How many 0's are at the end of 100! (100 factorial)?

The number of 0's at the end of a number is the same as the number of factors of 10 that the number has. So consider the prime factor decomposition of 100! and note that every factor of 10 has prime factors 2 and 5. Because both 2 and 5 are needed, the question can be rephrased as "How many factors of 2 and 5 has 100!, and which is the smaller number?". With this, we can start to work. Since 100! = 1 x 2 x... x 100, we see that prime factors of 2 come from 2, 4, 6, ..., 98, 100, and prime factors of 5 come from 5, 10, ..., 95, 100. Note that while we look for numbers divisible by 2, we must also remember that they can be divisible by 4, 8, and any other powers of 2, contributing more prime factors of 2. The same goes for factors of 5. With this, we can see that clearly it is the number of prime factors of 5 that will be smaller, and this the number of factors of 10. Now all that remains is to count the number of prime factors of 5. All multiples of 5 have a prime factor of 5, giving 20 (from 5, 10, 15, 20,... , 100). Also, recalling that multiples of 25 one extra factor of 5, we get 4 more factors of 5 (from 25, 50, 75, 100). Noting that 5^3 is larger than 100, we are done. Thus, 100! ends with 24 0's.