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

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For how many different pairs of positive integers (a, b) can  [#permalink]

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18 Oct 2013, 16:58
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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  [#permalink]

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18 Oct 2013, 23:07
7
11
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  [#permalink]

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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  [#permalink]

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20 Oct 2013, 05:53
1
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  [#permalink]

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13 Oct 2015, 06:37
1
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  [#permalink]

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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  [#permalink]

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12 Feb 2020, 11:48
arunavamunshi1988 wrote:
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.

the guy has already given 7 solution and said as only the same pair 15,15 is there, so the next solution will be in pairs, hence all the solution will be in count like 2k+1
above 7 only 9 answer choice is present ,as it cannot be 10 or 8 (they are even). hence without solving for further soln we can pick D

Kudos if this helped
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Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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23 Feb 2020, 23:23
Can anyone explain this further?
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Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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26 Feb 2020, 11:37
In addition, (a+b) must be even (odd+odd or even+even) due to ab=(15(a+b))/2
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Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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21 Apr 2020, 10:45
MentorTutoring

Can you please help me in solving this question?
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Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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21 Apr 2020, 14:23
2
1
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

$$\frac{2}{15} = \frac{1}{a} + \frac{1}{b}$$
$$\frac{2}{15} = \frac{a+b}{ab}$$
$$2ab = 15a+15b$$
$$4ab = 30a+30b$$
$$0 = 30a-4ab+30b$$
$$(2a-15)(2b-15) = (30a-4ab+30b) + (2a-15)(2b-15)$$
$$(2a-15)(2b-15) = (30a-4ab+30b) + (4ab-30a-30b+225)$$
$$(2a-15)(2b-15) = 225$$

The left side represents factor pairs of 225, implying the following options:
1*225 or 225*1 --> 2 options
3*75 or 75*3 --> 2 options
5*45 or 45*5 --> 2 options
9*25 or 25*9 --> 2 options
15*15 --> 1 option, where a=b
Total options = 9

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For how many different pairs of positive integers (a, b) can  [#permalink]

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21 Apr 2020, 15:39
1
1
Lucky1994 wrote:
MentorTutoring

Can you please help me in solving this question?

Hello, Lucky1994, and thank you for tagging me. I will admit to taking a brute force approach to this one, using a little number sense and leaning on the answers in a similar way to what mau5 had done. I will simply walk you through that process, even if, again, it may not be a great method to copy.

1) If

$$\frac{1}{a}+\frac{1}{b}=\frac{2}{15}$$,

then we can anticipate that our different integer combinations will play off of multiples of

$$\frac{2}{15}$$. I created a list of several to get started:

$$\frac{2}{15}$$,

$$\frac{4}{30}$$,

$$\frac{6}{45}$$,

$$\frac{8}{60}$$,

$$\frac{10}{75}$$,

$$\frac{12}{90}$$.

2) For

$$\frac{1}{a}+\frac{1}{b}=\frac{2}{15}$$,

if a and b were the same, it is clear that they will each be 15. This is our first solution: (15, 15).

3) To test

$$\frac{4}{30}$$,

we cannot resort to

$$\frac{2}{30}+\frac{2}{30}$$,

since each fraction would reduce to

$$\frac{1}{15}$$,

and we have already exhausted that combination. Thus, the only other positive integers available to us are 1 and 3:

$$\frac{1}{30}+\frac{3}{30}=\frac{4}{30}$$.

How do we derive the values of a and b? We simply divide the denominator of each fraction by its numerator and take the integer value.

$$\frac{30}{1}=30$$,

and

$$\frac{30}{3}=10$$.

Since it makes no difference which unknown takes the value of 30 or 10, we derive two more valid combinations: (10, 30) and (30, 10).

4) To test

$$\frac{6}{45}$$,

we cannot use 3 and 3, since, again, we have already exhausting the matching-value solution, and here is another important realization: neither can we use any even-integer combination, such as 2 and 4, since the denominator of our fraction will not be evenly divisible by these values. Hence, it only makes sense to try 1 and 5:

$$\frac{1}{45}+\frac{5}{45}=\frac{6}{45}$$.

Using our mental math approach of dividing the denominator by the numerator of each fraction, we get another combination: 45 and 9. Add these two solutions to your tally: (9, 45) and (45, 9). At this point, we have derived five solutions, and a quick check against the answers reveals that only 5 or 9 will make any sense. Why? Because, once more, we have already exhausted the solution in which our two unknowns equal each other, and we have already seen how to derive our other "mirrored" solutions. Thus, we can anticipate that we are either sitting on the answer already or need to find just one more winning combination.

5) To test

$$\frac{8}{60}$$,

we cannot use 4 and 4. The other combinations are 5/3, 6/2, and 7/1. I start with the larger number because it is easier to assess whether dividing the denominator by such a value will yield an integer. We can see that both 5/3 and 6/2 will work, but not 7/1, since 60 is not divisible evenly by 7 but has all the other numbers as factors.

$$\frac{5}{60}+\frac{3}{60}=\frac{8}{60}$$.

