If a, b, and c are positive integers such that 1/a + 1/b = 1 : GMAT Data Sufficiency (DS)
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# If a, b, and c are positive integers such that 1/a + 1/b = 1

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If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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18 Nov 2012, 12:06
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If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Nov 2012, 03:01, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Fraction and Inequality [#permalink]

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18 Nov 2012, 18:01
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monsoon1 wrote:
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

[Reveal] Spoiler:
OA-> B

given is c= ab/(a+b)
thus, since c is integer, ab/a+b must be an integer.

statement 1: b ≤ 4
No information can be drawn. Not sufficient
statement 2: ab ≤ 15
Only possible value of ab, such that ab/a+b is integer could be when a=2,b=2. Thus, c=1.
Sufficient.

Ans B it is.
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Re: Fraction and Inequality [#permalink]

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19 Nov 2012, 16:38
Vips0000 wrote:
monsoon1 wrote:
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

[Reveal] Spoiler:
OA-> B

given is c= ab/(a+b)
thus, since c is integer, ab/a+b must be an integer.

statement 1: b ≤ 4
No information can be drawn. Not sufficient
statement 2: ab ≤ 15
Only possible value of ab, such that ab/a+b is integer could be when a=2,b=2. Thus, c=1.
Sufficient.

Ans B it is.

How did you find that only these values would satisfy?
Did you test several numbers?

The question also doesn't give us the clue whether the numbers are the same or different.So, we have to test many numbers.right?

Can you please show the steps or any other way to get to the correct answer?
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Re: Fraction and Inequality [#permalink]

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19 Nov 2012, 17:01
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monsoon1 wrote:
How did you find that only these values would satisfy?
Did you test several numbers?

The question also doesn't give us the clue whether the numbers are the same or different.So, we have to test many numbers.right?
Can you please show the steps or any other way to get to the correct answer?

You actually dont need to test numbers. It could be purely algebric approach coupled with some logical deductions.

we have$$c= ab/(a+b)$$
or $$c = \frac{1}{(1/a+1/b)}$$

If you notice this expression and remember that c has to be integer, that would mean Denominator has to be 1 (Since numerator is already 1).
There is only one such possiblity of a and b that could give you 1/a+1/b =1
So you dont need to test any number.

Hope it helps. Lets kudos
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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07 Oct 2013, 11:38
I have a question here. If the solution given by Vips0000 is correct and to me it seems perfect, then neither of the statements is actually needed to solve the question there can be only one combination possible from the question stem itself; a=b=2 and consequently c=1. In such a scenario shouldn't the answer be D, since both statements independently will lead us to the answer.
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Re: Fraction and Inequality [#permalink]

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07 Oct 2013, 11:59
I can't see how you did the following:

Vips0000 wrote:
monsoon1 wrote:

we have$$c= ab/(a+b)$$
or $$c = \frac{1}{(1/a+1/b)}$$

Could you explain? thanks.

Edit: Think I got it now. Reciprocal of both sides of the original equation? or is there a different way?
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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07 Oct 2013, 12:19
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Nilohit wrote:
I have a question here. If the solution given by Vips0000 is correct and to me it seems perfect, then neither of the statements is actually needed to solve the question there can be only one combination possible from the question stem itself; a=b=2 and consequently c=1. In such a scenario shouldn't the answer be D, since both statements independently will lead us to the answer.

Well, lets say the ab<or= 15 restriction was not there. Then if a=10 and b=10 then ab/(a+b) would equal 100/20...which is an integer.

The only possible values when ab<or=15 are 2 and 2.

but for the b<or=4 statement, you could still have a case where say b=4 and a=12, and ab/a+b=48/16=3. Then you would have 2 (or more) possible values for a and b if you also include the possible value set of "a=2 and b=2".

Vips0000 reasoning isn't actually perfect. In the case of $$\frac{1}{1/a+1/b}$$ the denominator does not have to be "1". $$\frac{1}{1/a+1/b}$$ could equal $$\frac{1}{1/4+1/12}$$ which reduces to $$\frac{1}{4/12}$$ then $$1*\frac{12}{4}$$ which equals 3
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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27 Oct 2013, 10:42
monsoon1 wrote:
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

Updating with new solution by jlgdr

OK so we have that ab / a+b = C is an integer. Therefore let's hit the first statement.

