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: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
18 Jun 2014, 09:23
Dear Bunuel (or whoever sees this)
in this step : 250 * (5!)^2 = 250 x 120 x 120 = 5^2 x 12 x 12 x 1000, I don't get why you multiply by 1000 and also why is 250 turning to 5^2 , and the 120's turning to 12's?
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
18 Jun 2014, 09:35
Expert's post
sagnik2422 wrote:
Dear Bunuel (or whoever sees this)
in this step : 250 * (5!)^2 = 250 x 120 x 120 = 5^2 x 12 x 12 x 1000, I don't get why you multiply by 1000 and also why is 250 turning to 5^2 , and the 120's turning to 12's?
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
18 Aug 2014, 10:47
Bunuel wrote:
aalriy wrote:
If \(10! - 2*(5!)^2\) is divisible by \(10^n\), what is the greatest value of n?
A. 1 B. 2 C. 3 D. 4 E. 5
We should determine the # of trailing zeros of \(10! - 2*(5!)^2\). Trailing zeros are a sequence of 0s in the decimal representation of a number, after which no other digits follow. For example, 2000 has 3 trailing zeros.
Note that every 5 and 2 in prime factorization will give one more trailing zero.
\(10!=5!*6*7*8*9*10=5!*(2*3)*7*(4*2)*9*(5*2)=5!*(2*3*4*5)*(7*2*9*2)*=252*(5!)^2\); \(252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=5^2*12*12*1,000\) --> 5^2 with two 2-s from 12 will provide with 2 more zeros, so \(10! - 2*(5!)^2\) has total of 5 trailing zeros.
So, max value of n for which above is divisible by 10^n is 5.
Answer: E.
P.S. Final value is \(252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000\).
Hope it's clear.
Hi Bunuel,
Though I understood your approach completely. when I try to simplify the expression myself in a different manner, I don't arrive at the right conclusion. Could you please explain what is the flaw in what I am trying to do :
At this point, we have only 2 0s (contributed by 10^2) . We have 2^5 but no additional 5s (in 2, 81 o 7) Then how do we get 36000000 as the final result. (I verified : 10^2 * 2^5 (2*81*7 -9) = 3,600,000 which means I am not making any calculation errors) I am sure there is something silly that I am missing but I am not able to understand. Please explain.
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
18 Aug 2014, 11:26
Expert's post
1
This post was BOOKMARKED
apsForGmat wrote:
Bunuel wrote:
aalriy wrote:
If \(10! - 2*(5!)^2\) is divisible by \(10^n\), what is the greatest value of n?
A. 1 B. 2 C. 3 D. 4 E. 5
We should determine the # of trailing zeros of \(10! - 2*(5!)^2\). Trailing zeros are a sequence of 0s in the decimal representation of a number, after which no other digits follow. For example, 2000 has 3 trailing zeros.
Note that every 5 and 2 in prime factorization will give one more trailing zero.
\(10!=5!*6*7*8*9*10=5!*(2*3)*7*(4*2)*9*(5*2)=5!*(2*3*4*5)*(7*2*9*2)*=252*(5!)^2\); \(252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=5^2*12*12*1,000\) --> 5^2 with two 2-s from 12 will provide with 2 more zeros, so \(10! - 2*(5!)^2\) has total of 5 trailing zeros.
So, max value of n for which above is divisible by 10^n is 5.
Answer: E.
P.S. Final value is \(252*(5!)^2-2*(5!)^2=250*(5!)^2=250*120*120=1,000*25*12*12=1,000*300*12=12*300,000=3,600,000\).
Hope it's clear.
Hi Bunuel,
Though I understood your approach completely. when I try to simplify the expression myself in a different manner, I don't arrive at the right conclusion. Could you please explain what is the flaw in what I am trying to do :
At this point, we have only 2 0s (contributed by 10^2) . We have 2^5 but no additional 5s (in 2, 81 o 7) Then how do we get 36000000 as the final result. (I verified : 10^2 * 2^5 (2*81*7 -9) = 3,600,000 which means I am not making any calculation errors) I am sure there is something silly that I am missing but I am not able to understand. Please explain.
The point is that 2*81*7 - 9 = 3^2*5^3. 2^5 provides enough 2's for 5^3, which will result in additional 3 0's.
If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
18 Oct 2014, 00:01
1
This post received KUDOS
If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest value of n?
A. 1 B. 2 C. 3 D. 4 E. 5
The equation can be simplified as =>(5!)(10*9*8*7*6*5 - 2*(5!)) =>120(30240-240) =>(10^5) * K (No need to calculate the remaining part as we knew the power of 10)
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
21 Oct 2014, 01:26
usre123 wrote:
i did it this way and fail to see where I'm wrong: 10!/10^n - 2*(5!)^2/10^n
10! contains 2 trailing 0s.... 5!=120 so we have 120*120*2. Thats two zeros. So shouldn't the answer be 4?
We are allowed to split up fractions in this way, aren't we?
Yes, we are allowed to split the fractions.
