Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
Re: How many zeroes at the end of 60!? [#permalink]
26 Sep 2010, 12:15
15
This post received KUDOS
Expert's post
5
This post was BOOKMARKED
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end? 6 12 14 42 56
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n
It's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:
According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.
Answer: C.
For more on this issues check Factorials and Number Theory links in my signature.
Re: MGMAT Factorial 700+? [#permalink]
19 Jan 2011, 13:11
shashankp27 wrote:
devide 60! by a 5 as follows:
60/5 + 60/5^2 + .....= 14
Could you explain that a bit more? because 60/5 is 12, and adding 60/5^2 will be 14.4, so going further will result in an even larger number. Also why a 5?
Re: MGMAT Factorial 700+? [#permalink]
19 Jan 2011, 13:20
4
This post received KUDOS
During the test I would simply write down all the numbers between 1 and 60 that end with a 5 or a 0 (5*2 = 10), so we will have: 5, 10 , 15 , 20, 25 (5*5), 30, 35, 40, 45, 50 (5*5*2) , 55, 60. If we count all the numbers we will get 12, BUT we need to remember that 25 = 5*5 so we have 2 zeros and 50 = 5*5*2 so we have 2 more. Therefore, the correct answer is C - 14.
Re: MGMAT Factorial 700+? [#permalink]
19 Jan 2011, 18:13
1
This post received KUDOS
Whenever you want to find the number of zeros in a N! then do the following :
devide N by 5 such that (N/5) + (N/5^2) + (N/5^3) + ..... unless 5^p where p= 1, 2,3 ... is more than N .. Eg : let's say you want to find the number of Zero's in 125! so divide 125/5 = 25 then divide 125/5^2 =125/25= 5 then divide 125/5^3= 125/125=1 , so a total of 25 +5+1 trailing zeros will be present. Always consider the rounded figures .In the original example : 60/5 = 12 60/5^2 = 60/25=2.4 , however you are not concerned with the decimal values here, so take this as 2 next would be 60/5^3 = 60/125 , so this would be (.some number) so stop your division here. Whenever the denominator exceeds numerator , stop the process. Add the values to get the answer.
what would you do if the question asks to find the maximum power of 3 in 50! ?
Re: MGMAT Factorial 700+? [#permalink]
20 Jan 2011, 05:22
1
This post received KUDOS
shashankp27 wrote:
Whenever you want to find the number of zeros in a N! then do the following :
devide N by 5 such that (N/5) + (N/5^2) + (N/5^3) + ..... unless 5^p where p= 1, 2,3 ... is more than N .. Eg : let's say you want to find the number of Zero's in 125! so divide 125/5 = 25 then divide 125/5^2 =125/25= 5 then divide 125/5^3= 125/125=1 , so a total of 25 +5+1 trailing zeros will be present. Always consider the rounded figures .In the original example : 60/5 = 12 60/5^2 = 60/25=2.4 , however you are not concerned with the decimal values here, so take this as 2 next would be 60/5^3 = 60/125 , so this would be (.some number) so stop your division here. Whenever the denominator exceeds numerator , stop the process. Add the values to get the answer.
what would you do if the question asks to find the maximum power of 3 in 50! ?
Ok I understand the trailing zeros now. Ignoring the decimal helps, I was just adding. Make more sense thank you for the explanation.
Maximum power of 3 in 50!? Find the multiples of 3 from 1 to 50. Add them up. 3^21?
Re: How many zeroes at the end of 60!? [#permalink]
08 May 2012, 02:04
Expert's post
Bunuel wrote:
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end? 6 12 14 42 56
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n
It's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:
According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.
Answer: C.
For more on this issues check Factorials and Number Theory links in my signature.
Hope it helps.
a question Bunuel (just for sure): 32/25 is 1.28 ....BUT if the result was for instance a number / another number = 1,764 we round it to 2 (the next integer) ??
I hope to be clear with my question ..... _________________
Re: How many zeroes at the end of 60!? [#permalink]
08 May 2012, 10:03
carcass wrote:
a question Bunuel (just for sure): 32/25 is 1.28 ....BUT if the result was for instance a number / another number = 1,764 we round it to 2 (the next integer) ??
I hope to be clear with my question .....
no. U can check it
say we have 29!
29/5 -29/25=5+1=6
now check it for sure
29 has 25 (two 5's); 20 (one 5); 15(one 5);10(one 5);5(one 5) total 5^6 _________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth
Re: How many zeroes at the end of 60!? [#permalink]
08 May 2012, 10:15
Bunuel wrote:
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end? 6 12 14 42 56
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n
It's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:
According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.
