m12 q4

This isn't really a different approach, but it is a further explanation as to what is going on here.

The way to figure out how many zeros there will be at the end of 100! is to figure out how many times can you factor out 10 from 100! We know that the prime factorization of 10 is 2x5. There are certainly going to be more 2's factored out of 100! than there will be 5's, so we need to figure out how many times 5 would be found if we did the prime factorization of every single number in 100!

1 - 10: 2-5's (in 5 and 10)
11-20: 2-5's (in 15 and 20)
21 - 30: 3-5's (two in 25 and 1 in 30)
31-40: 2-5's (in 35 and 40)
41-50: 3-5's (in 45 and 2 in 50 [5*5*2])
51-60:2-5's (55 & 60)
61-70: 2-5's (65 & 70)
71-80: 3-5's (2 in 75 [3*5*5] and 1 in 80)
81-90: 2-5's (85 & 90)
91-100: 3-5's (1 in 95 and 2 in 100 [5*5*4])

now total them up: 2+2+3+2+3+2+2+3+2+3 = 24.

Again, we count the 5's because we know that there will be more than enough 2's in order to match up a "2" with each "5" and come up with 10's (2x5) that could be factored out of 100! which means there should be 24 zeros at the end of 100!.

I hope this makes sense. The answer explanation saying the largest n for 5^n that 100! is divisible by 5^n is saying the same thing I just illustrated.