As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer _________________
If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
21 Oct 2014, 01:50
PareshGmat wrote:
usre123 wrote:
i did it this way and fail to see where I'm wrong: 10!/10^n - 2*(5!)^2/10^n
10! contains 2 trailing 0s.... 5!=120 so we have 120*120*2. Thats two zeros. So shouldn't the answer be 4?
We are allowed to split up fractions in this way, aren't we?
Yes, we are allowed to split the fractions.
As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer
Oh haha. simple logic. Got it, thanks!! So this isn't the right way to solve. Although I tried to do 4/2- 3/2 which gives me 1/2, and (4-3)/2 also gives the same answer. I wonder why that doesnt work here
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
21 Oct 2014, 02:09
usre123 wrote:
PareshGmat wrote:
usre123 wrote:
i did it this way and fail to see where I'm wrong: 10!/10^n - 2*(5!)^2/10^n
10! contains 2 trailing 0s.... 5!=120 so we have 120*120*2. Thats two zeros. So shouldn't the answer be 4?
We are allowed to split up fractions in this way, aren't we?
Yes, we are allowed to split the fractions.
As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer
Oh haha. simple logic. Got it, thanks!! So this isn't the right way to solve. Although I tried to do 4/2- 3/2 which gives me 1/2, and (4-3)/2 also gives the same answer. I wonder why that doesnt work here
Actually in the problem in question, they are asking to count the zero's; that's why its better to calculate the numerator first.
Take this example:
\(\frac{2333}{10} - \frac{333}{10}\) ......... What is the power of 10 in this question... ??
Without solving, there is NO power of 10 in the numerator; however once you solve, its \(10^3\) in the numerator.......
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
21 Oct 2014, 13:05
10! contains 2 trailing 0s.... 5!=120 so we have 120*120*2. Thats two zeros. So shouldn't the answer be 4?
We are allowed to split up fractions in this way, aren't we?[/quote]
Yes, we are allowed to split the fractions.
As far as the computation of this problem is concerned, we require to get rid of the "subtraction" sign first to get the required answer[/quote]
perfect sense, thank you!
Oh haha. simple logic. Got it, thanks!! So this isn't the right way to solve. Although I tried to do 4/2- 3/2 which gives me 1/2, and (4-3)/2 also gives the same answer. I wonder why that doesnt work here[/quote]
Actually in the problem in question, they are asking to count the zero's; that's why its better to calculate the numerator first.
Take this example:
\(\frac{2333}{10} - \frac{333}{10}\) ......... What is the power of 10 in this question... ??
Without solving, there is NO power of 10 in the numerator; however once you solve, its \(10^3\) in the numerator.......
Does it makes sense?[/quote]
makes perfect sense, thank you. is this representative of a 750+ questions? or a 700+ questions
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
10 Mar 2015, 00:36
Given expression : 10!-2(5!)(5!) Since we have to find n in 10^n lets try and write it the given expression as a factor of 10 10!-2x5!x5! 5!(6.7.8.9.10-2x5!)=5!(120.7.4.9-2x5!) since 5!=120 5!x5!(7.4.9-2)=5!x5!x250=14400x250=36x100000=36x10^5 hence greats value of n=5 E _________________
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
02 Sep 2015, 01:56
The trick here is that it is not a product question. Addition and Subtraction can lead to formation of new numbers whose prime factorisation can give trailing zeroes.
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
19 Sep 2015, 07:37
10! - 2*(5!)^2 5!(6*7*8*9*10 -(2*5!)) 5!(6*7*8*9*10- (2*(1*2*3*4*5)) 5!((6*7*8*9*10)-(2*4*6*8*10)) 5!*6*8*10(7*9-(2*4)) 5!*6*8*10(55) 5!*6*8*10*55 1*2*3*4*5 *6*7*8*9*2*5 *11*5 i found only three 5... can some one plz tell me where did i went wrong?Bunuel _________________
kinaare paaon phailane lage hian, nadi se roz mitti kat rahi hai....
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest [#permalink]
19 Sep 2015, 12:02
1
This post received KUDOS
saroshgilani wrote:
10! - 2*(5!)^2 5!(6*7*8*9*10 -(2*5!)) 5!(6*7*8*9*10- (2*(1*2*3*4*5)) 5!((6*7*8*9*10)-(2*4*6*8*10)) 5!*6*8*10(7*9-(2*4)) 5!*6*8*10(55) 5!*6*8*10*55 1*2*3*4*5 *6*7*8*9*2*5 *11*5 i found only three 5... can some one plz tell me where did i went wrong?Bunuel
Highlighted step is wrong. You multiplied 2 with every term in the bracket which are in multiplication, not addition. 5!(6*7*8*9*10- (2*(1*2*3*4*5)) = 5!((6*7*8*9*10)-(10*8*3)) = 5!*10*8*3(9*7*2 - 1) = 12*10*10*3*\(2^3\)(125) = 12*10*10*3*\(2^3\)(\(5^3\)) = 12*10*10*3*\(10^3\) =12*3*\(10^5\)
n=5
gmatclubot
Re: If 10! - 2*(5!)^2 is divisible by 10^n, what is the greatest
[#permalink]
19 Sep 2015, 12:02
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...