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To find the number of zeros in the product of \((60^8 \times 90^{12} \times 25^{15})\), we need to determine how many factors of 10 are present in this product. A factor of 10 is created by multiplying a factor of 2 and a factor of 5. Therefore, we need to count the number of factors of 2 and 5 in the prime factorizations of \(60^8\), \(90^{12}\), and \(25^{15}\).

**Step 1: Prime Factorization of Each Number**
1. **60**:  
   \(60 = 2^2 \times 3 \times 5\)  
   Therefore, \(60^8 = (2^2 \times 3 \times 5)^8 = 2^{16} \times 3^8 \times 5^8\).

2. **90**:  
   \(90 = 2 \times 3^2 \times 5\)  
   Therefore, \(90^{12} = (2 \times 3^2 \times 5)^{12} = 2^{12} \times 3^{24} \times 5^{12}\).

3. **25**:  
   \(25 = 5^2\)  
   Therefore, \(25^{15} = (5^2)^{15} = 5^{30}\).

**Step 2: Combine All Factors**
Multiplying all the factorizations together:
\[
60^8 \times 90^{12} \times 25^{15} = (2^{16} \times 3^8 \times 5^8) \times (2^{12} \times 3^{24} \times 5^{12}) \times (5^{30})
\]

Combine like terms:
\[
= 2^{16 + 12} \times 3^{8 + 24} \times 5^{8 + 12 + 30} 
\]
\[
= 2^{28} \times 3^{32} \times 5^{50}
\]

**Step 3: Determine Number of Zeros**
The number of zeros is determined by the limiting factor between the counts of 2s and 5s. Since a zero is formed by a factor of \(10 = 2 \times 5\), the limiting factor is the smaller of the counts of 2s and 5s.

Here, we have:
- Factors of 2: 28
- Factors of 5: 50

Since factors of 2 are the limiting factor, the number of zeros is \( \min(28, 50) = 28\).

**Answer: The product has 28 zeros.**

Would you like more details or have any questions?

Here are some related questions you might find useful:
1. How do you find the number of trailing zeros in a factorial?
2. What is the prime factorization method, and how is it used in number theory?
3. How does multiplying large powers affect the number of trailing zeros in a product?
4. How do the powers of other primes affect the number of trailing zeros?
5. How would the number of zeros change if additional factors were introduced?

**Tip:** When finding the number of trailing zeros in a product, always look for pairs of 2s and 5s, as each pair forms a factor of 10, contributing to a trailing zero.

Find the number of zeros in the product of (60^8 × 90^12 × 25^15).

Learn how to determine the number of zeros in a complex product involving powers of numbers such as 60, 90, and 25, through the use of prime factorization and the counting of factor pairs.

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