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Find the number of trailing zeros in the expansion of 24 Aug 2019, 08:58
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Bunuel
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do we get such questions on gmat?

Trailing zeros concept is tested on the GMAT (check here: https://gmatclub.com/forum/everything-ab ... 85592.html), though this particular question is a bit out of scope.
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Hey Bunuel. Old post but, nevertheless.
What really makes a question out of scope? The concept tested here definitely is included on the GMAT. Difficulty? Time to solve a question? complexity? number of concepts related to the question?

Just for the sake of everybody's clarity.
I've read a lot of great things about you.
I'm a newcomer here. So, kindly enlighten me!
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Find the number of trailing zeros in the expansion of 26 Aug 2019, 01:37
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sharathnair14
Bunuel
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do we get such questions on gmat?

Trailing zeros concept is tested on the GMAT (check here: https://gmatclub.com/forum/everything-ab ... 85592.html), though this particular question is a bit out of scope.

Hey Bunuel. Old post but, nevertheless.
What really makes a question out of scope? The concept tested here definitely is included on the GMAT. Difficulty? Time to solve a question? complexity? number of concepts related to the question?

Just for the sake of everybody's clarity.
I've read a lot of great things about you.
I'm a newcomer here. So, kindly enlighten me!
Show more

I think a little bit of everything you've mentioned. If not the shortcut mentioned in my post, the question would require more time than 2 minutes on average. Also, you can see in our stats that 43% of users out of 1442, as of today, solve the question incorrectly. Which is quite a big percentage of wrong answers. Those who solved correctly spent 02:46 on average.

Plus the wording is not the one GMAC would use. They wouldn't use "trailing zeros" (you are not supposed to know what a trailing zero is), instead the'd say something like "the number of zeros after rightmost non-zero digit of ..."
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Find the number of trailing zeros in the expansion of 26 Aug 2019, 08:31
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feruz77
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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Asked: Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.

(20!*21!*22! ……… *33!)^3!.

Highest Power of 5 in 20! = 4
Highest Power of 5 in 21! = 4
Highest Power of 5 in 22! = 4
Highest Power of 5 in 23! = 4
Highest Power of 5 in 24! = 4
Highest Power of 5 in 25! = 5 + 1 = 6
Highest Power of 5 in 26! = 5 + 1 = 6
Highest Power of 5 in 27! = 5 + 1 = 6
Highest Power of 5 in 28! = 5 + 1 = 6
Highest Power of 5 in 29! = 5 + 1 = 6
Highest Power of 5 in 30! = 6 + 1 = 7
Highest Power of 5 in 31! = 6 + 1 = 7
Highest Power of 5 in 32! = 6 + 1 = 7
Highest Power of 5 in 33! = 6 + 1 = 7

Number of trailing zeros in (20!*21!*22! ……… *33!)^3!. = (4*5 + 6*5 + 4*7)*6 = (20 + 30 + 28)*6 = 78*6 = 468

IMO A
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Find the number of trailing zeros in the expansion of 20 Apr 2020, 16:55
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feruz77
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?
Show more

There seems to be a problem with the question as highlighted below:
(20!*21!*22! ……… *33!)^3!.

The 3! at the last should be multiplied instead of power.
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Find the number of trailing zeros in the expansion of 20 Apr 2020, 19:20
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feruz77
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?
Show more

It is not difficult question , rather it tests the concept of trailing zero's,

(20!) has 4
21! has 4
22! has 4
23! has 4
24! has 4
25! has (25!/5+25!/5^2\(\)) =6
26! has 6
27! has 6
28! has 6
29! has 6
30! has (30!/5+ 30!/5^2\(\))= 7
31! has 7
32! has 7
33! has 7

the above means

((10^4)^5+ (10^6)^5+ (10^ 7)^4)^3!

=(10^78)^6
=10^468

hence the answer
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Find the number of trailing zeros in the expansion of 11 Feb 2021, 03:55
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Thank you for the multiple explanations, I solved it by taking the long route of factorials, but nice to see easier explanations too. Next time I come across such a question, I would think about the easy way too. Thank you Bunuel and Gurpreet for alternative explanations - this would greatly reduce my calculation time!
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Find the number of trailing zeros in the expansion of 22 Feb 2025, 20:10
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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;

why there is 6+1 when it can be divided by 6 only ?
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Find the number of trailing zeros in the expansion of 23 Feb 2025, 00:44
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Check out Trailing Zeroes section in this article https://gmatclub.com/forum/everything-a ... 85592.html, it explains how are trailing zeroes calculated in n!. Hope it helps.
basanti7
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;

why there is 6+1 when it can be divided by 6 only ?
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Find the number of trailing zeros in the expansion of 02 Nov 2025, 22:27
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misread the question here - should be to the power of (3!), not the power of 3, and then factorial; the prompt is a bit misleading here

then, it's pretty straightforward

for trailing 0s, the restrictive factor here is 5, not 2

so we count 5s:

20-24 - 5*4 0s = 20
25-29 - 5*6 0s = 30
30-33 - 4*7 0s = 28

10^(20+30+28) = 10^78
(10^78)^3! = (10^78)^6 = 10^(78*6) = 10^468

(A) 468

feruz77
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
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