The greatest number of four digits which is divisible by s 12, 18, 21 & 28 each is  Select an answerA9848B9864C9828D9636

To find the greatest 4-digit number divisible by 12, 18, 21, and 28, we first need to find the least common multiple (LCM) of these numbers.

Step 1: Prime factorization of each number 12 = 2^2 * 3 18 = 2 * 3^2 21 = 3 * 7 28 = 2^2 * 7

Step 2: Find the LCM The LCM is found by multiplying the highest power of all prime numbers obtained from the factorization of each number. LCM = 2^2 * 3^2 * 7 = 252

Step 3: Find the greatest 4-digit number divisible by the LCM The greatest 4-digit number is 9999. Divide 9999 by 252 to find the remainder. 9999 ÷ 252 = 39 remainder 231

Subtract the remainder from 9999 to find the greatest 4-digit number that is divisible by 252. 9999 - 231 = 9768

So, the greatest 4-digit number divisible by 12, 18, 21, and 28 is 9768. However, this option is not available in the choices given. There might be a mistake in the question or the options provided.