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Re M1204 [#permalink]
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15 Sep 2014, 23:46
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Official Solution:How many zeros does 100! end with? A. 20 B. 24 C. 25 D. 30 E. 32 Trailing zeros in 100!: \(\frac{100}{5}+\frac{100}{5^2}=20+4=24\) THEORY: Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example 125,000 has 3 trailing zeros; The number of trailing zeros n!, the factorial of a nonnegative integer \(n\), can be determined with this formula: \(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} \gt n\) It's easier if we consider an example: How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\). So there are 7 zeros in the end of 32!. Another example, how many trailing zeros does 125! have? \(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\), The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero. Answer: B
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I think this is a poorquality question and the explanation isn't clear enough, please elaborate. When I used a calculator to get 100! I got = 93,326,215,443,944,100,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 which adds up to 143 zeros (i think). Now I am really confused.



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Re M1204 [#permalink]
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21 Oct 2015, 07:06
I think this is a poorquality question and I don't agree with the explanation. I am really confused. I am getting that 32! has about 20 trailing zeroes with a calculator.



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danjbon wrote: I think this is a poorquality question and the explanation isn't clear enough, please elaborate. When I used a calculator to get 100! I got = 93,326,215,443,944,100,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 which adds up to 143 zeros (i think). Now I am really confused. 100! is a huge number and ordinary calculators round it. BTW 100! has 24 trailing zeros: http://www.wolframalpha.com/input/?i=100%21For similar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory. Hope it helps.
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Re M1204 [#permalink]
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31 Oct 2015, 12:38
I think this is a highquality question and I agree with explanation.



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Re: M1204 [#permalink]
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11 Jan 2016, 10:07
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I didn't know that there was a formula for trailing zeros so this is how I got it though it took me a while and had I seen this on the gmat I probably would've just considered this question a time sucker and skipped it:
100*99= bunch of numbers ending with200 * 98=bunch of numbers ending with300 *97= you get the point. So you don't actually get another zero until you hit *95.
So every multiple of 5 adds a zero. But then this being the GMAT you know there is a trick and the tricky part is that multiples of 25 give an extra zero. I proved it by just picking 50*20 and 25*4. Since there are 4 multiples of 25 the answer is 24.



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Re: M1204 [#permalink]
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18 Mar 2016, 00:21
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Well, to find the number of zeros. we can use the following method.
Generally, we need to find the power of 5 that can divide the 100!. i.e 5^x ( we need to find x ).
Do as successive division of just 100 by 5, until you cann't divide further.
100/5 = 20 > Quotient 20/5 = 4 > Quotient
Here 4 is the last possible quotient. So the number of zero's are 20 + 4 = 24 [ add up all the quotients ].
Similarly, for 125 ! 125/5 = 25 25/5 = 5 5/5 = 1 25 + 5 + 1 = 31.
Here finding the number of zero's is an application of finding the highest power of 5 that can divide 100! or 125!. i.e this approach is used to find the highest power of any prime number that can divide a factorial without leaving any remainder. This is method applicable only if the number whose highest power is to be found is a prime number.



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Re: M1204 [#permalink]
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25 Jul 2016, 09:41
Hi Bunuel,
The formula mentioned lists to continue till 5 to power of k+1 to be greater than n.
Should that be less than n?



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Re: M1204 [#permalink]
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04 Aug 2016, 02:31
Please let me know if this approach and my understanding is correct.
To calculate number of zeroes in products of n numbers, we must consider pair of (2,5).
Now, from 110 we have two (5,2) pairs : 1st 2 & 5 itself and 2nd 2 X 5 (number 10). Similarly from 1100 (100!) we will have 2 X 20 pairs of (5,2). So, until now we have 20 zeroes.
Moving forward if we closely observe 100 it self has 2 (5,2) pairs so now the total is 20+1= 21. Also, similarly 25, 50 & 75 have extra 5 in them, which when multiplied by any even number will produce an extra zero. So, now total number of zeroes in 100! = 20+1+3=24.
Regards Yash



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Re M1204 [#permalink]
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17 Aug 2016, 06:28
I think this is a highquality question and I agree with explanation. I have marked the correct answer as mentioned in the explanation but it is still showing wrong answer in test.



