Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 26 Sep 2013
Posts: 182
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41

For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
18 Oct 2013, 16:58
Question Stats:
24% (02:13) correct 76% (02:24) wrong based on 310 sessions
HideShow timer Statistics
For how many different pairs of positive integers (a, b) can the fraction \(\frac{2}{15}\) be written as the sum \(\frac{1}{a}\) \(+\) \(\frac{1}{b}\) (A) 4 (B) 5 (C) 8 (D) 9 (E) 10
Official Answer and Stats are available only to registered users. Register/ Login.




Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 569

Re: For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
18 Oct 2013, 23:07
AccipiterQ wrote: For how many different pairs of positive integers (a, b) can the fraction \(\frac{2}{15}\) be written as the sum \(\frac{1}{a}\) \(+\) \(\frac{1}{b}\)
(A) 4
(B) 5
(C) 8
(D) 9
(E) 10 \(\frac{1}{a}+\frac{1}{b} =\frac{2}{15}\) \(I. \frac{1}{a}+\frac{1}{b} =\frac{1}{15}+\frac{1}{15} (a,b) = (15,15)\) \(II. \frac{1}{a}+\frac{1}{b} = \frac{2*2}{15*2}=\frac{1}{30}+\frac{3}{30} (a,b) = (10,30) and (30,10)\) \(III. \frac{1}{a}+\frac{1}{b} = \frac{2*3}{15*3}=\frac{1}{45}+\frac{5}{45} (a,b) = (45,9) and (9,45)\) \(IV. \frac{1}{a}+\frac{1}{b} = \frac{2*4}{15*4}=\frac{3}{60}+\frac{5}{60} (a,b) = (20,12) and (12,20)\) Now, we know that the no of different pairs can only be of the form 2k+1. We already have 7 such combinations. Thus, the only value which will satisfy from the given options is 9. D.
_________________




Intern
Joined: 07 Jan 2013
Posts: 32
Location: India
Concentration: Finance, Strategy
GMAT 1: 570 Q46 V23 GMAT 2: 710 Q49 V38
GPA: 2.9
WE: Information Technology (Computer Software)

Re: For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
20 Oct 2013, 05:46
mau5 wrote: AccipiterQ wrote: For how many different pairs of positive integers (a, b) can the fraction \(\frac{2}{15}\) be written as the sum \(\frac{1}{a}\) \(+\) \(\frac{1}{b}\)
(A) 4
(B) 5
(C) 8
(D) 9
(E) 10 \(\frac{1}{a}+\frac{1}{b} =\frac{2}{15}\) \(I. \frac{1}{a}+\frac{1}{b} =\frac{1}{15}+\frac{1}{15} (a,b) = (15,15)\) \(II. \frac{1}{a}+\frac{1}{b} = \frac{2*2}{15*2}=\frac{1}{30}+\frac{3}{30} (a,b) = (10,30) and (30,10)\) \(III. \frac{1}{a}+\frac{1}{b} = \frac{2*3}{15*3}=\frac{1}{45}+\frac{5}{45} (a,b) = (45,9) and (9,45)\) \(IV. \frac{1}{a}+\frac{1}{b} = \frac{2*4}{15*4}=\frac{3}{60}+\frac{5}{60} (a,b) = (20,12) and (12,20)\) Now, we know that the no of different pairs can only be of the form 2k+1. We already have 7 such combinations. Thus, the only value which will satisfy from the given options is 9. D. Hi Mau, Can you pls explain why no. of combinations can be of form 2k+1??



Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 569

Re: For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
20 Oct 2013, 05:53
adg142000 wrote: Hi Mau,
Can you pls explain why no. of combinations can be of form 2k+1?? The first pair which we get is (15,15) where the value of both a and b is the same. Thus, (15,15) counts as one such pair.However, any other pair will give 2 such resultant pairs : (a,b) and (b,a). Thus, for example, (12,20) and (20,12) are 2 different integer pairs. Thus, the total no of pairs would be 2k+1[this one is coming because of the identical pair of (15,15)] Hope this helps.
_________________



Intern
Joined: 03 Jul 2015
Posts: 27

Re: For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
13 Oct 2015, 06:37
mau5 wrote: AccipiterQ wrote: For how many different pairs of positive integers (a, b) can the fraction \(\frac{2}{15}\) be written as the sum \(\frac{1}{a}\) \(+\) \(\frac{1}{b}\)
(A) 4
(B) 5
(C) 8
(D) 9
(E) 10 \(\frac{1}{a}+\frac{1}{b} =\frac{2}{15}\) \(I. \frac{1}{a}+\frac{1}{b} =\frac{1}{15}+\frac{1}{15} (a,b) = (15,15)\) \(II. \frac{1}{a}+\frac{1}{b} = \frac{2*2}{15*2}=\frac{1}{30}+\frac{3}{30} (a,b) = (10,30) and (30,10)\) \(III. \frac{1}{a}+\frac{1}{b} = \frac{2*3}{15*3}=\frac{1}{45}+\frac{5}{45} (a,b) = (45,9) and (9,45)\) \(IV. \frac{1}{a}+\frac{1}{b} = \frac{2*4}{15*4}=\frac{3}{60}+\frac{5}{60} (a,b) = (20,12) and (12,20)\) Now, we know that the no of different pairs can only be of the form 2k+1. We already have 7 such combinations. Thus, the only value which will satisfy from the given options is 9. D. you just made 4 attempt and then how did u come up with the decision that the aans is 9?



Manager
Joined: 22 Mar 2014
Posts: 98
Location: United States
Concentration: Finance, Operations
GPA: 3.91
WE: Information Technology (Computer Software)

For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
21 Dec 2016, 08:12
Hi, I have a doubt as I think the explanation given above is incomplete. I guess (a,b) can take 9 pairs, so the ans is 9. PFB the pairs 
(a,b) = (15,15) (10,30), (30,10) (12,20), (20,12) (9,45), (45,9) (8,120), (120,8)
Please let me know if you think otherwise.



Intern
Joined: 07 Apr 2018
Posts: 13
Location: India
Concentration: General Management, Finance
GPA: 3.4
WE: Project Management (Consulting)

Re: For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
12 Feb 2020, 11:48
arunavamunshi1988 wrote: Hi, I have a doubt as I think the explanation given above is incomplete. I guess (a,b) can take 9 pairs, so the ans is 9. PFB the pairs 
(a,b) = (15,15) (10,30), (30,10) (12,20), (20,12) (9,45), (45,9) (8,120), (120,8)
Please let me know if you think otherwise. the guy has already given 7 solution and said as only the same pair 15,15 is there, so the next solution will be in pairs, hence all the solution will be in count like 2k+1 above 7 only 9 answer choice is present ,as it cannot be 10 or 8 (they are even). hence without solving for further soln we can pick D Kudos if this helped



Manager
Joined: 21 Aug 2019
Posts: 118

Re: For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
23 Feb 2020, 23:23
Can anyone explain this further?
_________________
I don't believe in giving up!



Intern
Joined: 04 Feb 2018
Posts: 2

Re: For how many different pairs of positive integers (a, b) can
[#permalink]
Show Tags
26 Feb 2020, 11:37
In addition, (a+b) must be even (odd+odd or even+even) due to ab=(15(a+b))/2




Re: For how many different pairs of positive integers (a, b) can
[#permalink]
26 Feb 2020, 11:37






