What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?

### Concept & Formula Let the required number be $x$. The problem states that when $x$ is doubled ($2x$), it becomes exactly divisible by $12, 18, 21,$ and $30$. The smallest number exactly divisible by a set of numbers is their Least Common Multiple (LCM). Therefore, we have the relationship: $$2x = \text{LCM of } (12, 18, 21, 30)$$ $$x = \frac{\text{LCM of } (12, 18, 21, 30)}{2}$$ ### Step-by-Step Solution **Given:** * Divisors: $12, 18, 21, 30$ **Calculation:** * Step 1: Find the prime factorization of each divisor. * $12 = 2^2 \times 3$ * $18 = 2 \times 3^2$ * $21 = 3 \times 7$ * $30 = 2 \times 3 \times 5$ * Step 2: Collect the highest power of each prime factor to find the LCM. * Highest power of $2 = 2^2$ * Highest power of $3 = 3^2$ * Highest power of $5 = 5^1$ * Highest power of $7 = 7^1$ $$\text{LCM} = 2^2 \times 3^2 \times 5 \times 7$$ $$\text{LCM} = 4 \times 9 \times 5 \times 7 = 1260$$ * Step 3: Since this LCM represents the *doubled* value of our target number, divide it by $2$. $$x = \frac{1260}{2} = 630$$ ### Exam Strategy & Shortcut Use option elimination by tracking the condition "when doubled". Double the options and check basic divisibility rules (like divisibility by $10$ or $9$): * Option (b) is $630$. When doubled, it becomes $1260$. * $1260$ ends in $0$, so it is divisible by $30$. * Sum of digits is $1+2+6+0 = 9$, so it is divisible by $18$. * $1260 \div 12 = 105$ (Passes) * $1260 \div 21 = 60$ (Passes) This satisfies all parameters instantly without executing a complete prime factorization tree. ### Common Pitfall The most frequent mistake is finding the LCM ($1260$) and matching it directly to option (c) without dividing by $2$. Always double check if you have performed the inverse operation required by the wording "when doubled". ### Final Answer **Therefore, the correct answer is 630.**