The least five digit number which is exactly divisible by 12, 18, and 21 is:Options :A] 10010B] 10015C] 10080D] 10020

To find the least five digit number which is exactly divisible by 12, 18, and 21, we need to find the least common multiple (LCM) of these three numbers.

Step 1: Prime factorization of the numbers 12 = 2^2 * 3 18 = 2 * 3^2 21 = 3 * 7

Step 2: Find the LCM The LCM is the product of the highest powers of all the factors that appear in any of the numbers. So, LCM = 2^2 * 3^2 * 7 = 252

Step 3: Find the least five digit number divisible by the LCM The smallest five digit number is 10000. Divide this by the LCM. If it divides evenly, then 10000 is the number we're looking for. If not, we need to add the remainder to 10000 to get the smallest five digit number that is divisible by the LCM.

10000 divided by 252 gives a quotient of 39 and a remainder of 208.

So, the smallest five digit number that is divisible by 12, 18, and 21 is 10000 + (252 - 208) = 10044.

However, this is not in the options given. The next multiple of 252 after 10044 is 10044 + 252 = 10296, which is also not in the options.

The next multiple of 252 after 10296 is 10296 + 252 = 10548, which is also not in the options.

Continuing in this way, the first option we reach that is in the options given is 10080.

So, the answer is C] 10080.