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 find 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 Least Common Multiple (LCM)
We need to find the LCM of the numbers 12, 18, 21, and 30.

### Step 2: Prime Factorization
Let's perform the prime factorization of each number:
- **12** = 2² × 3¹
- **18** = 2¹ × 3²
- **21** = 3¹ × 7¹
- **30** = 2¹ × 3¹ × 5¹

### Step 3: Identify the Highest Powers of Each Prime
Now, we will take the highest power of each prime number that appears in 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 4: Calculate the LCM
Now we can calculate the LCM using these highest powers:
\[ \text{LCM} = 2² × 3² × 5¹ × 7¹ \]
Calculating this step-by-step:
- \(2² = 4\)
- \(3² = 9\)
- \(5¹ = 5\)
- \(7¹ = 7\)

Now, multiply these together:
\[ \text{LCM} = 4 × 9 = 36 \]
\[ 36 × 5 = 180 \]
\[ 180 × 7 = 1260 \]

### Step 5: Find the Required Number
The least number we are looking for is half of the LCM, since we need the number which when doubled gives us the LCM:
\[ \text{Required Number} = \frac{\text{LCM}}{2} = \frac{1260}{2} = 630 \]

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

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