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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.

Please follow the link given in my first post.

The # of trailing zeros of 310! is 310/5 + 310/5^2 + 310/5^3 = 62 + 12 + 2 = 76.

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

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10 Jan 2014, 08:58

madn800 wrote:

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.

The formula for the finding the number of zeroes in N! is N/5 + N/5^2 + N/5^3 + ....+ N/5^r where N < 5^(r+1). So for 310, we need to keep continuing for 125 as well so 310/5 + 310/25 + 310/125 = 62 + 12 + 2 = 76 trailing zeroes.
_________________

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

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05 May 2014, 05:24

How does one know that the "something" cannot contribute to adding further trailing zeros? If the "something" captures a 2 and a 5 couldn't it add further 0s?

How does one know that the "something" cannot contribute to adding further trailing zeros? If the "something" captures a 2 and a 5 couldn't it add further 0s?

We used all 5's available, so that "something" won't have any.
_________________

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.

As you mentioned that any extra 2 or 5 left out from any of the factorial will result in One more Trailing Zero. How can we know that there is no extra 2 or 5 left out ? Also if possible please direct me to such a question where we have to take care of extra 2 or 5 being left out .

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.

As you mentioned that any extra 2 or 5 left out from any of the factorial will result in One more Trailing Zero. How can we know that there is no extra 2 or 5 left out ? Also if possible please direct me to such a question where we have to take care of extra 2 or 5 being left out .

And can such a concept be tested in GMAT ?

Thankssss A LOT !

We used all 5's when counting trailing zeros for 20!, 21!, 22!, ..., and 33!. So, there are no more left.

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

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21 Jul 2014, 01:16

All factorials from 20! --> 24! have four 5-factors (5,10,15,20). This means that\(4 * 5 = 20\) All factorials from 25! --> 29! have six 5-factors (5,10,15,20,25) where 25 has two 5 factors. This means that \(6 * 5 = 30\) All factorials from 30! --> 34! have seven 5-factors (5,10,15,20,25,30) where 25 has two 5 factors. This means that \(7 * 4 = 28\)

\(20+30+28=78\) trailing zeroes

Given the \(3!\) exponent (translating to the power of 6), We multiply this power with the number trailing zeroes

Why is this term "something" in the equation above still there? Why is it needed, I don't understand why you put that in there. Could you explain that?
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Why is this term "something" in the equation above still there? Why is it needed, I don't understand why you put that in there. Could you explain that?

Each term has many factors other than the 5s and a corresponding number of 2s. For example, 20! = 1*2*3*4*5*6*7...*20

"something" specifies the rest of the factors (3s, 7s, 11s, some 2s etc) after you segregate the 5s and 2s in the form of 10s. We don't care what is leftover since we just need the number of 0s so whatever is leftover is written as "something". It's just a placeholder and you can very well write "x" or "n" or whatever you please there.
_________________

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

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30 Aug 2016, 16:21

madn800 wrote:

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.

Count the number of fives and the number of two's. A ten is formed as follows => 5 multiplied by 2

P.S => Explanation by Bunuel regarding => trailing zeroes must be some multiple of 6
_________________

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

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19 Oct 2016, 10:19

madn800 wrote:

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.

using your concept 310/5+310/5^2=62+12=74 trailing zeroes

You must take into account the multiples of 5^3 => 310/5^3 = 2

So the correct function would be 310/5 + 310/5^2 + 310/5^3 =62+12+2=76 trailing zeroes

If you get to 625 factorial, you would have to consider 656/5^4

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

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15 Oct 2017, 07:37

feruz77 wrote:

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?

It is not that difficult.. It will take you a minute to solve. 20! is having 4 5's; multiply the same with 6 as you are give 3! as a power. Same should go for 21,22,23 and 24!. 25! will have 2 more 5's adding up to the solution.

So basically. 4*6=24*5= 120 6*6= 36: 36*5=180 7*6=42: 42*4=168 (we are stopping at 33!)

Brunel, this factorial hyperlink is not working.Also some time back I read the formula in terms of 5^k for finding trailing zeros in n!. Couldn't recall. Can you please share?

Brunel, this factorial hyperlink is not working.Also some time back I read the formula in terms of 5^k for finding trailing zeros in n!. Couldn't recall. Can you please share?