How many distinct factors of 8000 are perfect cubes.

Options

A. 12
B. 6
C. 8
D. 4

Answer

To find how many distinct factors of 8000 are perfect cubes, we need to follow these steps:

### Step 1: Prime Factorization of 8000
First, we need to find the prime factorization of 8000.
- 8000 can be expressed as \( 8000 = 8 \times 1000 \).
- We know that \( 8 = 2^3 \) and \( 1000 = 10^3 = (2 \times 5)^3 = 2^3 \times 5^3 \).
- Therefore, we can combine these:
\[ 8000 = 2^3 \times (2^3 \times 5^3) = 2^{3+3} \times 5^3 = 2^6 \times 5^3 \]

### Step 2: Finding Perfect Cube Factors
A factor is a perfect cube if all the exponents in its prime factorization are multiples of 3.

From the prime factorization \( 8000 = 2^6 \times 5^3 \):
- The exponent of 2 can be \( 0, 3, \) or \( 6 \) (these are the multiples of 3 that are less than or equal to 6).
- The exponent of 5 can be \( 0 \) or \( 3 \) (these are the multiples of 3 that are less than or equal to 3).

### Step 3: Counting the Combinations
Now we can count the combinations of these exponents:
- For \( 2^x \): \( x \) can be \( 0, 3, 6 \) (3 choices).
- For \( 5^y \): \( y \) can be \( 0, 3 \) (2 choices).

To find the total number of distinct factors that are perfect cubes, we multiply the number of choices for \( x \) and \( y \):
\[ \text{Total perfect cube factors} = \text{(choices for } x\text{)} \times \text{(choices for } y\text{)} = 3 \times 2 = 6 \]

### Final Answer
Thus, the number of distinct factors of 8000 that are perfect cubes is **6**.

---