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# Find the number of trailing zeros in the expansion of

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Find the number of trailing zeros in the expansion of  [#permalink]

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Updated on: 06 Jul 2013, 03:04
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57% (02:47) correct 43% (02:33) wrong based on 1544 sessions

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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?

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.
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Find the number of trailing zeros in the expansion of  [#permalink]

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25 Jan 2011, 08:01
50
56
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 5-s 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.

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/everything-ab ... 85592.html
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Re: trailing zeros question (complicated one)  [#permalink]

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29 Jan 2011, 02:04
54
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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.

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##### General Discussion
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Re: trailing zeros question (complicated one)  [#permalink]

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Updated on: 19 Feb 2011, 08:34
1
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)  [#permalink]

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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)  [#permalink]

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19 Feb 2011, 03:15
7
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)  [#permalink]

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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)  [#permalink]

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01 Apr 2011, 08:40
1
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.

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)  [#permalink]

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03 Apr 2011, 07:48
2

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  [#permalink]

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21 Dec 2012, 01:47
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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.

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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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Updated on: 03 Jan 2013, 19:57
2
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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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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: everything-about-factorials-on-the-gmat-85592.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  [#permalink]

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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  [#permalink]

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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  [#permalink]

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08 Jul 2013, 01:08
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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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  [#permalink]

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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.php
please suggest a fool proof method for calculating trailing zeroes of any +ve integer.
Re: Find the number of trailing zeros in the expansion of   [#permalink] 12 Oct 2013, 04:08

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