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Re: M1204 [#permalink]
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06 Sep 2016, 14:42
Like we are considering the factors of 5 in 100!, is it okay to say that anything with a 2*5 will produce one 0. so basically we are looking for 2*5's in that big number.



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Re: M1204 [#permalink]
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06 Sep 2016, 23:42
vivgpta007 wrote: Like we are considering the factors of 5 in 100!, is it okay to say that anything with a 2*5 will produce one 0. so basically we are looking for 2*5's in that big number. Hi, Yes you are almost right. I suggest that you practice the method described by Bunuel in the thread above. Although, it is based on the same concept but it is precise and easy to remember. Regards Yash "Give Kudos Take Kudos isn't it simple "



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Re M1204 [#permalink]
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29 Nov 2016, 14:35
I think this is a highquality question. I think it is a very good question and fantastic explanation.



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Re: M1204 [#permalink]
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22 Dec 2016, 19:00
I hope this is high quality question.
So simple if I memorize, which is not gmat looking for.



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Re: M1204 [#permalink]
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05 Feb 2017, 14:56
In the explanation, I think it should be "k must be chosen such that n5(k+1)<n", which it makes more sense.



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The answer is correct, but I didn't know the method. I did it this way:
There are 20 (100/5) numbers ending 5 or 0 between 1 and 100. Among this numbers:
1. 4 numbers contain 25 (25, 50, 75, 100)
For each of these numbers we can make a multiple of 100, using the several even numbers in the 100! So, we have 2 zeroes for each of these numbers. n1= 4x2 = 8 zeros
2. 16 numbers contain only one 5 in its factors (5, 10, 15, 20, 30,.. )
For each of these numbers we can make only a multiple of 10. So, we have only 1 zero from each of these numbers. n2 = 16x1 = 16 zeros
Total = 8 + 16 = 24 zeros
If the factorial would be greater, I would look for numbers that contain 125(5^3), 625(5^4), and so on. Interestingly, the method does it in a very simple way, dividing. First, by dividing by 5 the method takes all the 5s contained in every number from 1 to 100 but do not capture any repeated 5 in any number. Second, by dividing by 5^2 they capture the second 5 of all the numbers that contain a second five. The method continue doing so until it captures all the fives. For each five there can be made a zero, since the number of 2s contained is much more greater. Ahead I will use the method in order to save time.



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Re: M1204 [#permalink]
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16 Apr 2017, 12:23
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Bunuel wrote: The number of trailing zeros n!, the factorial of a nonnegative integer \(n\), can be determined with this formula: \(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} \gt n\) It's easier if we consider an example: How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\). So there are 7 zeros in the end of 32!
Sigh, another formula to memorize. Great explanation!



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Re: M1204 [#permalink]
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22 Aug 2017, 15:59
Bunuel wrote: Official Solution:
How many zeros does 100! end with?
A. 20 B. 24 C. 25 D. 30 E. 32
Trailing zeros in 100!: \(\frac{100}{5}+\frac{100}{5^2}=20+4=24\) THEORY: Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow. For example 125,000 has 3 trailing zeros; The number of trailing zeros n!, the factorial of a nonnegative integer \(n\), can be determined with this formula: \(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where \(k\) must be chosen such that \(5^{(k+1)} \gt n\) It's easier if we consider an example: How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\). Notice that the last denominator (\(5^2\)) must be less than 32. Also notice that we take into account only the quotient of the division, that is \(\frac{32}{5}=6\). So there are 7 zeros in the end of 32!. Another example, how many trailing zeros does 125! have? \(\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31\), The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
Answer: B I guess there is a typo in the explanation. Shouldn't the highlighted part be 5^(k+1) less than or equal to N?










