To find the number \( x \) that is divisible by 8, 12, 30, 36, and 45 and is also a perfect square, we will follow these steps:
### Step 1: Find the prime factorization of each number.
- **8**: \( 2^3 \)
- **12**: \( 2^2 \times 3^1 \)
- **30**: \( 2^1 \times 3^1 \times 5^1 \)
- **36**: \( 2^2 \times 3^2 \)
- **45**: \( 3^2 \times 5^1 \)
### Step 2: Determine the least common multiple (LCM).
To find the LCM, we take the highest power of each prime factor that appears in the factorizations:
- For \( 2 \): The highest power is \( 2^3 \) (from 8).
- For \( 3 \): The highest power is \( 3^2 \) (from 36 and 45).
- For \( 5 \): The highest power is \( 5^1 \) (from 30 and 45).
Thus, the LCM is:
\[
\text{LCM} = 2^3 \times 3^2 \times 5^1
\]
### Step 3: 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, 12, 30, 36, and 45 is \( 360 \).
### Step 4: Ensure \( x \) is a perfect square.
Since \( x \) must be a perfect square, we need to adjust the prime factorization of the LCM to ensure all exponents are even:
- The current factorization is \( 2^3 \times 3^2 \times 5^1 \).
- To make \( 2^3 \) even, we need one more \( 2 \) (making it \( 2^4 \)).
- \( 3^2 \) is already even.
- To make \( 5^1 \) even, we need one more \( 5 \) (making it \( 5^2 \)).
Thus, we adjust the LCM to:
\[
x = 2^4 \times 3^2 \times 5^2
\]
### Step 5: Calculate the value of \( x \).
Calculating \( x \):
\[
2^4 = 16, \quad 3^2 = 9, \quad 5^2 = 25
\]
Now, calculate:
\[
x = 16 \times 9 \times 25
\]
Calculating step-by-step:
\[
16 \times 9 = 144
\]
\[
144 \times 25 = 3600
\]
Thus, the value of \( x \) is \( 3600 \).
### Final Answer:
\[
\boxed{3600}
\]
