How many trailing zeros are there in the product \(20 \times 25 \times 30 \times 35 \times 40\)?
Answer: C
To find the number of trailing zeros, we count the number of pairs of prime factors 2 and 5 in the product.
Let's find the prime factorization of each number:
\(20 = 2^2 \times 5^1\)
\(25 = 5^2\)
\(30 = 2 \times 3 \times 5^1\)
\(35 = 5^1 \times 7\)
\(40 = 2^3 \times 5^1\)
Total number of factors of 2 = \(2 + 1 + 3 = 6\).
Total number of factors of 5 = \(1 + 2 + 1 + 1 + 1 = 6\).
Since we have 6 factors of 2 and 6 factors of 5, we can form 6 pairs of (2, 5). Each pair creates a trailing zero. Therefore, there are 6 trailing zeros.