Find the smallest perfect square number, which is divisible by `8` and `12`.

Options

A. `121`
B. `144`
C. `169`
D. `196`

Answer

To find the smallest perfect square number that is divisible by both 8 and 12, we can follow these steps:

### Step 1: Find the LCM of 8 and 12
To find the least common multiple (LCM), we can use the prime factorization method.

- The prime factorization of 8 is \(2^3\).
- The prime factorization of 12 is \(2^2 \times 3^1\).

Now, we take the highest power of each prime factor:
- For \(2\), the highest power is \(2^3\).
- For \(3\), the highest power is \(3^1\).

Thus, the LCM is:
\[ \text{LCM} = 2^3 \times 3^1 = 8 \times 3 = 24 \]

### Step 2: Check if the LCM (24) is a perfect square
A perfect square must have even powers for all prime factors. The prime factorization of 24 is:
\[ 24 = 2^3 \times 3^1 \]
Here, the powers of both \(2\) and \(3\) are odd, which means 24 is not a perfect square.

### Step 3: Make the LCM a perfect square
To make 24 a perfect square, we need to adjust the powers of the prime factors to be even. We can do this by multiplying by the necessary factors:

- For \(2^3\), we need one more \(2\) to make it \(2^4\) (even).
- For \(3^1\), we need one more \(3\) to make it \(3^2\) (even).

Thus, we multiply 24 by \(2 \times 3 = 6\):
\[ \text{Perfect Square} = 24 \times 6 = 144 \]

### Step 4: Conclusion
The smallest perfect square number that is divisible by both 8 and 12 is:
\[ \boxed{144} \]

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