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

Options

A. 2520
B. 1260
C. 630
D. 196

Answer

To solve the problem of finding the least number which, when doubled, will be exactly divisible by 12, 18, 21, and 30, we can follow these steps:

### Step 1: Find the LCM of the given numbers
To find the least common multiple (LCM) of 12, 18, 21, and 30, we can use the prime factorization method.

- **Prime Factorization:**
- 12 = 2² × 3¹
- 18 = 2¹ × 3²
- 21 = 3¹ × 7¹
- 30 = 2¹ × 3¹ × 5¹

### Step 2: Determine the highest powers of each prime factor
Now, we take the highest power of each prime factor from the factorizations:

- For 2: the highest power is 2² (from 12)
- For 3: the highest power is 3² (from 18)
- For 5: the highest power is 5¹ (from 30)
- For 7: the highest power is 7¹ (from 21)

### Step 3: Calculate the LCM
Now we can calculate the LCM by multiplying these highest powers together:

\[ LCM = 2² × 3² × 5¹ × 7¹ = 4 × 9 × 5 × 7 \]

Calculating step by step:

- \(4 × 9 = 36\)
- \(36 × 5 = 180\)
- \(180 × 7 = 1260\)

So, the LCM of 12, 18, 21, and 30 is **1260**.

### Step 4: Find the required number
The problem states that the least number, when doubled, should be divisible by the LCM found. Therefore, we need to find the number \(x\) such that:

\[ 2x = 1260 \]

To find \(x\), we divide 1260 by 2:

\[ x = \frac{1260}{2} = 630 \]

### Conclusion
Thus, the least number which, when doubled, will be exactly divisible by 12, 18, 21, and 30 is **630**.

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