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.