Answer:
13
Explanation:
A trailing zero results from a factor of 10 in the product. Such a factor is a combination of a factor of 5 and a factor of 2. At least one factor of 2 will be contributed by each even number. At least one factor of 5 wil be contributed by each number divisible by 5.
There are many more even numbers among the numbers 1–55 than there are numbers divisible by 5, so the latter will determine the number of trailing zeros.
Each number divisible by 5 will contribute 1 factor of 5. There are 55/5 = 11 such numbers. Each number divisible by 25 will contribute an additional factor of 5. There are floor(55/25) = 2 of those. So, there are a total of 11+2 = 13 factors of 5 in the number 55!. This means there are 13 trailing zeros in the number 55!.