Discussion
Find the number of trailing zeros in \(2^5 \times 3^2 \times 4^8 \times 5^{10}\)
Answer: B
We need to find the number of pairs of 2 and 5. First, express all numbers in their prime factors. \(2^5 \times 3^2 \times (2^2)^8 \times 5^{10} = 2^5 \times 3^2 \times 2^{16} \times 5^{10} = 2^{21} \times 3^2 \times 5^{10}\). We have 21 factors of 2 and 10 factors of 5. The number of pairs of (2,5) is limited by the smaller power, which is 10. So there are 10 trailing zeros.
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