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Find the number of trailing zeros in the expansion of
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Updated on: 06 Jul 2013, 03:04
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Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!. A. 468 B. 469 C. 470 D. 467 E. 471 Can someone help me how to solve this question? I think, there must be more than one solution method. Do questions of such a level of difficulty appear on the actual GMAT?
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Originally posted by feruz77 on 25 Jan 2011, 06:48.
Last edited by Bunuel on 06 Jul 2013, 03:04, edited 1 time in total.
Edited the answer choices.




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Find the number of trailing zeros in the expansion of
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25 Jan 2011, 08:01
feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
a) 10^468 b) 10^469 c) 10^470 d) 10^467 e) 10^471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actual GMAT? # of trailing zeros in 20!, 21!, 22!, 23!, and 24! will be 4 (20/5=4. For 21!, 22!, 23! and 24!, instead of 20 you'll have 21, 22, ... but the result will be the same) > total of 4*5=20 trailing zeros for these 5 terms. (Note here that this won't always be correct: for example 20 and 50 have one trailing zero each but 20*50=1,000 has three trailing zeros not two. That's because extra 2 in 20 and extra 5 in 50 "produced" one more trailing zero. In our case though, we won't have any extra 5s in any factorial, as all are already used for existing trailing zeros); # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; So, \((20!*21!*22!*...*33!)^{3!}=(10^{20}*10^{30}*10^{28}*something)^{3!}=(10^{78}*something)^6=10^{468}*something^6\). Total of 468 trailing zeros. Answer: A. OR: as we have (something)^6 then the # of trailing zeros must be multiple of 6 only answer choice A satisfies this. Theory on this topic: http://gmatclub.com/forum/everythingab ... 85592.html
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Re: trailing zeros question (complicated one)
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29 Jan 2011, 02:04
feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
a) 10^468 b) 10^469 c) 10^470 d) 10^467 e) 10^471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actual GMAT? Excellent explanation by Bunuel. Let me give you one quick way to solve such questions. 3! = 6 => the answer must be of the form \(10^{6n}\) Only A and C have even powers. Hence B C E are out Out of A and C only A's power is divisible by 3, hence C is out. Hence A is the answer.
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Re: trailing zeros question (complicated one)
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Updated on: 19 Feb 2011, 08:34
Thank You Bunuel.
I have one more remark: This series is not a series of numbers, it is a factorial and in this respect your approach makes sense because in a factorial, as a ganeral rule, a number of trailing zeros will depend on the highest power of 5.
Thank You one more time.
Originally posted by feruz77 on 29 Jan 2011, 01:57.
Last edited by feruz77 on 19 Feb 2011, 08:34, edited 1 time in total.



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Re: trailing zeros question (complicated one)
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18 Feb 2011, 23:06
Bunuel  i understand the rule of trailing zeroes, but how did u deduct from that that if its ^6  it have be a multiply of 6? thanks.
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Re: trailing zeros question (complicated one)
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19 Feb 2011, 03:15
144144 wrote: Bunuel  i understand the rule of trailing zeroes, but how did u deduct from that that if its ^6  it have be a multiply of 6?
thanks. For example: 100 has 2 trailing zeros, 100^6=(10^2)^6=10^12 will have 2*6 trailing zeros. Now, we have (something)^6: if # of trailing zeros of that something is x then # of trailing zeros of (something)^6 will be 6x, so multiple of 6.
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Re: trailing zeros question (complicated one)
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19 Feb 2011, 03:21
oh... thanks. +1 do u ever sleep btw?
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Re: trailing zeros question (complicated one)
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01 Apr 2011, 08:40
gurpreetsingh wrote: feruz77 wrote: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.
a) 10^468 b) 10^469 c) 10^470 d) 10^467 e) 10^471
Can someone help me how to solve this question? I think, there must be more than one solution method.
Do questions of such a level of difficulty appear on the actual GMAT? Excellent explanation by Bunuel. Let me give you one quick way to solve such questions. 3! = 6 => the answer must be of the form \(10^{6n}\) Only A and C have even powers. Hence B C E are out Out of A and C only A's power is divisible by 3, hence C is out. Hence A is the answer. Hi gurpreet can u pls tell me , out of A,C ,why did u check whether the ans is div by 3 I know their is smthng silly i am missing ,....but pls tell me what is that.....



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Re: trailing zeros question (complicated one)
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03 Apr 2011, 07:48
read my solution again. he given impression is raised to the power 3! which is 6. => the answer should be of the form  something raised to the power 6n. out of all the given powers,which are raised on 10, only A and C are eligible to be divisible by 6 as all others are odd numbers. Now out of A and C, 470 is not divisible by 6n, hence it can not be the answer. let me know if you are still not able to make it.
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Re: Find the number of trailing zeros in the expansion of
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21 Dec 2012, 01:47
I love trailing zeroes. 20! has 4 factors of 5 = 5^4 21! to 24! also have 4 factors of 5 each = 5^16 25! has 6 factors of 5 = 5^6 26! to 29! also have 6 factors of 5 each = 5^24 30! has 7 factors of 5 = 5^7 31! to 33! has 7 factors of 5 = 5^21 There are 5^78 then raised to 3!=6 so we have 5^468. Obviously we have more than 468 factors of 2 so the count of 5 is our limiting factor. Answer: A
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Re: Find the number of trailing zeros in the expansion of
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Updated on: 03 Jan 2013, 19:57
I cannot seem to understand how A is the correct answer. I do understand that should be 468 trailing zeros. But 10^468 is not correct. For example 6! has 1 trailing zero (and not 10^1=10, trailing zeros). I guess the answer choices should only be 468,...,471 (without the 10^), Isn't it? Maybe I am missing something.
Originally posted by joe123 on 03 Jan 2013, 17:39.
Last edited by joe123 on 03 Jan 2013, 19:57, edited 1 time in total.



