Find all School-related info fast with the new School-Specific MBA Forum

It is currently 23 Oct 2014, 10:10

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

M12 # 21

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
2 KUDOS received
Intern
Intern
avatar
Joined: 19 Jun 2008
Posts: 20
Followers: 0

Kudos [?]: 12 [2] , given: 0

M12 # 21 [#permalink] New post 13 Nov 2008, 12:47
2
This post received
KUDOS
If x is a positive integer, is the number of its divisors smaller than 2* \lceil \sqrt{x} \rceil - 1 ?

\lceil a \rceil is defined as the smallest integer that is not less than a .

1. x is not a square of an integer
2. x is prime

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Sufficiency of S2 is easier to prove, and is a good starting point. A prime number only has 2 divisors. For the smallest prime x=2 , the equation gives 2*2-1=3 and 2 < 3 is satisfied. For larger primes, it will also be satisfied, as the right hand side grows, but the left hand side (the number of divisors) remains 2.

Proving the sufficiency of S1 is a lot more difficult and may not be worth the time. If you attempt it and see the time running out, consider doing a guess. It will be a good educated guess with a 50% chance of being right, as the sufficiency of S2 eliminates the choices A, C, E.

To prove that S1 is sufficient, group all divisors of x into (a, \frac{x}{a}) pairs. You have to count the pairs for which a \lt \frac{x}{a} (S1 says that a \ne \frac{x}{a} ) and double the result to get the number of divisors.

Consider the worst possible case when x is divisible by 2, and 3, and 4, and 5, and so on, then the pairs will be (1, x) , (2, \frac{x}{2}) , (3, \frac{x}{3}) and up until (s, \frac{x}{s}) where s is the greatest integer less than \sqrt{x} . We can express it as s = \lceil \sqrt{x} \rceil - 1 . Then, if the number of divisors of x is N(x) , we obtain N(x) \le 2s = 2\lceil\sqrt{x}\rceil - 2 \lt 2\lceil\sqrt{x}\rceil - 1 .
The correct answer is D.



Would someone feeling up to it please explain the explanation of statement 1 in plain English?
Kaplan GMAT Prep Discount CodesKnewton GMAT Discount CodesVeritas Prep GMAT Discount Codes
CEO
CEO
User avatar
Joined: 29 Aug 2007
Posts: 2500
Followers: 54

Kudos [?]: 512 [0], given: 19

Re: M12 # 21 [#permalink] New post 13 Nov 2008, 20:30
snowy2009 wrote:
If x is a positive integer, is the number of its divisors smaller than 2* \lceil \sqrt{x} \rceil - 1 ?

\lceil a \rceil is defined as the smallest integer that is not less than a .

1. x is not a square of an integer
2. x is prime

(C) 2008 GMAT Club - m12#21

Sufficiency of S2 is easier to prove, and is a good starting point. A prime number only has 2 divisors. For the smallest prime x=2 , the equation gives 2*2-1=3 and 2 < 3 is satisfied. For larger primes, it will also be satisfied, as the right hand side grows, but the left hand side (the number of divisors) remains 2.

Proving the sufficiency of S1 is a lot more difficult and may not be worth the time. If you attempt it and see the time running out, consider doing a guess. It will be a good educated guess with a 50% chance of being right, as the sufficiency of S2 eliminates the choices A, C, E.

To prove that S1 is sufficient, group all divisors of x into (a, \frac{x}{a}) pairs. You have to count the pairs for which a \lt \frac{x}{a} (S1 says that a \ne \frac{x}{a} ) and double the result to get the number of divisors.

Consider the worst possible case when x is divisible by 2, and 3, and 4, and 5, and so on, then the pairs will be (1, x) , (2, \frac{x}{2}) , (3, \frac{x}{3}) and up until (s, \frac{x}{s}) where s is the greatest integer less than \sqrt{x} . We can express it as s = \lceil \sqrt{x} \rceil - 1 . Then, if the number of divisors of x is N(x) , we obtain N(x) \le 2s = 2\lceil\sqrt{x}\rceil - 2 \lt 2\lceil\sqrt{x}\rceil - 1 .
The correct answer is D.

