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Math Expert V
Joined: 02 Sep 2009
Posts: 59229
Re: Find the number of trailing zeros in the expansion of  [#permalink]

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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.

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

Questions to practice:
if-n-is-the-greatest-positive-integer-for-which-2n-is-a-fact-144694.html
what-is-the-largest-power-of-3-contained-in-103525.html
if-n-is-the-product-of-all-positive-integers-less-than-103218.html
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html
if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187.html
if-p-is-the-product-of-integers-from-1-to-30-inclusive-137721.html
what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html
if-6-y-is-a-factor-of-10-2-what-is-the-greatest-possible-129353.html
if-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html
if-p-is-a-natural-number-and-p-ends-with-y-trailing-zeros-108251.html
if-73-has-16-zeroes-at-the-end-how-many-zeroes-will-147353.html
find-the-number-of-trailing-zeros-in-the-expansion-of-108249.html
how-many-zeros-are-the-end-of-142479.html
how-many-zeros-does-100-end-with-100599.html
find-the-number-of-trailing-zeros-in-the-product-of-108248.html
if-60-is-written-out-as-an-integer-with-how-many-consecuti-97597.html
if-n-is-a-positive-integer-and-10-n-is-a-factor-of-m-what-153375.html
if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html

Hope it helps.
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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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.

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

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Intern  Joined: 17 Feb 2014
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 59229
Re: Find the number of trailing zeros in the expansion of  [#permalink]

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renedo87 wrote:
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.
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Re: trailing zeros question (complicated one)  [#permalink]

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Bunuel 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?

# 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.

Hello Bunuel

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 ! Math Expert V
Joined: 02 Sep 2009
Posts: 59229
Re: trailing zeros question (complicated one)  [#permalink]

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niyantg wrote:
Bunuel 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?

# 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.

Hello Bunuel

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.

When using trailing zero approach shown in my post we use all 5's to get the number of 0's, so I don't know what kind of problems you are talking about. As for similar questions check here: find-the-number-of-trailing-zeros-in-the-expansion-of-108249-20.html#p1277367

Hope it helps.
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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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

$$78* 6 =468$$

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

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20! to 24! give us 4*5=20 zeros
25! to 29! give us 6*5=30 zeros
30! to 33! give us 7*4=28 zeros

20+30+28=78 zeros in ^3! means 78*6=468, so answer is A
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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Bunuel wrote:
feruz77 wrote:
Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.

# 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.

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

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reto wrote:
Bunuel wrote:
feruz77 wrote:
Find the number of trailing zeros in the expansion of (20!*21!*22! ……… *33!)^3!.

# 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.

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.
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GRE 1: Q169 V154 Re: Find the number of trailing zeros in the expansion of  [#permalink]

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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.

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

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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.

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

I hope this helps
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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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!)

Add them up and your solution arrives. Intern  B
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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Bunuel wrote:
Sachin9 wrote:
do we get such questions on gmat?

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

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?

Posted from my mobile device
Math Expert V
Joined: 02 Sep 2009
Posts: 59229
Re: Find the number of trailing zeros in the expansion of  [#permalink]

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Saps wrote:
Bunuel wrote:
Sachin9 wrote:
do we get such questions on gmat?

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

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?

Posted from my mobile device

Here it is again:

For more check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope this helps.
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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Thanks a lot Brunel, for super quick help Posted from my mobile device
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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Shortest solution:

We need to find trailing zeroes in (a)^3! or (a)^6

The trailing zeros will be of form 10^6n

B, D, E are odd numbers hence out.
C is not a multiple of 3 hence out.

A is a multiple of 2 and 3 and so 6. Hence correct answer.

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

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Bunuel 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?

# 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

Amazing explanation but I couldn't understand why did you add 1 to the zeros of 25 to 33? You didn't do that here https://gmatclub.com/forum/how-many-zer ... 00599.html. What did I miss?
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Re: Find the number of trailing zeros in the expansion of  [#permalink]

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The number of trailing zeroes at the end of a number depends on the highest power of 10 in that number. This in turn depends on the highest power of 5 in that number.

The highest power of 5 in 20! is 4. Therefore, the highest power of 5 in all numbers between 20 and 25 will also be 4. This gives us $$5^{20}$$.

The highest power of 5 in 25! is 6 (remember, 25 gives us $$5^2$$). So, the highest power of 5 in all numbers between 25 and 30 will also be 6. This gives us $$5^{30}$$.
The highest power of 5 in 30! is 7. The highest power of 5 in all numbers from 30 to 33 will also be 7. This gives us $$5^{28}$$.
Since the expression inside the bracket is a product, the highest power of 5 inside the brackets will be $$5^{20}$$ * $$5^{30}$$ * $$5^{28}$$, which gives $$5^{78}$$. The exponent is 3! which is nothing but 6.

$$(5^{78})^6$$ = $$5^{468}$$. Clearly, this will be the highest power of 10 and hence the number of trailing zeroes at the end of the number given. The correct answer option is A.
Although this is not exactly a GMAT type of question, it’s still worth investing your time for consolidating the concepts.
Hope this helps!
_________________ Re: Find the number of trailing zeros in the expansion of   [#permalink] 10 Jul 2019, 07:28

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