II think first find hcf of those three number then divide greatest 6 digit no. by it and then aubtract remainder from 6 digit number then answer will be your final sol

Problem: Find the greatest number of 6 digits exactly divisible by 24, 15, and 36.

Solution:
To find the greatest number of 6 digits that is exactly divisible by 24, 15, and 36, we need to find the least common multiple (LCM) of these three numbers and then find the largest multiple of the LCM that is within the range of 6 digits.

Finding the LCM:
To find the LCM of 24, 15, and 36, we can break down each number into its prime factors and then take the maximum power of each prime factor.

Prime factors of 24 = 2 * 2 * 2 * 3 = 2^3 * 3
Prime factors of 15 = 3 * 5 = 3 * 5^1
Prime factors of 36 = 2 * 2 * 3 * 3 = 2^2 * 3^2

Taking the highest powers of prime factors, the LCM of 24, 15, and 36 is:
LCM = 2^3 * 3^2 * 5^1 = 8 * 9 * 5 = 360

Finding the largest multiple within the range:
Now that we have the LCM as 360, we need to find the largest multiple of 360 that is within the range of 6 digits.

To find the largest multiple, we divide the largest 6-digit number (999,999) by 360 and find the quotient. Then we multiply the quotient by 360 to get the largest multiple.

999,999 ÷ 360 = 2777.775

Since we are looking for a whole number, we take the floor value of 2777.775, which is 2777.

The largest multiple of 360 within the range of 6 digits is:
360 * 2777 = 999,720

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