The Zeroes Problem

For each zero at the end, there must be at least one 2 and 5, as these when multiplied become 10, which is equivalent to adding a zero on the end (in base 10). There will clearly be more 2s so we do not need to worry about these and instead think about the numbers of 5. Any multiples of 5 will contain at least one 5 in the prime factorisation. There are 20 multiples of 5 in 1-100 so the number of 5s is at least 20. Furthermore, there are 4 multiples of 25 (25,50,75,100) these contain an extra prime in the factorisation and so give us 4 extra 5s. There are no more additions of 5s to make as $5^{3}=125>100$ so the total is 20+4=24 trailing zeros.