Would someone feeling up to it please explain the explanation of statement 1 in plain English?


You put big effort to explain it. But I am wondering whether the question really needs any statement to answer it.

Just wanted to say that, for me, if a = 5, \lceil 5 \rceil should be 6. If this not true what exactly meant by \lceil a \rceil?
_________________

Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html
Math: new-to-the-math-forum-please-read-this-first-77764.html
Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

CIO
CIO
avatar
Joined: 02 Oct 2007
Posts: 1218
Followers: 87

Kudos [?]: 691 [0], given: 334

GMAT ToolKit User
Re: M12 # 21 [#permalink] New post 27 Nov 2008, 06:10
This is not very important but I think you're wrong. It's said that
snowy2009 wrote:
\lceil a \rceil is defined as the smallest integer that is not less than a .

So if you have \lceil 5 \rceil, it equals 5, not 6 as you said. It has to be the SMALLEST integer which is not less than. In your example 5 is not less than 5. It is the smallest integer from all possible integers that are not less than 5. This might be easier to explain as rounding but in a special way. Even if the number is slightly greater than some integer (say, 6.05), \lceil 6.05 \rceil will equal 7. Whereas rounding 6.05 to units results into 6, \lceil 6.05 \rceil results into 7. In other words, the number after this operation is the next greater integer on the number line. I know it's confusing. Just wanted to check that you're not mistaken.

As to the confusing explanation for this problem, I'm not sure I'll be able to explain better. These concepts won't be tested in the real GMAT.
GMAT TIGER wrote:
You put big effort to explain it. But I am wondering whether the question really needs any statement to answer it.

Just wanted to say that, for me, if a = 5, \lceil 5 \rceil should be 6. If this not true what exactly meant by \lceil a \rceil?

_________________

Welcome to GMAT Club! :)
Facebook TwitterGoogle+LinkedIn
Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Please read this before posting in GMAT Club Tests forum
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

Get the best GMAT Prep Resources with GMAT Club Premium Membership

1 KUDOS received
Manager
Manager
User avatar
Joined: 22 Jul 2009
Posts: 192
Followers: 4

Kudos [?]: 196 [1] , given: 18

Re: M12 # 21 [#permalink] New post 04 Oct 2009, 14:05
1
This post received
KUDOS
I tested a few numbers.

1 -> 2* \lceil \sqrt{x} \rceil - 1 = 1 -> factors of 1 = 1
4 -> 2* \lceil \sqrt{x} \rceil - 1 = 3 -> factors of 4 = 3
9 -> 2* \lceil \sqrt{x} \rceil - 1 = 5 -> factors of 9 = 3
16 -> 2* \lceil \sqrt{x} \rceil - 1 = 7 -> factors of 16 = 4

2 -> 2* \lceil \sqrt{x} \rceil - 1 = 3 -> factors of 2 = 2
3 -> 2* \lceil \sqrt{x} \rceil - 1 = 3 -> factors of 3 = 2
5 -> 2* \lceil \sqrt{x} \rceil - 1 = 5 -> factors of 5 = 2
6 -> 2* \lceil \sqrt{x} \rceil - 1 = 5 -> factors of 6 = 4

It seems that number of divisors is smaller for all positive integers with the exceptions of 1 and 4.
_________________

Please kudos if my post helps.

Intern
Intern
avatar
Joined: 15 Nov 2009
Posts: 16
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: M12 # 21 [#permalink] New post 18 Apr 2010, 17:58
mmm, I understood the answer but takes a lot of time. Couldn't do it myself.
Manager
Manager
avatar
Joined: 15 Sep 2009
Posts: 139
Followers: 1

Kudos [?]: 16 [0], given: 2

Re: M12 # 21 [#permalink] New post 18 Apr 2010, 22:56
I think the best starting point for this q could be to start with option 2.

We can answer the query considering the prime nos.

then comes statement 1.I did the question by taking some examples like 12,36 which has lot of divisors.

