If x, y and z are positive integers, such that 1/x + 1/y + 1/z = 1

I was thinking about just putting random values to check, but I was really hoping to find out a pattern. When I had this ex in my mock test, I found 9 to fit and had a strong feeling about 11 as well, but I tried like 4 4 3 and one more and it didn't work, so I concluded that only the second one works. Is there any way to find it easier or just trial and error?

Number guessing is the easiest way in these type of questions. We can eliminate most of the sets directly.

1/5 + .... cannot make a combination.

1/2 + 1/4 + 1/4
1/3+1/3+1/3
1/6 + 1/2 + 1/3

Only possible sets.

Hi Bunuel chetan2u, is there any other way to solve this than number guessing? Thank you in advance.

Use the pattern: 
x, y and z are positive integers so their minimum value is 1 each. None of the three can be 1 because then the sum of reciprocals will become 1 with just that number. 
So each of them must be at least 2. 
2, 2, 2
If one number is 2, no other number can be 2 since the two of them will add to give 1 (1/2 + 1/2 = 1).
So one number can be 2 but then the other two need to be at least 3. So sum cannot be 7
2, 3, 3
But their reciprocals will not add up to 1. If two numbers are defined, the third is automatically fixed.
2, 3, 6 - Sum = 11 (possible)

9 is simple to see - 3, 3, 3

Hence answer (E)



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­samarpan.g28, the moment I read your question, my instinct told me to simplify the equation, add number properties to it and look for a solution. Normally, using options would be the best way forward.
(I)
\(\frac{1}{x} + \frac{1}{y }+\frac{ 1}{z} = 1 \)
Observation: The greatest term (1/a) of the three cannot be less than the average 1/3, so one of the three integers has to be 2 or 3.
So if one is 3: \(\frac{1}{3} + \frac{1}{y }+\frac{ 1}{z} = 1 \)......\( \frac{1}{y }+\frac{ 1}{z} = \frac{2}{3} =\frac{1}{3} +\frac{1}{3}\)....3+3+3=9
If one is 2: \(\frac{1}{2} + \frac{1}{y }+\frac{ 1}{z} = 1 \)......\( \frac{1}{y }+\frac{ 1}{z} = \frac{1}{2} =\frac{1}{4} +\frac{1}{4}=\frac{1}{3} +\frac{1}{6}\)
....2+4+4=10 and 2+3+4 = 11
Clearly taking the two largest as 1/2 and 1/3 will require 1/6. Thus none can be less than 6.

(II)
x, y and z are positive integers, so let us take three cases

1. All three are equal: x=y=z, so \(\frac{1}{x} + \frac{1}{y }+\frac{ 1}{z} = 1 \)
\(\frac{1}{x} + \frac{1}{x }+\frac{ 1}{x} = 1.......\frac{3}{x}=1....x=3 \)
\(\frac{1}{3} + \frac{1}{3 }+\frac{ 1}{3} = 1 \)
So, solution, 3,3,3 and x+y+z=9
2. Exactly two are equal: y=z, so \(\frac{1}{x} + \frac{1}{y }+\frac{ 1}{z} = 1 \)
\(\frac{1}{x} + \frac{1}{y }+\frac{ 1}{y} = 1.......\frac{1}{x}+\frac{2}{y}=1......\frac{1}{x}+/frac{1}{\frac{y}{2}} =1\)
Only possibility = \(\frac{1}{2}+\frac{1}{2}=1........x=2, y=z=4\) 
\(\frac{1}{2} + \frac{1}{4}+\frac{ 1}{4} = 1 \)
So, solution, 2, 4, 4 and x+y+z=10
3. All three are different:  \(\frac{1}{x} + \frac{1}{y }+\frac{ 1}{z} = 1 \)
\(\frac{xy+yz+xz}{xyz}=1........... xy+yz+xz=xyz.........xyz-xy=xz+yz.......xy(z-1)=z(x+y)\)
As z and z-1 are co-prime, we can take z=xy and z-1 = x+y. Thus, xy = x+y+1
Only possibility = > x=2 and y=3.
\(\frac{1}{2} + \frac{1}{3}+\frac{ 1}{6} = 1 \)
So, solution, 2, 3, 6 and x+y+z=11
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KarishmaB gmatophobia, Can I face any problem if I follow this approach ? This looks much easier to follow and quick also.

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Yes. This is actually the mathematical way to solve this problem.

AM>=HM
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For equal values of x,y,z; the value of expression will have min of 9. Now only make sure integers work here

­I think choosing numbers and noticing the constraints on the numbers is ultimately the best way to go. That harmonic mean thing doesn't help you much and its more work. Karishmab explains it nicely.

Here's me working through it:

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