**Finding the Greatest Number Divisible by 24, 15, and 36**

To find the greatest number of 6 digits that is divisible by 24, 15, and 36, we need to consider the prime factors of each of these numbers and find their least common multiple (LCM). The LCM will give us the required number.

**Prime Factors of 24:**
The prime factorization of 24 is 2 * 2 * 2 * 3 = 2^3 * 3.

**Prime Factors of 15:**
The prime factorization of 15 is 3 * 5.

**Prime Factors of 36:**
The prime factorization of 36 is 2 * 2 * 3 * 3 = 2^2 * 3^2.

**Finding the LCM:**
To find the LCM, we need to consider the highest power of each prime factor present in the given numbers.

- The highest power of 2 is 2^3.
- The highest power of 3 is 3^2.
- The highest power of 5 is 5^1.

Therefore, the LCM of 24, 15, and 36 is 2^3 * 3^2 * 5 = 8 * 9 * 5 = 360.

**Finding the Greatest 6-Digit Number Divisible by 360:**
To find the greatest 6-digit number divisible by 360, we need to divide the largest 6-digit number (999,999) by 360 and find the largest quotient.

999,999 ÷ 360 = 2777.775.

Since the quotient should be an integer, we need to round down the decimal value to the nearest whole number.

Therefore, the largest quotient is 2777.

To find the greatest 6-digit number divisible by 360, we multiply the largest quotient (2777) by 360.

2777 * 360 = 999,720.

Hence, the greatest number of 6 digits exactly divisible by 24, 15, and 36 is 999,720.