My take would be option D
Manager
Manager
User avatar
Status: On my way !!!
Joined: 11 Apr 2011
Posts: 115
Location: France
Concentration: Operations, Strategy
GMAT 1: 760 Q50 V44
GPA: 3.1
WE: Manufacturing and Production (Energy and Utilities)
Followers: 2

Kudos [?]: 14 [0], given: 4

Re: M12 # 21 [#permalink] New post 20 Apr 2011, 04:32
The second statement is easy to test as prime numbers will have only two divisors and 2*[sqrt[x] - 1 is always greater than or equal to 3 for prime numbers.

For the first statement I put in some smart numbers to (10, 25) to see that it holds true

So my best guess would be (D)

--------------------------------
http://pyarapopat.wordpress.com

Seeking kudos :)
_________________

http://pyarapopat.wordpress.com

SVP
SVP
avatar
Joined: 16 Nov 2010
Posts: 1691
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 31

Kudos [?]: 299 [0], given: 36

Premium Member Reviews Badge
Re: M12 # 21 [#permalink] New post 20 Apr 2011, 04:51
(1) x = 2,3,6, 8

2 * [root(2)] - 1 = 2 * [1.41] - 1 = 2 *2 - 1 = 3

2 * [root(8)] - 1 = 2 * [2.somthing] - 1 = 2 *3 - 1 = 5

Sufficient

(2)

x = 11 (One more prime example)

2 * [root(11)] - 1 = 2 * [3.something] - 1 = 2 *4 - 1 = 7

Sufficient

Answer - D
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 15 Apr 2011
Posts: 65
Location: Bangalore India
Schools: LBS, HBS, ISB, Kelloggs, INSEAD
Followers: 2

Kudos [?]: 4 [0], given: 2

Re: M12 # 21 [#permalink] New post 20 Apr 2011, 05:10
I did it by plugging in the number for both the statements and found out that D is the answer.
_________________

Thanks,
AM

Manager
Manager
avatar
Joined: 24 Aug 2008
Posts: 232
Location: India
WE 1: 3.75 IT
WE 2: 1.0 IT
Followers: 2

Kudos [?]: 32 [0], given: 5

Re: M12 # 21 [#permalink] New post 20 Apr 2011, 22:27
Plugging number was useful and easy way to solve it..
Answer is D.
_________________

Cheers,
Varun


If you like my post, give me KUDOS!!

Director
Director
avatar
Joined: 01 Feb 2011
Posts: 768
Followers: 14

Kudos [?]: 58 [0], given: 42

CAT Tests
Re: M12 # 21 [#permalink] New post 23 Apr 2011, 13:26
1. Sufficient.

plugging in numbers we can see this is sufficient

2. Sufficient.

prime number will only have 2 divisors - 1 and the number itself

min value of 2 *[sqrt(x}] -1 is 3

hence this is always true.

Answer is D.
Manager
Manager
avatar
Joined: 20 Jan 2011
Posts: 65
Followers: 1

Kudos [?]: 1 [0], given: 8

Re: M12 # 21 [#permalink] New post 24 Apr 2011, 11:40
Extremely difficult one
Intern
Intern
avatar
Status: Preparing again for second attempt....
Joined: 11 Dec 2010
Posts: 23
WE 1: 6 years
Followers: 0

Kudos [?]: 9 [0], given: 2

Re: M12 # 21 [#permalink] New post 24 Apr 2012, 07:53
Statement 1: x isnt a square of an integer.

I thought of attempting this statement this way.
Since x is a positive integer, it can take values 1,2,3,4....so on. If x were 1, it will have to be a square of an integer, which is 1 itself. Therefore x cannot be 1. Let's say x were 2. This is possible since 2 isnt a square of an integer. So the minimum value x can take is 2. Lets put 2 in the given equation now.

2 x [sqrt2] - 1 = 2 x 2 - 1 = 4 - 1 = 3 > No. of factors of 2. Thus sufficient. I am assuming that all integers that are not a square of any integer and are on the right side of 2 on the number line would follow the same. I ain't very sure about this though. please comment.
Manager
Manager
User avatar
Status: Married
Joined: 30 Oct 2010
Posts: 62
Location: India
Concentration: Technology, Social Entrepreneurship
GMAT 1: 710 Q49 V38
GPA: 2.9
WE: Information Technology (Computer Software)
Followers: 2

Kudos [?]: 14 [0], given: 40

Re: M12 # 21 [#permalink] New post 24 Apr 2012, 19:28
I like the first advice as the best...do the s-2 part, and then guess with 50% chances..

