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Manager  Joined: 26 Sep 2013
Posts: 190
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]

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4
36 00:00

Difficulty:   95% (hard)

Question Stats: 23% (02:19) correct 77% (02:29) wrong based on 248 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
Verbal Forum Moderator B
Joined: 10 Oct 2012
Posts: 604
Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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5
6
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.
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Intern  Joined: 07 Jan 2013
Posts: 35
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]

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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??
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Joined: 10 Oct 2012
Posts: 604
Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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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.
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Intern  Joined: 03 Jul 2015
Posts: 30
Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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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  S
Joined: 22 Mar 2014
Posts: 120
Location: United States
Concentration: Finance, Operations
GMAT 1: 530 Q45 V20 GPA: 3.91
WE: Information Technology (Computer Software)
For how many different pairs of positive integers (a, b) can  [#permalink]

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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.
Non-Human User Joined: 09 Sep 2013
Posts: 11719
Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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_________________ Re: For how many different pairs of positive integers (a, b) can   [#permalink] 28 Mar 2019, 10:50
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