What is the least number which is exactly divisible by 8, 9, 12, 15 and 18 and is also a perfect square?

Options

A. 3600
B. 7200
C. 5200
D. 6500

Answer

To find the least number which is exactly divisible by 8, 9, 12, 15, and 18 and is also a perfect square, we can follow these steps:

### Step 1: Find the prime factorization of each number.
- **8** = \(2^3\)
- **9** = \(3^2\)
- **12** = \(2^2 \times 3^1\)
- **15** = \(3^1 \times 5^1\)
- **18** = \(2^1 \times 3^2\)

### Step 2: Determine the highest power of each prime factor.
- For **2**: The highest power is \(2^3\) (from 8).
- For **3**: The highest power is \(3^2\) (from 9 and 18).
- For **5**: The highest power is \(5^1\) (from 15).

### Step 3: Write the LCM using the highest powers.
The LCM of the numbers is:
\[ \text{LCM} = 2^3 \times 3^2 \times 5^1 \]

### Step 4: Calculate the LCM.
Calculating the LCM:
\[ 2^3 = 8, \quad 3^2 = 9, \quad 5^1 = 5 \]
\[ \text{LCM} = 8 \times 9 \times 5 \]
Calculating step-by-step:
\[ 8 \times 9 = 72 \]
\[ 72 \times 5 = 360 \]
Thus, the LCM of 8, 9, 12, 15, and 18 is **360**.

### Step 5: Adjust for perfect square.
To make the LCM a perfect square, we need to ensure that all prime factors have even powers:
- The prime factorization of 360 is \(2^3 \times 3^2 \times 5^1\).
- For \(2\), we have \(2^3\) (we need one more \(2\) to make it \(2^4\)).
- For \(3\), we have \(3^2\) (already even).
- For \(5\), we have \(5^1\) (we need one more \(5\) to make it \(5^2\)).

### Step 6: Multiply by the necessary factors to make it a perfect square.
To make it a perfect square, we need to multiply by \(2^1\) and \(5^1\):
\[ \text{Perfect Square} = 360 \times 2^1 \times 5^1 = 360 \times 2 \times 5 \]

Calculating:
\[ 360 \times 2 = 720 \]
\[ 720 \times 5 = 3600 \]

### Final Answer:
The least number which is exactly divisible by 8, 9, 12, 15, and 18 and is also a perfect square is **3600**.

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