SOLUTION :  The greatest number of  6 digits  =  999999First, we need to find the LCM of 24, 15 and 36 and then divide 6 digits number 999999by the LCM and subtract the remainder  from 6 digit number 99999924, 15 and 3624 = 2 x 2 x2 x315 = 3 x 536 = 2 x 2 x 3 x 3L.C.M of 24, 15 and 36 = 360Now, 999999 /  360quotient =  2777Remainder =  279Therefore, the remainder is 279. Hence the required number is = 999999 – 279 = 999720Hence, 999720 is the greatest number of 6 digits exactly divisible by 24, 15 and 36.

Introduction:
To find the greatest number of 6-digits which is exactly divisible by 15, 24, and 36, we need to find the highest common multiple (HCM) of these three numbers. The HCM is the largest number that divides all the given numbers evenly.

Step 1: Prime Factorization
- Prime factorize each of the given numbers: 15, 24, and 36.
- The prime factorization of 15 is 3 * 5.
- The prime factorization of 24 is 2^3 * 3.
- The prime factorization of 36 is 2^2 * 3^2.

Step 2: Find Common Factors
- Identify the common factors among the prime factorizations of the given numbers.
- The common factors among 3 * 5, 2^3 * 3, and 2^2 * 3^2 are 2^2 and 3.

Step 3: Determine the Highest Common Multiple (HCM)
- Multiply the common factors identified in Step 2.
- HCM = 2^2 * 3 = 12.

Step 4: Find the Greatest 6-Digit Number Divisible by 12
- To find the greatest 6-digit number divisible by 12, we need to divide 999,999 (the largest 6-digit number) by 12 and find the largest whole number quotient.
- Divide 999,999 by 12: 999,999 ÷ 12 = 83,333.25.
- The largest whole number quotient is 83,333.
- Multiply 83,333 by 12: 83,333 * 12 = 999,996.

Conclusion:
The greatest number of 6-digits which is exactly divisible by 15, 24, and 36 is 999,996.