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

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Archit3110

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

Updated on: 01 Nov 2019, 09:12

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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]

01 Nov 2019, 09:00

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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]

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]

01 Nov 2019, 09:15

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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]

01 Nov 2019, 09:33

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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"!!

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]

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]

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]

29 Dec 2019, 05:11

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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]

25 Feb 2020, 18:57

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I have been finding that for these types of problems there is a shortcut. If you simplify the fraction in question, then take the denominator of that fraction, square it, and then count the number of factors of that squared denominator, you will arrive at the answer.

The simplified fraction here is (2/3)
The denominator is 3
3^2 = 9
9 has 3 factors
The answer will always be odd if this shortcut has any merit

Can someone please let me know if this will ever not work?

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

25 Feb 2020, 19:16

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Here's my take.

$\frac{1}{a}+\frac{1}{b}=\frac{2}{3}$ which is obviously $\frac{1}{3}+\frac{1}{3}$

So (3,3) is a possible pair

Now for each other sets of solutions we get 2 pairs (a,b)=(x,y) and (a,b)=(y,x) and so the total number of possible pairs must be odd which already leaves us with 2 answer choices B and D

Now since we have already tested with 3 for one of a or b, we only need to test 2 and 1 since we know that a and b are positive integers.

If a=2, we find that b=6 and so this gives us 2 more pairs (2,6) and (6,2)

And now we have 3 pairs in total - (3,3) (2,6) and (6,2)

So the answer must be B

Hit Kudos if this helped!

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

28 Mar 2020, 02:42

Bunuel wrote:

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

Asked: How many different pairs of positive integers (a, b) satisfy the equation $\frac{1}{a}+\frac{1}{b}=\frac{34}{51}$ ?

1/a + 1/b = 34/51 = 2/3;
1/a = 2/3 - 1/b = (2b-3)/3b
a = 3b/(2b-3)

b=1; a = -3; Not feasible
b=2; a = 6; Feasible
b=3; a = 3; Feasible
b=4; a = 12/5; Not feasible
b=5; a = 15/7; Not feasible
b=6; a =2; Feasible

(a,b) = {(2,6),(3,3),(6,2)}

IMO B

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

31 Mar 2020, 05:56

Bunuel wrote:

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

Asked: How many different pairs of positive integers (a, b) satisfy the equation $\frac{1}{a}+\frac{1}{b}=\frac{34}{51}$ ?

1/a + 1/b = 34/51 = 2/3

1/a = 2/3 - 1/b = (2b - 3)/3b
a = 3b/(2b-3) ; a, b <>0

b=2; a =6; is a solution
b=1; a= -3; is NOT a solution since a is a positive integer
b=3; a= 3; is a solution

(a, b) = {(3,3),(2,6),(6,2)}

3 solutions are possible

IMO B

B

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

12 May 2020, 21:36

We have 1/a +1/b=34/51=2/3
Simplifying we have (a+b)/ab=2/3
Cross multipying we have 3(a+b)=2(ab)
This means that we have an even multiple of 3 on the left side...since the right side is a multiple of 2
Let's start with 2
3(2) means a=b=1 but 3(1+1) not=2(1)
With the next even integer 4, it can be 1&3 or 2&2
But 3(1+3)not=2(3)
3(2+2) not=2(4)
Next 6 can be 1&5 or 3&3 or 2&4
With 1&5 and 2&4 it doesn't work but it does work with 3&3 as3(3+3)=2(9)
With 8, it can be 1&7,2&6, 3&5 or 4&4 and it works for
3(2+6)=2(12)
Beyond this number we can see that the product ab will become very large for the equation to be true.
So it works for(3,3), (2,6) & (6,2)..three sets of values
Hence B

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

03 Jun 2021, 09:31

once you factorize the equation to 1/a+1/b=2/3, its pretty self explanatory that 3,3 will be a pair.
the only other way to get a 3 in denominator will be if 4/6, 6/9, 8/12...
and to get 4/6,6/9,8/12, on of the fractions has to be 1/6,1/9,1/12, which means the other would have to be 3/6,5/9,7/12. here since the numerator has to be 1, 3/6 works since that can get factored to 1/2, but the others cant.

i can't prove this but you need realize that every fractions numerator is going up by 2 and denominator by 3 so you'll just keep getting 9/15,11/18,13/21... none of can be factored to a one on top. so you're stuck with three pairs.

The other intuitive (i'm sure someone figured that out) would be to know that you got one pair 3,3 and then 2,6, 6,2. so your answers can only be A or B, but it's not possible to get another identical pair like 3,3, so you can't have 6 pairs.

B

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

11 Jun 2021, 06:12

Given: 1/a + 1/b = 2/3
+ if a=b then a=b=3
+ if a and b are distinct, suppose that a<b --> 2/a > 2/3 >2/b --> a < 3 < b
Since a is a positive integer, then a = 1 (so b= -1, impossible), or a = 2 (then b=6)
The answer is 3: (3,3), (2,6) and (6,2).

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

26 Jun 2021, 10:01

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Disappointed to see lot of explanations directly going by trial and error method. So i thought I'll put a concrete approach.

Equation says.

$(a+b)/ab = 2/3$

Which means a+b = 2x and ab = 3x. (

Substituting a =3x/b. you get

$b² -2xb + 3x$

If you put x = 1,2,3 you get 3 different equations which you can solve to get b =3,2,6

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

14 Apr 2022, 11:51

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Actually it is a very easy question. $\frac{1}{a}$ +$\frac{1}{b}$ = 0.6666666....= $\frac{2}{3}$
You can write it as 0.5+0.16666666666..... this is = $\frac{1}{2}$+$\frac{1}{6}$ - thus we have two pairs (6,2) & (2,6), also $\frac{2}{3}$=$\frac{1}{3}$+$\frac{1}{3}$ , hence another pair is (3,3). We are done without much algebra.

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

14 Apr 2022, 11:51

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