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

Question

If x, y and z are positive integers, such that 1/x + 1/y + 1/z = 1, which of the following could be the value of x + y + z?

I. 7
II. 9
III. 11

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only

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Attachment:

2024-01-29_19-36-12.png

Jahnavi_M · Accepted Answer

Here, we have to use Arithmetic mean >= Harmonic Mean,

(X+Y+Z)/3 >= 3/(1/X + 1/Y + 1/Z)

(X+Y+Z) >= 3 * 3/1

(X+Y+Z) >= 9

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KarishmaB · Answer

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)

ankitmba95 · Answer

By trying to form sum:

7 -> 1 cannot be any of the 3 numbers. Next (2,3,2) is again >1. so not possible
9 -> (3,3,3) fits well. possible
11 -> trying (2,3,6); It fits well as 1/2+1/3 = 5/6. possible

Thus answer is E.

Toma13 · Answer

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?

HarshaBujji · Answer

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.

chetan2u · Answer

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 equa l: 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

sayan640 · Answer

KarishmaB   gmatophobia , Can I face any problem if I follow this approach ? This looks much easier to follow and quick also.

Oppenheimer1945 · Answer

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

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

DanTheGMATMan · Answer

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:

https://www.youtube.com/watch?v=NmVmpDILrEU

KarishmaB · Answer

If you know it and it comes to mind during the test, you can of course. GMAT doesn't care what method you use to arrive at the answer. Though it doesn't question you on GP and HP directly. There would always be other methods (or logic or estimation) by which you can get the answer. Otherwise there will be no end to math concepts you must know for GMAT. After all, it is a test of logic.

egmat · Answer

Dive into this challenging question that tests your ability to think outside the box. While there's no predefined method to solve, this video demonstrates how to approach such problems by:
 
 Carefully analyzing the given information 
 Drawing key inferences from the data 
 Developing a step-by-step solving process

Watch as we break down the question, highlighting crucial insights that lead to the solution. This video emphasizes the importance of:
 
 Patience in tackling complex problems 
 The value of spending time on challenging questions 
 How strong inference skills can unlock seemingly difficult scenarios

https://www.youtube.com/watch?v=XQ8eOMhTabM

While this question may take longer than others, mastering such problems can significantly boost your problem-solving abilities and exam performance.
Can you spot the critical inferences before they're pointed out in the solution?

Dbrunik · Answer

try this case to make sure you understand the concept tested here.

it also helps to break them out the answer choices into its factors.

for example, 18 has 18,1,2,9,3,6 as factors.

this tells you the fractions have to include the numbers above, and we can prove this, since 1/3+1/3+1/6+1/6 = 18

rachanarc · Answer

Dbrunik Answer is (A) is correct? works only for 18.. But this approach took me sometime to solve

Can you please elaborate the approach with the concept that you are suggesting here. Thanks in advance! This brings a new perspective to these kindof questions:)

Dbrunik · Answer

No a is not correct, but you could also see that there are other numbers not listed that could work. The obvious one being 16.

Even if you try 24 you’ll find the factors to be 24 and one 12 and 2, and three and 8, and six and four. If you play around with these numbers. You’ll find that you cannot make one with any organization of their reciprocal.

Beyond 16, there are numbers beyond (18, 20, 24) that can serve as the sum of four positive integers whose reciprocals add up to 1. There’s no inherent uniqueness to 18; it just happens to be a convenient example that works out nicely among the given options.

One is equal to 1/2+1/3+1/7+1/42

If this isn’t Immediately obvious to you, notice how 2×3×7 = 42. Try this tactic with other numbers or with other sets of fractions.

54 works.

Dbrunik · Answer

20 also works here

This becomes clear because 1/6+1/6+1/6+1/2=1

Kinshook · Answer

If x, y and z are positive integers, such that 1/x + 1/y + 1/z = 1, which of the following could be the value of x + y + z?

1/z = 1 - 1/x - 1/y = (xy - x - y)/xy

z = xy/(xy-x-y)

x = 1; y = 1; z = 1/(1-1-1) = -1; Not feasible since x, y and z are positive integers
x = 1; y = 2; z = 2/(2-1-2) = -2; Not feasible since x, y and z are positive integers
x= 2; y = 2; z = 4/(4-2-2); Not feasible since x, y and z are positive integers
x = 2; y = 3; z = 6/(6-2-3) = 5; x + y + z = 2 + 3 + 6 = 11
x = 3; y = 3; z = 9/(9-3-3) = 3; x + y + z = 3 + 3 + 3 = 9
x = 3; y = 4; z = 12/(12-3-4) = 12/5; Not feasible since x, y and z are positive integers
x = 4; y = 4; x + y = 8 > 7; x+y+z can not be 7

I. 7
II. 9
III. 11

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) II and III only

IMO E

terminatork07 · Answer

Solution without involving any knowledge of Harmonic Mean concepts
 
If x = y = z = k , then k = 3 (then x + y + z = 9 is one solution).

Now let's keep of one the variables (z) = 3; which implies the sum of reciprocals of x and y should be equal to 2/3. (1/x + 1/y = 2/3)

From the above equation we can deduce (x+y)/xy = 2/3

From I and III lets back-solve -

For I to be correct - x + y = 7 - 3 = 4 AND x.y = 6

The solutions to these 2 equations is non-integeral so I is out (it's given in the question prompt that x and y can only be positive integers)

For III to be correct x + y = 8 AND x.y = 12

BINGO - x = 6 and y = 2 fits, so 11 is a possible solution.

RETRO -

Concepts tested in the question -
1.  Averages  (three number's sum - 1/x,1/y,1/z is 1 then average is 1/3, which means that  if you decrease one number others should increase in the same proportion )
2. How fast are how in plugging in values and validating the answers.
3. Back-Solve from given data.

XLmafia21 · Answer

i interpreted this as %s adding upto 100%
drew a quick table for my reference starting from 2,3,4... till 10
1/2 = 50%
1/3 = 33.33%
1/4 = 25%
1/5 = 20%
1/6 = 16.67%
1/7 = 14.28%
1/8 = 12.5%
1/9 = 11.11%
1/10 = 10%
Now, can i think of combinations which would yield 100% as sum:
1. 33.33 % + 33.33% + 33.33% = 100% => x=y=z = 3 => x+y+z = 9
2. 50% + 25% + 25% = 100% => x=2, y=z= 4 => x+y+z = 10 (ignore)
3. 50% + 33.33% + 16.67% => x=2, y=3, z=6 => x+y+z = 11 
Hence, option E
I think the best way is the AM>= HM property suggested by others.

GMATGuruNY · Answer

\frac{1}{x} + \frac{1}{y} = 1 - \frac{1}{z} 
 \frac{1}{x} + \frac{1}{y} = \frac{z-1}{z} 
Implication:
1/x + 1/y = 1/2, 2/3, 3/4, 4/5...

Case 1: z=2, implying that 1/x + 1/y = 1/2
If 1/x = 1/3, then 1/y = 1/2 - 1/3 = 1/6
Thus, it's possible that 1/x = 1/3 and 1/y = 1/6, with the result that x=3 and y=6.
In this case:
x+y+z = 3+6+2 = 11
Since III is possible, eliminate A, B and D, none of which includes III.

Case 2: z=3, implying that 1/x + 1/y = 2/3
Since 1/3 + 1/3 = 2/3, it's possible that 1/x = 1/3 and 1/y = 1/3, with the result that x=3 and y=3.
In this case:
x+y+z = 3+3+3 = 9
Since II is possible, eliminate C, which does not include II.

E

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

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

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