What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?
A.196
B.630
C.1260
D.2520

Hint: Here we will first find the LCM of these numbers and for finding the LCM, we will first find the factors by first finding the factors of the given numbers individually. Then we will divide the obtained LCM by 2 to get the required number.

Complete step-by-step answer:
Here we need to find the least number which when doubled will be completely divisible by the given numbers and the given numbers are 12, 18, 21 and 30.
We will now first find the factors of the given numbers.
Factors of the number 12 \[ = 2 \times 2 \times 3\]
Factors of the number 18 \[ = 2 \times 3 \times 3\]
Factors of the number 21 \[ = 3 \times 7\]
Factors of the number 30 \[ = 2 \times 2 \times 3 \times 5\]
Now, we know that LCM of these will be the least number which will be the multiple of all these numbers.
Therefore, LCM of 12, 18, 21 and 30 \[ = 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 1260\]
But here, we need the least number which when doubled will be completely divisible by the given numbers. So for that, we will divide the obtained LCM by 2.
Therefore
Required number \[ = \dfrac{{LCM}}{2}\]
Substituting \[LCM = 1260\] in the above equation, we get
\[ \Rightarrow \] Required number \[ = \dfrac{{1260}}{2} = 630\]
Hence, the required least number is equal to 630.
Thus, the correct option is option B.

Note: Here, we have divided the obtained LCM by 2 because this will give us the number which when multiplied by 2 will give us the number divisible by all the given numbers. The Least Common Multiple (L.C.M) of two numbers is defined as the smallest number which is divisible by both the numbers. Common Factor is the factor that is common to all the numbers whereas the prime factors are the factors, which is the product of the powers of the prime numbers.