What is the least natural number which leaves no remainder when divided by all the digits from 1 to 9?

### Concept & Logic The problem asks for the least natural number that is exactly divisible by all the digits from $1$ to $9$. This requires finding the Least Common Multiple (LCM) of the numbers $1, 2, 3, 4, 5, 6, 7, 8,$ and $9$. ### Step-by-Step Solution **Given:** * Set of divisors: $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ **Calculation:** * Step 1: Write down the prime factorization of each number from $1$ to $9$. * $1 = 1$ * $2 = 2^1$ * $3 = 3^1$ * $4 = 2^2$ * $5 = 5^1$ * $6 = 2^1 \times 3^1$ * $7 = 7^1$ * $8 = 2^3$ * $9 = 3^2$ * Step 2: To find the LCM, take the highest power of each prime factor present across all numbers. * Highest power of $2$ is $2^3 = 8$ * Highest power of $3$ is $3^2 = 9$ * Highest power of $5$ is $5^1 = 5$ * Highest power of $7$ is $7^1 = 7$ * Step 3: Multiply these values together to compute the LCM. $$\text{LCM} = 2^3 \times 3^2 \times 5^1 \times 7^1$$ $$\text{LCM} = 8 \times 9 \times 5 \times 7$$ $$\text{LCM} = (8 \times 5) \times (9 \times 7)$$ $$\text{LCM} = 40 \times 63 = 2520$$ ### Exam Strategy & Shortcut Instead of long-form LCM extraction, use divisibility rules to quickly eliminate wrong options: * The number must be divisible by $9$ (sum of digits must be a multiple of $9$). Let's check the options: * (a) $1800 \rightarrow 1+8+0+0 = 9$ (Passes) * (b) $1920 \rightarrow 1+9+2+0 = 12$ (Fails) * (c) $2520 \rightarrow 2+5+2+0 = 9$ (Passes) * (d) $5040 \rightarrow 5+0+4+0 = 9$ (Passes) * The number must be divisible by $8$ (last three digits divisible by $8$). * Check $2520$: The last three digits are $520$. Since $520 \div 8 = 65$, it passes. * The number must be divisible by $7$. * Check $1800$: $1800 \div 7 = 257.14$ (Fails) * Check $2520$: $2520 \div 7 = 360$ (Passes) Since $2520$ is smaller than $5040$, it is the least number. ### Common Pitfall Students often mistake the phrase "all the digits from 1 to 9" and forget to include $8$ or $9$ in their mental math, accidentally computing the LCM up to $6$ or $7$ instead. Always write down the upper boundary explicitly. ### Final Answer **Therefore, the correct answer is 2520.**