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Find the number of trailing zeros in the expansion of

1 2 3

sharathnair14

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24 Aug 2019, 08:58

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Bunuel

Sachin9

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!

Bunuel

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26 Aug 2019, 01:37

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sharathnair14

Bunuel

Sachin9

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.

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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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?

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

ImranBinkhaled

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20 Apr 2020, 16:55

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feruz77

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.

sachinsavailable

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20 Apr 2020, 19:20

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feruz77

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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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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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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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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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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28 Dec 2025, 08:24

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Division by 5:
=4 for 20-24
=5 for 25-29
=6 for 30-33
Partial solution:
4*5 + 5*5 + 6*4 =69

Division by 5^2:
=1 for 25-33
Partial solution 1*9

We have inside the parenthesis some number with 69 + 9 = 78 zeroes at the end of it.
We may factor a power of 10^78 that represents those zeroes, necessary for the next step to work logically.
Now we distribute the power outside the parentheses to the factors of the product inside.
We only care about the power of 10.

The power of 10 becomes 10^(78*6), meaning the number has 78*6 = 468 trailing zeroes.