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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.

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.

_________________

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

what we need are a pair of 2 and 5 to cunt the number of zeros in 100!. so lets find the number of 2's and 5's but remember there are always more 2's than 5's in any factorial >1!. so focus on no. of 5's.

No of 5's in 100 = 100/5 = 20 doing that we are missing 5's in 25, 50, 75 and 100. therefore,

No of another 5's (of 25) in 100 = 100/25 = 4 now we have covered all 5's. if the factorial were >125!, then we need to find the no. of 5's in 125 but in this case 125 is out of scope. so we are done.

total 5's in 100! = 24 for no. of 2's, there are 100/2 + 100/4 + 100/8 + 100/16 + 100/32 + 100/64 (and not 100/128) = 50 + 25 + 12 + 6 + 3 + 1 = 97

only twenty-four (24) 2's and 5's make zeros. so there are 24 zeros in 100!.

Nice question. Remember, this can be rephrased as "What is the largest power of 10 that gives a remainder of zero dividing 100!".

Similarly, we can also be asked to find: 1) What is the largest power of 7 that divides 100! 2) How many zeroes does 100! end with, when it is expressed in base 7?

Both these questions are essentially the same. Just find out how many seven's are in 100!

Hi guys! I've just tried to calculate factorial of 100 using Excel formula fact and came up with a lot more than 24 zeros. So what am I getting wrong? The figure is 933262154439442000000000000...

Nice post but one quesiton.. How did you get '3's...

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])

CONSIDER THE INTERVAL 1-5=1, 6-10=1,11-15=1,16-20=1,21-25=1+2(FOR MULTIPLES OF 25 WE HAVE ALWAYS TWO ZEROE'S WHICH ARE 25,50,75,100).....SUM UP WE GET 24. IMO=24

I was really confused at first when looking at the solutions because I haven't really covered this concept yet in my Veritas GMAT prep. In fact. I don't think they ever mentioned using factorial likes this lol, but once I looked at the factorial review page posted in this thread it all made sense. To anyone confused, use this formula

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File comment: where k must be chosen such that 5^(k+1)>n

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.

This is how I solved ..knowing how many zeroes is basically how many 5's using the standard formula i get -