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Re: Find the number of trailing zeros in the expansion of
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03 Jan 2013, 18:49
joe123 wrote: I cannot seem to understand how A is the correct answer. I do understand that should be 468 trailing zeros. But 10^468 is not correct. For example 6! has 1 trailing zero (and not 10^1=10, trailing zeros). I guess the answer choices should only be 468,...,471, Isn't it? Maybe I am missing something. 20! = 20*19*18*17*...*4*3*2*1 We know that from the above has 5^4. We know that from the above it has 10 even numbers and some of them like 8 = 2^3. Thus, there are at least 10 factors of 2 or 2^17 to be exact. To get the trailing zero, you have to capture a pair of 5 and 2. Choose the limiting factor. Thus, we have 5^4*2^17=(5^4)(2^4)(2^13) giving 10^4... Continue to do this in the other factorials. 21!,22!,23!,24! will have a total of 10^16 25! will have 10^6 since 25 has two factors of 5. Do it until 33! and we will have 78 factors of 10. But we have to raise by 3! = 6. 78*6= 468
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Re: Find the number of trailing zeros in the expansion of
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09 Jan 2013, 01:42
do we get such questions on gmat?
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Re: Find the number of trailing zeros in the expansion of
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09 Jan 2013, 02:23
joe123 wrote: I cannot seem to understand how A is the correct answer. I do understand that should be 468 trailing zeros. But 10^468 is not correct. For example 6! has 1 trailing zero (and not 10^1=10, trailing zeros). I guess the answer choices should only be 468,...,471 (without the 10^), Isn't it? Maybe I am missing something. Lets say 20! = 10^x 21! = 10^y 22! = 10^z Then (20!*21!*22! ……… *33!)^3! = (10^x * 10^y * 10^z....10^n)^3! = (10)^(3!(x*y*z*...n) = (10)^6(xyz...n) In given options, 6 is the multiple of 468 only. So Ans (A)
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Re: Find the number of trailing zeros in the expansion of
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09 Jan 2013, 03:00
Sachin9 wrote: do we get such questions on gmat? Trailing zeros concept is tested on the GMAT (check here: everythingaboutfactorialsonthegmat85592.html), though this particular question is a bit out of scope.
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Re: Find the number of trailing zeros in the expansion of
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14 Jan 2013, 00:27
joe123 wrote: I cannot seem to understand how A is the correct answer. I do understand that should be 468 trailing zeros. But 10^468 is not correct. For example 6! has 1 trailing zero (and not 10^1=10, trailing zeros). I guess the answer choices should only be 468,...,471 (without the 10^), Isn't it? Maybe I am missing something. You are correct. The answer is 468, not \(10^{468}\). The problem statement is wrong.
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Re: Find the number of trailing zeros in the expansion of
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07 Feb 2013, 01:28
joe123 wrote: I cannot seem to understand how A is the correct answer. I do understand that should be 468 trailing zeros. But 10^468 is not correct. For example 6! has 1 trailing zero (and not 10^1=10, trailing zeros). I guess the answer choices should only be 468,...,471 (without the 10^), Isn't it? Maybe I am missing something. That threw me for a loop as well. I was going crazy trying to figure out how there would be so many damn zeros. The answers should be 468  471 or the problem should written differently.



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Re: Find the number of trailing zeros in the expansion of
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08 Jul 2013, 01:08
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
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Re: Find the number of trailing zeros in the expansion of
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25 Sep 2013, 09:09
Superb question. Captured the essence of GMAT in a single shot. As soon as I saw the ^ 3! I made a note (some number) x6 on my scrap paper. Started going through the choices for anything dividing by 6 and voila! 468 was the first choice and was divisible. Saw that the rest are consecutive so marked it directly. +1 Cheers for the poster



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Re: Find the number of trailing zeros in the expansion of
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12 Oct 2013, 04:08
Buneul, here's my doubt: # of trailing zeros in 25!, 26!, 27!, 28!, and 29! will be 5+1=6 (25/5+25/5^2=6) > total of 6*5=30 trailing zeros for these 5 terms; # of trailing zeros in 30!, 31!, 32!, and 33! will be 6+1=7 (30/5+30/5^2=7) > total of 7*4=28 trailing zeros for these 5 terms; for calculating trailing zeros up til 24! you did just 20/5=4. but above those numbers i.e., from 25! on wards you did (25/5+25/5^2=6) and (30/5+30/5^2=7) Suppose I want # of trailing zeros in 310! using your concept 310/5+310/5^2=62+12=74 trailing zeroes BUT using the factorial calculator below I am getting 76 trailing zeroes http://www.nitrxgen.net/factorialcalc.phpplease suggest a fool proof method for calculating trailing zeroes of any +ve integer.




Re: Find the number of trailing zeros in the expansion of
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