Seriously, in a timed environment as a GMAT, do we get these kinda questions? and if we do, does'nt make any sense solve it through.
_________________

KUDOS-ing does'nt cost you anything, but might just make someone's day!!!

"Musings of a Muddled Muggle Mind"
http://felinesmile.blogspot.in/
...comments Welcome :)

2 KUDOS received
Current Student
avatar
Joined: 09 Mar 2012
Posts: 97
Location: India
GMAT 1: 740 Q50 V39
GPA: 3.4
WE: Business Development (Manufacturing)
Followers: 2

Kudos [?]: 17 [2] , given: 12

Reviews Badge
Re: M12 # 21 [#permalink] New post 12 May 2012, 23:21
2
This post received
KUDOS
Just came across this thread. Let me try

I dont think anyone has problem with the statement 2.

For statement 1:

The maximum no. of factors any integer can have is 2*\sqrt{x} -1.
This is because any integer x can be expressed a the product of 2 factors in the following ways: (1,x), (2,x/2), (3, x/3)........ (\sqrt{x},\sqrt{x}) all of which add up to 2\sqrt{x} -1 no of factors. (Note that this is the best possible case, we cannot have more factors than this)
(-1 because in case of the last term, \sqrt{x} will be counted twice, if it exists)

For eg: 36: No. of factors = (1,36), (2,18),(3,12),(4,9),(6,6) === add up to 9 factors which is less than 11 (2\sqrt{36}-1)
Note that we counted 6 only once.

From statement 1: is x is not a square of an integer, than the ceiling function of x is greater than square root of x
Is is equal only iff x is a perfect square.
for eg: ceiling function of 12 is 4, which is greater than \sqrt{12}.

From this it follows that the no. of factors is necessarily less than 2*(celing fn of x)-1

PS: Sorry i dont know how to post the image of the step function, it is also known as the ceiling function of x.
Current Student
avatar
Joined: 09 Mar 2012
Posts: 97
Location: India
GMAT 1: 740 Q50 V39
GPA: 3.4
WE: Business Development (Manufacturing)
Followers: 2

Kudos [?]: 17 [0], given: 12

Reviews Badge
Re: M12 # 21 [#permalink] New post 19 May 2012, 23:01
If you know this concept, it wont take you more than 20 secs to solve the problem. Now I know, how easy it is to understand a concept but how difficult to teach the same to some one else.I'm probably not the best person to explain this concept.

Unless I cant find a way to solve a question algebraically, I normally avoid substituting values, for there are numerous cases & you miss one, chances are that you might zero down on the wrong answer.

I believe that in a timed & a stressful environment, whatever method suits you to zero in on the right answer is the best way to solve the problem. So, if you cannot find a way to solve the question under 2 mins, go ahead with substitution/calculated guessing. To each his own.

Maybe bb can come up with a better explanation to this problem ( maybe??......of course he will :-D )
Current Student
avatar
Joined: 09 Mar 2012
Posts: 97
Location: India
GMAT 1: 740 Q50 V39
GPA: 3.4
WE: Business Development (Manufacturing)
Followers: 2

Kudos [?]: 17 [0], given: 12

Reviews Badge
Re: M12 # 21 [#permalink] New post 19 May 2012, 23:18
Also, pls note that I was just trying to explain how to understand/solve the problem, and not show the most optimum & fastest way to get to the correct answer. Of course, the latter is what you should aim during the gmat, but it is always beneficial, especially during your preps to understand how to approach each question, rather than just mark the correct answer without understanding the concept underlying the question.

PS:
I didn't get your question you asked. What i mean by the celing function is:
the smallest integer not less than a, i.e, ceiling function of 5.2 == 6, ceiling function of 6 == 6.

