Find the greastest number of 5 digits exactly divisible by 24 , 15 and 36 .

Answer

To find the greatest number of 5 digits exactly divisible by 24, 15, and 36, we can follow these steps:

### Step 1: Find the LCM of 24, 15, and 36
To find the LCM, we first need to find the prime factorization of each number:

- **24** = \(2^3 \times 3^1\)
- **15** = \(3^1 \times 5^1\)
- **36** = \(2^2 \times 3^2\)

Now, we take the highest power of each prime number that appears in the factorizations:

- For \(2\): highest power is \(2^3\) (from 24)
- For \(3\): highest power is \(3^2\) (from 36)
- For \(5\): highest power is \(5^1\) (from 15)

So, the LCM is:
\[ LCM = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360 \]

### Step 2: Find the greatest 5-digit number
The greatest 5-digit number is \(99999\).

### Step 3: Check divisibility by 360
Now, we need to check if \(99999\) is divisible by \(360\). We can do this by dividing \(99999\) by \(360\) and finding the remainder.

\[ 99999 \div 360 = 277.775 \quad \text{(approximately)} \]
Calculating \(360 \times 277\):
\[ 360 \times 277 = 99720 \]

### Step 4: Calculate the remainder
Now, we find the remainder:
\[ 99999 - 99720 = 279 \]

### Step 5: Subtract the remainder from the greatest 5-digit number
To find the greatest 5-digit number that is divisible by \(360\), we subtract the remainder from \(99999\):
\[ 99999 - 279 = 99720 \]

### Conclusion
Thus, the greatest number of 5 digits exactly divisible by \(24\), \(15\), and \(36\) is **99720**.

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