Let x be the least number divisible by 8, 12, 30, 36 and 45 and x is also a perfect square. What is the sum of the digits of the value of x?

Options

A. 6
B. 7
C. 4
D. 9

Answer

To find the least 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 LCM of the numbers
We start by finding the least common multiple (LCM) of the numbers 8, 12, 30, 36, and 45.

**Prime Factorization:**
- \( 8 = 2^3 \)
- \( 12 = 2^2 \times 3^1 \)
- \( 30 = 2^1 \times 3^1 \times 5^1 \)
- \( 36 = 2^2 \times 3^2 \)
- \( 45 = 3^2 \times 5^1 \)

**LCM Calculation:**
To find the LCM, we take the highest power of each prime factor:
- For \( 2 \): highest power is \( 2^3 \) (from 8)
- For \( 3 \): highest power is \( 3^2 \) (from 36 and 45)
- For \( 5 \): highest power is \( 5^1 \) (from 30 and 45)

Thus, the LCM is:
\[ \text{LCM} = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 \]

Calculating this:
\[ 8 \times 9 = 72 \]
\[ 72 \times 5 = 360 \]

So, the LCM of the numbers is \( 360 \).

### Step 2: Ensure \( x \) is a perfect square
Next, we need to adjust \( 360 \) to make it a perfect square. The prime factorization of \( 360 \) is:
\[ 360 = 2^3 \times 3^2 \times 5^1 \]

To be a perfect square, all prime factors must have even powers:
- \( 2^3 \) needs one more \( 2 \) to become \( 2^4 \)
- \( 3^2 \) is already even
- \( 5^1 \) needs one more \( 5 \) to become \( 5^2 \)

Thus, we need to multiply \( 360 \) by \( 2^1 \times 5^1 = 10 \):
\[ x = 360 \times 10 = 3600 \]

### Step 3: Calculate the sum of the digits of \( x \)
Now, we find the sum of the digits of \( 3600 \):
\[ 3 + 6 + 0 + 0 = 9 \]

### Final Answer
The sum of the digits of \( x \) is \( 9 \).

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