Answer: C.
For more on this issues check Factorials and Number Theory links in my signature.
Hope it helps.
Hello Brunuel Thanks for this great answer but I am not familiar at all with trailing zeros How did you determined the limit to raise the power up to K ? how did you get the K
Re: How many zeroes at the end of 60!? [#permalink]
08 May 2012, 10:18
keiraria wrote:
Hello Brunuel Thanks for this great answer but I am not familiar at all with trailing zeros How did you determined the limit to raise the power up to K ? how did you get the K
Re: How many zeroes at the end of 60!? [#permalink]
09 May 2012, 00:05
Expert's post
carcass wrote:
Bunuel wrote:
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end? 6 12 14 42 56
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n
It's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:
According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.
Answer: C.
For more on this issues check Factorials and Number Theory links in my signature.
Hope it helps.
a question Bunuel (just for sure): 32/25 is 1.28 ....BUT if the result was for instance a number / another number = 1,764 we round it to 2 (the next integer) ??
I hope to be clear with my question .....
We take into account only the quotient of the division, that is 32/5=6.
keiraria wrote:
Hello Brunuel Thanks for this great answer but I am not familiar at all with trailing zeros How did you determined the limit to raise the power up to K ? how did you get the K
BEST regards
keiraria
The last denominator (5^2) must be less than numerator (60).
Re: How many zeroes at the end of 60!? [#permalink]
20 May 2012, 20:03
Bunuel wrote:
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end? 6 12 14 42 56
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n
It's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:
According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.
Answer: C.
For more on this issues check Factorials and Number Theory links in my signature.
Hope it helps.
Thanks for the detailed explanation... new cocnept for me _________________
Best Vaibhav
If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks
During the test I would simply write down all the numbers between 1 and 60 that end with a 5 or a 0 (5*2 = 10), so we will have: 5, 10 , 15 , 20, 25 (5*5), 30, 35, 40, 45, 50 (5*5*2) , 55, 60. If we count all the numbers we will get 12, BUT we need to remember that 25 = 5*5 so we have 2 zeros and 50 = 5*5*2 so we have 2 more. Therefore, the correct answer is C - 14.
Re: How many zeroes at the end of 60!? [#permalink]
18 May 2014, 13:25
Bunuel wrote:
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end? 6 12 14 42 56
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n
It's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:
According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.
Answer: C.
For more on this issues check Factorials and Number Theory links in my signature.
Hope it helps.
Hi Bunuel,
This method makes complete sense but question for you regarding the accounting for "2" part.
You say that "The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero."
How do we know that there aren't more factors of 2 vs. 5? If there were more factors of 2, would we modify the equation to account for powers of 2 in the denominator?
Re: How many zeroes at the end of 60!? [#permalink]
18 May 2014, 20:35
russ9 wrote:
Bunuel wrote:
Orange08 wrote:
If 60! is written out as an integer, with how many consecutive 0’s will that integer end? 6 12 14 42 56
Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation of a number, after which no other digits follow.
125000 has 3 trailing zeros;
The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:
\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n
It's more simple if you look at an example:
How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)
So there are 7 zeros in the end of 32!
The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.
BACK TO THE ORIGINAL QUESTION:
According to above 60! has \(\frac{60}{5}+\frac{60}{25}=12+2=14\) trailing zeros.
Answer: C.
For more on this issues check Factorials and Number Theory links in my signature.
Hope it helps.
Hi Bunuel,
This method makes complete sense but question for you regarding the accounting for "2" part.
You say that "The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero."
How do we know that there aren't more factors of 2 vs. 5? If there were more factors of 2, would we modify the equation to account for powers of 2 in the denominator?
Thanks!
How do we know that there aren't more factors of 2 vs. 5? If there were more factors of 2, would we modify the equation to account for powers of 2 in the denominator?
As Bunuel has shown above, NUMBER of 2's can be found in the same way as we did it for 5
so we have 60/2 = 30 60/4=15 60/8 =7 60/16= 3 60/32=1
So number of 2's in 60! will be : 30+15+7+3+1= 55
Hope it helps _________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
Re: How many zeroes at the end of 60!? [#permalink]
18 May 2014, 21:19
Expert's post
russ9 wrote:
How do we know that there aren't more factors of 2 vs. 5? If there were more factors of 2, would we modify the equation to account for powers of 2 in the denominator?
Thanks!
The point is that you need both a 2 and a 5 to make a 10. If I have 100 2s but only 3 5s, I can make only 3 10s. No number of 2s alone can make a 10. So even if there are many more 2s, they are useless to us because we have limited number of 5s. _________________
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...