Hope that helps.
Senior Manager
Senior Manager
User avatar
Joined: 13 Jan 2012
Posts: 303
Weight: 170lbs
GMAT 1: 730 Q48 V42
GMAT 2: 740 Q48 V42
WE: Analyst (Other)
Followers: 9

Kudos [?]: 75 [0], given: 37

Re: M12 # 21 [#permalink] New post 25 Apr 2013, 11:53
Excited to see Bunuel's response to this question.
Manager
Manager
avatar
Joined: 05 Nov 2012
Posts: 72
Concentration: International Business, Operations
Schools: Foster '15 (S)
GPA: 3.65
Followers: 1

Kudos [?]: 48 [0], given: 8

GMAT ToolKit User
Number of divisors [#permalink] New post 29 Apr 2013, 17:24
If x is a positive integer, is the number of its divisors smaller than 2*<sqrt{x}> - 1 ?

<a> is defined as the smallest integer that is not less than a .

1. x is not a square of an integer
2. x is prime

Source: m12-72874.html

Bunuel any help on this - how to go about solving this question
_________________

___________________________________________
Consider +1 Kudos if my post helped

Expert Post
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 10 Oct 2012
Posts: 628
Followers: 43

Kudos [?]: 598 [0], given: 135

Premium Member
Re: Number of divisors [#permalink] New post 29 Apr 2013, 21:49
Expert's post
pikachu wrote:
If x is a positive integer, is the number of its divisors smaller than 2*<sqrt{x}> - 1 ?

<a> is defined as the smallest integer that is not less than a .

1. x is not a square of an integer
2. x is prime

Source: m12-72874.html

Bunuel any help on this - how to go about solving this question



We know that 2*<sqrt{x}> - 1 will always take on odd-values. Let x be a perfect square and x = a^{2k}, where a is a prime number. For any value of k, x will have an odd number of factors. It is for this very reason that we are given x is not a square of an integer, so that the number of factors of x and any value of 2*<sqrt{x}> - 1 don't equal each other, as the latter can only take on odd values. By design from F.S 1, x can not have any even powers and thus, we will always get even no of factors for x. If they had not mentioned this point, we would not be able to give a definitive answer from F.S 1.

So we know that if not anything, 2*<sqrt{x}> - 1 and no of divisors for x WILL never be equal.Thus, if we can prove that there is a general rule where either the no of divisors are ALWAYS less than 2*<sqrt{x}> - 1 OR are ALWAYS more than 2*<sqrt{x}> - 1 , we will get a definitive answer.

No of factors for x can only be even:

Minimum x with 2 factors = 2 Value of 2*<sqrt{2}> - 1= 3.
Minimum x with 4 factors = 6 Value of 2*<sqrt{6}> - 1 = 5.
Minimum x with 6 factors = 12 Value of 2*<sqrt{12}> - 1 = 7.
Minimum x with 8 factors = 24 Value of 2*<sqrt{24}> - 1 = 9. And so on..

What we get is that if for 12, the value of 2*<sqrt{12}> - 1 is 9, then for numbers that have 6 factors and are bigger than 12 say 28, the value of 2*<sqrt{28}> - 1 will be even more than 9(11). Thus, we can say that the no of factors will always be less than the value of 2*<sqrt{x}> - 1 .

If anyone has a doubt as to whether this pattern is uniform, notice that the no of divisors and the value of 2*<sqrt{x}> - 1 are even and odd respectively. Thus, assuming that some value of x with say 8 factors has 2*<sqrt{x}> - 1 as 7 or 5 or 3,it won't be possible because we have proved that the MINIMUM value of2*<sqrt{x}> - 1 for any x with 8 factors is 9.

From F.S 2, it is very clear that the no of factors is only 2 for any prime value of x and the minimum value for 2*<sqrt{2}> - 1 is 3.Sufficient.
_________________

All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

Re: Number of divisors   [#permalink] 29 Apr 2013, 21:49
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic M12-21 Bunuel 1 15 Sep 2014, 23:47
10 Experts publish their posts in the topic m12, #29 ritula 14 16 Nov 2008, 22:22
6 Experts publish their posts in the topic m12 q4 CrushTheGMAT 16 13 Sep 2008, 12:53
23 Experts publish their posts in the topic M12 Q17 durgesh79 29 09 Aug 2008, 06:21
19 Experts publish their posts in the topic M12-10 sondenso 21 14 May 2008, 02:17
Display posts from previous: Sort by

M12 # 21

  Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: WoundedTiger, Bunuel



cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.