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How many different pairs of positive integers (a, b) satisfy the equat
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01 Nov 2019, 05:36
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How many different pairs of positive integers (a, b) satisfy the equation \(\frac{1}{a}+\frac{1}{b}=\frac{34}{51}\) ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
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Re: How many different pairs of positive integers (a, b) satisfy the equat
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Updated on: 01 Nov 2019, 09:41
Bunuel wrote: How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/51 ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level QuestionsGIVEN: 1/a + 1/b = 34/51 Simplify right side: 1/a + 1/b = 2/3 Eliminate fractions by multiplying both sides by 3ab to get: 3b + 3a = 2ab Rewrite as: 2ab  3a  3b = 0At this point, it would be nice to be able to factor the above equation. ASIDE: Notice that (2a  3)(b  1.5) = 2ab  3a  3b + 4.5, which looks A LOT like our above equation. So, take: 2ab  3a  3b = 0Add 4.5 to both sides to get: 2ab  3a  3b + 4.5= 4.5 Factor: (2a  3)(b  1.5) = 4.5 Multiply both sides by 2 to get: (2a  3)(2b  3) = 9 The KEY here is that a and b are POSITIVE INTEGERS. This means (2a  3) and (2b  3) are integers. So, we need only examine the different ways that the product 2 INTEGERS can equal 9 We have 3 cases: case i: (1)(9) = 9 In other words, (2a  3) = 1 and (2b  3) = 9 Solve to get: a = 2 & b = 6case ii: (9)(1) = 9 In other words, (2a  3) = 9 and (2b  3) = 1 Solve to get: a = 6 & b = 2case iii: (3)(3) = 9 In other words, (2a  3) = 3 and (2b  3) = 3 Solve to get: a = 3 & b = 3There are 3 solutions Answer: B Cheers, Brent
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Originally posted by GMATPrepNow on 01 Nov 2019, 07:20.
Last edited by GMATPrepNow on 01 Nov 2019, 09:41, edited 2 times in total.



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Re: How many different pairs of positive integers (a, b) satisfy the equat
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Updated on: 01 Nov 2019, 09:12
Bunuel wrote: How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/51 ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level Questionsgiven 1/a+1/b = 2/3 or say a+b/ab = 2/3 3(a+b)=2ab (6, 2) (3, 3) 2,6 3such pairs possible IMO B
Originally posted by Archit3110 on 01 Nov 2019, 07:42.
Last edited by Archit3110 on 01 Nov 2019, 09:12, edited 3 times in total.



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Re: How many different pairs of positive integers (a, b) satisfy the equat
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Updated on: 01 Nov 2019, 09:17
Bunuel wrote: How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/51 ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level Questions1/a + 1/b = 34/51 > (a + b)/ab = 34/51 = 2/3 > 3a + 3b = 2ab > a(2b  3) = 3b > a = 3b/(2b  3) b = 2, a = 3*2/(4  3) = 6 > (a, b) = (6, 2) b = 3, a = 3*3/(6  3) = 3 > (a, b) = (3, 3) b = 6, a = 3*6/(12  3) = 2 > (a, b) = (2, 6) > 3 different pairs are possible IMO Option B
Originally posted by Dillesh4096 on 01 Nov 2019, 08:53.
Last edited by Dillesh4096 on 01 Nov 2019, 09:17, edited 1 time in total.



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Re: How many different pairs of positive integers (a, b) satisfy the equat
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01 Nov 2019, 09:00
GMATPrepNow wrote: Bunuel wrote: How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/51 ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level QuestionsGIVEN: 1/a + 1/b = 34/51 Simplify right side: 1/a + 1/b = 2/3 Eliminate fractions by multiplying both sides by 3ab to get: 3b + 3a = 2Factor: 3(b + a) = 2 If a and b are POSITIVE INTEGERS, then (b+a) ≥ 2, so it's impossible to have any solutions to the equation 3(b + a) = 2 Answer: E Cheers, Brent Hi GMATPrepNow, Pls See the highlighted part, you should be getting 3b + 3a = 2 ab



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Re: How many different pairs of positive integers (a, b) satisfy the equat
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01 Nov 2019, 09:11
Dillesh4096 you have mentioned 6,2 twice** Dillesh4096 wrote: Bunuel wrote: How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/51 ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level Questions1/a + 1/b = 34/51 > (a + b)/ab = 34/51 = 2/3 > 3a + 3b = 2ab > a(2b  3) = 3b > a = 3b/(2b  3) b = 2, a = 3*2/(4  3) = 6 > (a, b) = (6, 2) b = 3, a = 3*3/(6  3) = 3 > (a, b) = (3, 3) b = 6, a = 3*6/(12  3) = 2 > (a, b) = (6, 2) > 3 different pairs are possible IMO Option B Posted from my mobile device



