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If a, b, and c are positive integers such that 1/a + 1/b = 1

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sharkr
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink] New post   07 Mar 2017, 22:26
VeritasPrepKarishma wrote:
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.


What are the numerators, the sum, and the denominator referring to?

Thanks in advance.
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink] New post   08 Mar 2017, 03:58
VeritasPrepKarishma wrote:
sharkr wrote:
VeritasPrepKarishma wrote:
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.


What are the numerators, the sum, and the denominator referring to?

Thanks in advance.



1/a + 1/b = 1/c

You have to look for values of a and b (denominators) such that when you add 1/a and 1/b, the numerator you get is divisible by the denominator of the fraction (which you get after adding)

e.g. if a = 3, b = 6

1/a + 1/b = 1/3 + 1/6 = (2 + 1)/6 = 3/6
The numerator is 3 and the denominator is 6. The numerator is divisible by the denominator such that you get 1 in the numerator after cancelling.

1/a + 1/b = 1/3 + 1/6 = 1/2

When you get 1 in the numerator, the denominator of the resulting fraction is the value of c (i.e. 2 above)


Thanks for the reply.
For clarification, don't you mean the denominator (ie. 6) is divisible by the numerator (ie. 3)?
3 / 6 = 0*6 + 3 whereas 6/3 = 3*2 (no remainder).
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink] New post   05 Apr 2017, 15:49
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




Is there any algebric approach to be sure that B is sufficient
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink] New post   04 Jan 2018, 10:03
mau5 wrote:

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.


Whoa! That's some really interesting point you are making. Never heard of this... that the average of two average of two numbers is always greather than the square root of the product of two numbers! Does this property hold always?
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink] New post   26 Sep 2021, 06:32
Vips0000 wrote:
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 ;-)

This might seem true. But not so. Possible solutions for (ab)in the above equation are (2,2) (4,4) (8,8).. (2,2) is not the only solution
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink] New post   25 May 2022, 03:59
talismaaniac wrote:
mau5 wrote:

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.


Whoa! That's some really interesting point you are making. Never heard of this... that the average of two average of two numbers is always greather than the square root of the product of two numbers! Does this property hold always?

I think this property holds for non-negative integers only. mau5 KarishmaB please confirm.
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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink] New post   06 Jun 2023, 04:50
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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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