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13 Sep 2008, 13:53
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How many zeros does 100! end with?
(A) 20
(B) 24
(C) 25
(D) 30
(E) 32
Source: GMAT Club Tests - hardest GMAT questions
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.
(A) 20
(B) 24
(C) 25
(D) 30
(E) 32
B
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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13 Sep 2008, 20:52
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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.
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
15 Nov 2008, 11:25
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ritula
what we need are a pair of 2 and 5 to *** 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!.
