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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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]
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]
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Kushchokhani wrote:
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.


Yes, assuming we have positive integers, Arithmetic mean >= Geometric Mean
Equality holds when the integers are equal i.e. say all are equal to N. Then AM = GM = N
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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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Re: If a, b, and c are positive integers such that 1/a + 1/b = 1 [#permalink]
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