tool is the Least Common Multiple (LCM). This concept helps determine the smallest number satisfying divisibility conditions, especially when working with real-world constraints such as digit patterns or range limits.", "In this article, we explore a key number theory insight: if a number is divisible by both 12 and 15, it must be divisible by their least common multiple. This principle not only simplifies problem-solving but also enables precise calculations in number theory, programming, and applied mathematics.", "---", "### Why Use LCM to Combine Divisibility Conditions?", "A number divisible by both 12 and 15 must be a multiple of LCM(12, 15). The least common multiple is defined as the smallest positive number that both input numbers divide without remainder. Calculating LCM(12, 15) reveals a clear path:", "- Prime factors of 12: $ 2^2 \ imes 3 $\n- Prime factors of 15: $ 3 \ imes 5 $\n- LCM takes the highest powers: $ 2^2 \ imes 3 \ imes 5 = 60 $", "Thus, $ \ ext{lcm}(12, 15) = 60 $. Any number divisible by both 12 and 15 is divisible by 60. This insight becomes invaluable when searching for minimal or constrained numbers, such as the smallest three-digit number divisible by 60 that also ends in 0.", "---", "### The Search: Smallest Three-Digit Number Divisible by 60 Ending in 0", "Three-digit numbers range from 100 to 999. We seek the smallest number in this range divisible by 60 and ending in 0.", "Why must it end in 0?\nNumbers ending in 0 are divisible by 10 — adding the restriction that the number be divisible by 60 imposes additional conditions. Since $ 60 = 6 \ imes 10 $, ending in 0 ensures divisibility by 10, and checking divisibility by 6 ensures evenness and divisibility by 3.", "Instead of checking every multiple of 60, we can directly generate three-digit multiples of 60 ending in 0.", "The smallest three-digit multiple of 60 is:\n$ 60 \ imes 2 = 120 $ → ends in 0 ✅", "This satisfies both conditions: divisible by 60 (hence by 12 and 15), and ends in 0.", "---", "### Verification", "- $ 120 \div 12 = 10 $ → divisible\n- $ 120 \div 15 = 8 $ → divisible\n- $ 120 \div 60 = 2 $ → divisible by LCM(12, 15)\n- Ends in 0 → satisfies additional constraint", "No smaller three-digit multiple of 60 ends in 0, since $ 60 \ imes 1 = 60 $ is only two digits.", "---", "### Conclusion", "This problem illustrates a practical application of the LCM concept: when multiple divisibility rules apply, the smallest solution lies in their LCM. In this case, the smallest three-digit number divisible by both 12 and 15—and ending in 0—is 120 — a direct result of $ \ ext{lcm}(12, 15) = 60 $ and careful checking within digit constraints.", "Such logic supports elegant and efficient solutions in number theory, algorithm design, and real-world problem-solving. Next time you face similar divisibility challenges, remember: start with the LCM, and refine toward your constraints.", "---", "Keywords: LCM, divisibility, three-digit number, divisible by 12 and 15, least common multiple, number theory, solving multiples, mathematical logic, smallest number divisible"]