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

Question

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

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BrentGMATPrepNow · Accepted Answer

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

CareerGeek · Answer

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

Archit3110 · Answer

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

CareerGeek · Answer

Hi  GMATPrepNow ,

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

Archit3110 · Answer

Dillesh4096  you have mentioned 6,2 twice**

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CareerGeek · Answer

Thanks Archit,

Mistyped. Edited

BrentGMATPrepNow · Answer

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

ShankSouljaBoi · Answer

Hi brent,

How have you factored out the equation ? How did you get 4.5 ????

Regards

BrentGMATPrepNow · Answer

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

AnirudhaS · Answer

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?

Kody · Answer

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?

firas92 · Answer

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

Kinshook · Answer

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

Kinshook · Answer

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

Avysar · Answer

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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yahya729 · Answer

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.

apretty · Answer

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).

Thelegend2631 · Answer

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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beeblebrox · Answer

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.

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

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