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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.

Answer: A.

Or as we have (something)^6 then the # of trailing zeros must be multiple of 6 only answer choice A satisfies this.

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.

Last edited by feruz77 on 19 Feb 2011, 08:34, edited 1 time in total.

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. _________________

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

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21 Dec 2012, 01:47

2

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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.

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

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03 Jan 2013, 17:39

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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.

Last edited by joe123 on 03 Jan 2013, 19:57, edited 1 time in total.

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 _________________

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.

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. _________________

Sergey Orshanskiy, Ph.D. I tutor in NYC: http://www.wyzant.com/Tutors/NY/New-York/7948121/#ref=1RKFOZ

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.

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

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.

gmatclubot

Re: Find the number of trailing zeros in the expansion of
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12 Oct 2013, 04:08

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