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

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How many different pairs of positive integers (a, b) satisfy the equat  [#permalink]

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New post 01 Nov 2019, 05:36
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A
B
C
D
E

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  95% (hard)

Question Stats:

16% (02:28) correct 84% (02:27) wrong based on 122 sessions

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

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New post Updated on: 01 Nov 2019, 09:41
1
Top Contributor
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 Questions


GIVEN: 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 = 0

At 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 = 0
Add 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 = 6

case ii: (9)(1) = 9
In other words, (2a - 3) = 9 and (2b - 3) = 1
Solve to get: a = 6 & b = 2

case iii: (3)(3) = 9
In other words, (2a - 3) = 3 and (2b - 3) = 3
Solve to get: a = 3 & b = 3

There 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  [#permalink]

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New post 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 Questions


given
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  [#permalink]

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New post Updated on: 01 Nov 2019, 09:17
2
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 Questions


1/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  [#permalink]

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New post 01 Nov 2019, 09:00
1
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 Questions


GIVEN: 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 = 2
Factor: 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 = 2ab
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Re: How many different pairs of positive integers (a, b) satisfy the equat  [#permalink]

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New post 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 Questions


1/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


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

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New post 01 Nov 2019, 09:15
1
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 Questions


1/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  [#permalink]

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New post 01 Nov 2019, 09:33
1
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Dillesh4096 wrote:

Hi GMATPrepNow,

Pls See the highlighted part, you should be getting 3b + 3a = 2ab



Ha! I was wondering what made the question so "challenging"!! :-D
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  [#permalink]

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New post 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 Questions


GIVEN: 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 = 0

At 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 = 0
Add 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 = 6

case ii: (9)(1) = 9
In other words, (2a - 3) = 9 and (2b - 3) = 1
Solve to get: a = 6 & b = 2

case iii: (3)(3) = 9
In other words, (2a - 3) = 3 and (2b - 3) = 3
Solve to get: a = 3 & b = 3

There 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  [#permalink]

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New post 26 Dec 2019, 08:20
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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 = 0
I 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  [#permalink]

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New post 29 Dec 2019, 05:11
1
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 = 0
I 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..

\(a = \frac{3b}{2b-3}\)
Plug in values of b,
b=1, NO
b=2, a=6 YES
b=3, a=3 YES
b=4, NO
b=5, NO
b=6, a=2 YES
b=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?
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Re: How many different pairs of positive integers (a, b) satisfy the equat   [#permalink] 29 Dec 2019, 05:11
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