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Re: How many different pairs of positive integers (a, b) satisfy the equat
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01 Nov 2019, 09:15
Archit3110 wrote: Dillesh4096 you have mentioned 6,2 twice** Dillesh4096 wrote: Bunuel wrote: How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/51 ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level Questions1/a + 1/b = 34/51 > (a + b)/ab = 34/51 = 2/3 > 3a + 3b = 2ab > a(2b  3) = 3b > a = 3b/(2b  3) b = 2, a = 3*2/(4  3) = 6 > (a, b) = (6, 2) b = 3, a = 3*3/(6  3) = 3 > (a, b) = (3, 3) b = 6, a = 3*6/(12  3) = 2 > (a, b) = (6, 2) > 3 different pairs are possible IMO Option B Posted from my mobile device Thanks Archit, Mistyped. Edited



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Re: How many different pairs of positive integers (a, b) satisfy the equat
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01 Nov 2019, 09:33
Dillesh4096 wrote: Hi GMATPrepNow, Pls See the highlighted part, you should be getting 3b + 3a = 2 abHa! I was wondering what made the question so "challenging"!! Looks like I need another cup of coffee. I've edited my solution. Thanks for the heads up! KUDOS for you! Cheers, Brent
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Re: How many different pairs of positive integers (a, b) satisfy the equat
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26 Dec 2019, 06:29
GMATPrepNow wrote: Bunuel wrote: How many different pairs of positive integers (a, b) satisfy the equation 1/a+1/b=34/51 ? A. 6 B. 3 C. 2 D. 1 E. 0 Are You Up For the Challenge: 700 Level Questions: 700 Level QuestionsGIVEN: 1/a + 1/b = 34/51 Simplify right side: 1/a + 1/b = 2/3 Eliminate fractions by multiplying both sides by 3ab to get: 3b + 3a = 2ab Rewrite as: 2ab  3a  3b = 0At this point, it would be nice to be able to factor the above equation. ASIDE: Notice that (2a  3)(b  1.5) = 2ab  3a  3b + 4.5, which looks A LOT like our above equation. So, take: 2ab  3a  3b = 0Add 4.5 to both sides to get: 2ab  3a  3b + 4.5= 4.5 Factor: (2a  3)(b  1.5) = 4.5 Multiply both sides by 2 to get: (2a  3)(2b  3) = 9 The KEY here is that a and b are POSITIVE INTEGERS. This means (2a  3) and (2b  3) are integers. So, we need only examine the different ways that the product 2 INTEGERS can equal 9 We have 3 cases: case i: (1)(9) = 9 In other words, (2a  3) = 1 and (2b  3) = 9 Solve to get: a = 2 & b = 6case ii: (9)(1) = 9 In other words, (2a  3) = 9 and (2b  3) = 1 Solve to get: a = 6 & b = 2case iii: (3)(3) = 9 In other words, (2a  3) = 3 and (2b  3) = 3 Solve to get: a = 3 & b = 3There are 3 solutions Answer: B Cheers, Brent Hi brent, How have you factored out the equation ? How did you get 4.5 ???? Regards



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Re: How many different pairs of positive integers (a, b) satisfy the equat
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26 Dec 2019, 08:20
ShankSouljaBoi wrote: Hi brent,
How have you factored out the equation ? How did you get 4.5 ????
Regards Good question. I started with 2ab  3a  3b = 0I started with (2a )(b ), which allows me to get 2ab when we expand. From here, it's just a matter of testing a few values. In order to get 3a (after expanding), let's try: (2a )(b  1.5 ) In order to get 3b (after expanding), let's try: (2a  3)(b  1.5 ) When we expand this, we get: 2ab  3a  3b + 4.5 Does that help? Cheers, Brent
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Re: How many different pairs of positive integers (a, b) satisfy the equat
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29 Dec 2019, 05:11
GMATPrepNow wrote: ShankSouljaBoi wrote: Hi brent,
How have you factored out the equation ? How did you get 4.5 ????
Regards Good question. I started with 2ab  3a  3b = 0I started with (2a )(b ), which allows me to get 2ab when we expand. From here, it's just a matter of testing a few values. In order to get 3a (after expanding), let's try: (2a )(b  1.5 ) In order to get 3b (after expanding), let's try: (2a  3)(b  1.5 ) When we expand this, we get: 2ab  3a  3b + 4.5 Does that help? Cheers, Brent Even for you, a club legend, this is going to be tricky under 2 minutes in test conditions? Is there a better way to solve this, a trick or something? What I did was do it the stupid way and hope you don't have more factors than you already found, but its not a very good way to be honest. I am looking for a quick trick.. \(a = \frac{3b}{2b3}\) Plug in values of b, b=1, NO b=2, a=6 YESb=3, a=3 YESb=4, NO b=5, NO b=6, a=2 YESb=7, NO b=8, NO b=9, NO At this point you have got 3 roots, the next answer up is 6. Which means you have to find another 3 roots. Is it worth the time? Nope. But again this is not convincing at all. In fact your solution is a much more elegant brent, but would you have enough time to do it as you suggested?




Re: How many different pairs of positive integers (a, b) satisfy the equat
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