From this, we can derive

$$\frac{60}{5}=12$$

and

$$\frac{60}{3}=20$$.

(12, 20) and (20, 12) represent two more valid solutions, and that ticks the tally beyond 5. We can thus conclude that the answer will be 9, without doing any more work. What about the other combination, though, 6 and 2? If you are curious, you can run it through for a quick check:

$$\frac{6}{60}+\frac{2}{60}=\frac{8}{60}$$.

From this, we can derive

$$\frac{60}{6}=10$$

and

$$\frac{60}{2}=30$$.

Although this is a valid combination, it is an extraneous solution, since we have already worked out these values above. Notice that we do not even need to test either of our other fractions, but as an added bonus, I will show you how I checked them anyway.

---UNNECESSARY EXTRA WORK---

6) To test

$$\frac{10}{75}$$,

we cannot use 5 and 5. The other combinations are 6/4, 7/3, 8/2, and 9/1. We can eliminate the even-numbered ones, since 75 is not divisible evenly by an even number. We can also see that neither 7 nor 9 will fit into 75 without getting into decimal territory. This is how I checked:

$$7*10=70$$,

so 75 is not evenly divisible by 7;

$$9*8=72$$,

so 75 is not evenly divisible by 9.

This one was an easy fraction to test. No valid combination of a and b values exists.

7) To test

$$\frac{12}{90}$$,

we cannot use 6 and 6. The other combinations are 7/5, 8/4, 9/3, 10/2, and 11/1. We can run the larger numbers through in a similar manner to what we had just done with 75:

$$7*10=70$$,

and since

$$70+7*3=91$$,

90 is not evenly divisible by 7;

$$8*10=80$$,

and since

$$80+8=88$$,

it is clear that 90 will not be evenly divisible by 8;

$$\frac{90}{9}=10$$,

and

$$\frac{90}{3}=30$$.

Thus, 9 and 3 will work.

$$\frac{90}{10}=9$$,

and

$$\frac{90}{2}=45$$.

Thus, 10 and 2 will also work.

$$11*8=88$$,

so it is clear that 90 will not be evenly divisible by 11.

Now we can test our first combination, 9 and 3:

$$\frac{9}{90}+\frac{3}{90}=\frac{12}{90}$$.

From this, we can derive

$$\frac{90}{9}=10$$

and

$$\frac{90}{3}=30$$.

Of course, we know already that these values work. How about our other combination, 10 and 2?

$$\frac{10}{90}+\frac{2}{90}=\frac{12}{90}$$.

From this, we can derive

$$\frac{90}{10}=9$$

and

$$\frac{90}{2}=45$$.

We came up short again, since we know already that these values work.

Honestly, I look to do as little work as I can think to do on Quant questions, but this one took me a good chunk of time, about 3 minutes, using the above approach. Please note that my approach above is not the quickest way to solve the problem, but the number sense you can apply may prove helpful in other types of questions, so I would not say that the above is a wasted effort.

I hope that helps. Phew! That has to be my longest post ever!

- Andrew
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Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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22 Apr 2020, 06:32
MentorTutoring wrote:
Lucky1994 wrote:
MentorTutoring

Can you please help me in solving this question?

Hello, Lucky1994, and thank you for tagging me. I will admit to taking a brute force approach to this one, using a little number sense and leaning on the answers in a similar way to what mau5 had done. I will simply walk you through that process, even if, again, it may not be a great method to copy.

1) If

$$\frac{1}{a}+\frac{1}{b}=\frac{2}{15}$$,

then we can anticipate that our different integer combinations will play off of multiples of

$$\frac{2}{15}$$. I created a list of several to get started:

$$\frac{2}{15}$$,

$$\frac{4}{30}$$,

$$\frac{6}{45}$$,

$$\frac{8}{60}$$,

$$\frac{10}{75}$$,

$$\frac{12}{90}$$.

2) For

$$\frac{1}{a}+\frac{1}{b}=\frac{2}{15}$$,

if a and b were the same, it is clear that they will each be 15. This is our first solution: (15, 15).

3) To test

$$\frac{4}{30}$$,

we cannot resort to

$$\frac{2}{30}+\frac{2}{30}$$,

since each fraction would reduce to

$$\frac{1}{15}$$,

and we have already exhausted that combination. Thus, the only other positive integers available to us are 1 and 3:

$$\frac{1}{30}+\frac{3}{30}=\frac{4}{30}$$.

How do we derive the values of a and b? We simply divide the denominator of each fraction by its numerator and take the integer value.

$$\frac{30}{1}=30$$,

and

$$\frac{30}{3}=10$$.

Since it makes no difference which unknown takes the value of 30 or 10, we derive two more valid combinations: (10, 30) and (30, 10).

