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.

CrushTheGMAT wrote:

Can anyone offer another explanation? I'm having trouble with this one:

How many zeros does 100! end with?

* 20

* 24

* 25

* 30

* 32

We have to find how many times factor 5 is contained in 100!. That is, we have to find the largest n such that 100! is divisible by 5^n . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by 25 = 5^2. So, the answer is 24.

The correct answer is B.

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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