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For how many different pairs of positive integers (a, b) can
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18 Oct 2013, 16:58
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25% (02:03) correct 75% (02:31) wrong based on 233 sessions
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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
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Re: For how many different pairs of positive integers (a, b) can
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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.
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Re: For how many different pairs of positive integers (a, b) can
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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??
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Re: For how many different pairs of positive integers (a, b) can
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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.
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Re: For how many different pairs of positive integers (a, b) can
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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?



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For how many different pairs of positive integers (a, b) can
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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.



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Re: For how many different pairs of positive integers (a, b) can
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28 Jan 2018, 06:46
Bunuel can you please help with this question Posted from my mobile device
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