4) To test

$$\frac{6}{45}$$,

we cannot use 3 and 3, since, again, we have already exhausting the matching-value solution, and here is another important realization: neither can we use any even-integer combination, such as 2 and 4, since the denominator of our fraction will not be evenly divisible by these values. Hence, it only makes sense to try 1 and 5:

$$\frac{1}{45}+\frac{5}{45}=\frac{6}{45}$$.

Using our mental math approach of dividing the denominator by the numerator of each fraction, we get another combination: 45 and 9. Add these two solutions to your tally: (9, 45) and (45, 9). At this point, we have derived five solutions, and a quick check against the answers reveals that only 5 or 9 will make any sense. Why? Because, once more, we have already exhausted the solution in which our two unknowns equal each other, and we have already seen how to derive our other "mirrored" solutions. Thus, we can anticipate that we are either sitting on the answer already or need to find just one more winning combination.

5) To test

$$\frac{8}{60}$$,

we cannot use 4 and 4. The other combinations are 5/3, 6/2, and 7/1. I start with the larger number because it is easier to assess whether dividing the denominator by such a value will yield an integer. We can see that both 5/3 and 6/2 will work, but not 7/1, since 60 is not divisible evenly by 7 but has all the other numbers as factors.

$$\frac{5}{60}+\frac{3}{60}=\frac{8}{60}$$.

From this, we can derive

$$\frac{60}{5}=12$$

and

$$\frac{60}{3}=20$$.

(12, 20) and (20, 12) represent two more valid solutions, and that ticks the tally beyond 5. We can thus conclude that the answer will be 9, without doing any more work. What about the other combination, though, 6 and 2? If you are curious, you can run it through for a quick check:

$$\frac{6}{60}+\frac{2}{60}=\frac{8}{60}$$.

From this, we can derive

$$\frac{60}{6}=10$$

and

$$\frac{60}{2}=30$$.

Although this is a valid combination, it is an extraneous solution, since we have already worked out these values above. Notice that we do not even need to test either of our other fractions, but as an added bonus, I will show you how I checked them anyway.

---UNNECESSARY EXTRA WORK---

6) To test

$$\frac{10}{75}$$,

we cannot use 5 and 5. The other combinations are 6/4, 7/3, 8/2, and 9/1. We can eliminate the even-numbered ones, since 75 is not divisible evenly by an even number. We can also see that neither 7 nor 9 will fit into 75 without getting into decimal territory. This is how I checked:

$$7*10=70$$,

so 75 is not evenly divisible by 7;

$$9*8=72$$,

so 75 is not evenly divisible by 9.

This one was an easy fraction to test. No valid combination of a and b values exists.

7) To test

$$\frac{12}{90}$$,

we cannot use 6 and 6. The other combinations are 7/5, 8/4, 9/3, 10/2, and 11/1. We can run the larger numbers through in a similar manner to what we had just done with 75:

$$7*10=70$$,

and since

$$70+7*3=91$$,

90 is not evenly divisible by 7;

$$8*10=80$$,

and since

$$80+8=88$$,

it is clear that 90 will not be evenly divisible by 8;

$$\frac{90}{9}=10$$,

and

$$\frac{90}{3}=30$$.

Thus, 9 and 3 will work.

$$\frac{90}{10}=9$$,

and

$$\frac{90}{2}=45$$.

Thus, 10 and 2 will also work.

$$11*8=88$$,

so it is clear that 90 will not be evenly divisible by 11.

Now we can test our first combination, 9 and 3:

$$\frac{9}{90}+\frac{3}{90}=\frac{12}{90}$$.

From this, we can derive

$$\frac{90}{9}=10$$

and

$$\frac{90}{3}=30$$.

Of course, we know already that these values work. How about our other combination, 10 and 2?

$$\frac{10}{90}+\frac{2}{90}=\frac{12}{90}$$.

From this, we can derive

$$\frac{90}{10}=9$$

and

$$\frac{90}{2}=45$$.

We came up short again, since we know already that these values work.

Honestly, I look to do as little work as I can think to do on Quant questions, but this one took me a good chunk of time, about 3 minutes, using the above approach. Please note that my approach above is not the quickest way to solve the problem, but the number sense you can apply may prove helpful in other types of questions, so I would not say that the above is a wasted effort.

I hope that helps. Phew! That has to be my longest post ever!

- Andrew

Hello Andrew,

You have made this question looks so simple.
Thankyou so much for above explanation.
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Posts: 896
Re: For how many different pairs of positive integers (a, b) can  [#permalink]

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22 Apr 2020, 10:14
Lucky1994 wrote:
Hello Andrew,

You have made this question looks so simple.
Thankyou so much for above explanation.

Hello, Lucky1994. I am glad to hear my thoughts were helpful to you. I may not have gotten the answer in the best way, but I will always aim to explain my process in the clearest, most useful manner I can think of when assisting others.

Keep up your studying.

- Andrew
Re: For how many different pairs of positive integers (a, b) can   [#permalink] 22 Apr 2020, 10:14

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

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