Statement 1 says that b<=4. We have two choices here (actually 3). Let's begin with (2,2) C would equal 2. Now if we pick (4,4), C is again two so same answer. But if we pick (6,6) then C= 3. So not sufficient.

From statement 2 we know that ab<=15. Hence ab has to be 2,2 since both a,b are positive integers and of course C = 2.

Therefore B stands

Gimme kudos
Cheers
J

Last edited by jlgdr on 29 Mar 2014, 06:09, edited 1 time in total.
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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27 Oct 2013, 12:01
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jlgdr wrote:
monsoon1 wrote:
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

Hey all,

This one was a bit tricky indeed. Is there some other way we can notice quickly which number satisfies the ab/a+b constraint giving 'c' as an integer value

Cheers!
J

From F.S 1, we know that for a to be positive, c<b. The given equation is valid for b=2,c=1 and also for b=3,c=2. Insufficient.

Now, back to your question.

We know that$$\frac{a+b}{2}\geq{\sqrt{ab}}$$

Also, from the question stem, we know that $$\frac{a+b}{ab} =\frac{1}{c}$$

Thus, $$(a+b) = \frac{ab}{c}$$. Replacing this in the first equation, we get $$\frac{ab}{2c}\geq{\sqrt{ab}}$$

Or,$$c\leq{\frac{\sqrt{ab}}{2}} \to c\leq{\frac{\sqrt{15}}{2}} \to c<{2}$$. Thus, the only positive integer less than 2 is One and thus c=1.Sufficient.
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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13 Nov 2013, 22:07
Doesn't mau5's solution, if correct, bring us back to Nilohit's point of 'why do we need know statement 1 and 2 if we already know C equals 1'? Regardless, this is definitely a challenging question. I would like to know if there is a shortcut to this problem, especially for statement #2. Just plugging in some numbers would be impossible given the time limit...
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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13 Nov 2013, 22:22
JepicPhail wrote:
Doesn't mau5's solution, if correct, bring us back to Nilohit's point of 'why do we need know statement 1 and 2 if we already know C equals 1'? Regardless, this is definitely a challenging question. I would like to know if there is a shortcut to this problem, especially for statement #2. Just plugging in some numbers would be impossible given the time limit...

I don't understand how you can get the answer from the First fact statement.

Also, how can you get that c=1, without fact statement 2?
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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13 Nov 2013, 22:44
mau5 wrote:
JepicPhail wrote:
Also, how can you get that c=1, without fact statement 2?

Oh, I see now why you need 15. $$sqrt(15)$$ is less than 4, so C is less than or equal to something like 1/2, 2/2, 3/2, etc... and since only integer here is 1, C equals 1.
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]

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13 Nov 2013, 23:22
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JepicPhail wrote:
Doesn't mau5's solution, if correct, bring us back to Nilohit's point of 'why do we need know statement 1 and 2 if we already know C equals 1'? Regardless, this is definitely a challenging question. I would like to know if there is a shortcut to this problem, especially for statement #2. Just plugging in some numbers would be impossible given the time limit...

Actually, this is not correct. c needn't be 1 in every case.

Take a = 2, b = 2. In this case c = 1

Take a = 4, b = 4. In this case,
1/4 + 1/4 = 1/2
c = 2

Take a = 3, b = 6. In this case,
1/3 + 1/6 = 1/2
c = 2

Take a = 15, b = 30. In this case
1/15 + 1/30 = 1/10
etc

Basically, you have to look for values such that when the numerators add up, the sum is divisible by the denominator. a and b cannot be 1 since we need the sum to be less than 1.

Statement 1: b <= 4
This gives you different values of c. c could be 1 or 2. Not sufficient.

Statement 2: ab <= 15
a and b could be 2 each. There is no other set of values. Try the small pairs (2, 4), (3, 3).
Hence (B) alone is sufficient.
(Note the algebraic solution provided by mau5 for statement 2.)
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