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

Question

If x, y and z are positive integers, such that 1/x + 1/y + 1/z = 4/3, which of the following could be the value of x + y + z?

I. 5 
 II. 7 
 III. 13

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

HarshavardhanR · Answer

x, y, and z --- positive integers.

Condition:  1/x + 1/y + 1/z = 4/3

Question:   W hich of the following  could be  the value of x + y + z?

So, if. for a particular option (say II. 7), we can find even one case meeting the above condition, that particular option "could be" the value of x + y + z.

I. 5

There are only 2 ways to get x + y + z = 5 here.

a) (1,1,3)
b) (1,2,2)

In both cases, 1/x + 1/y + 1/z = 4/3 ---is not met. So, "5" cannot be the value of x + y + z.

Also,  a sidenote  --- we need a multiple of 3 in the denominator to get 4/3. So, at least one among the 3 integers should be a 3-multiple. Good to keep in mind.

Reject choices A and D .

II. 7

Quick trial and error with 3-multiples ->

(1,3,3) --- does not work
 (2,2,3)  --- works. 1/2 + 1/2 + 1/3 = 1 + 1/3 = 4/3.

We have proven that "7" can be the value of x + y + z.

Reject choice C.

III. 13

Here too, I relied on quick trial and error with 3-multiples ->

(2,2,9) --- 1/2 + 1/2 + 1/9 = 1 + 1/9 ---does not work
(3,1,9) --- does not work 
  (6,6,1)  --- 1/6 + 1/6 + 1 = 1 + 1/3 = 4/3 ---works

We have proven that "13" can be the value of x + y + z.

Reject choice B.
  
The answer is  choice E  (II and III only).

---
Harsha

paragw · Answer

We can only get a sum of 4/3 in the given condition as 1+1/3, but here, there are two numbers only, so we've to split one of them into half.
For this, we have 2 options
1 can be written as 1/2 + 1/2
OR
1/3 can be written as 1/6 + 1/6
Therefore, we have 2 cases here
Case 1: 1/2 + 1/2 + 1/3 = 4/3
x+y+z = 7
Case 2: 1/6 + 1/6 + 1/1 = 4/3
 x+y+z = 13
Hence, option E

Deep1208 · Answer

Can someone elaborte,how they are getting x+y+z= 
I am stuck on that portion

egmat · Answer

Deep1208  I can see you're stuck on how to find values of  x ,  y , and  z  that satisfy  \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{4}{3} , and then determine  x + y + z . This is a very common challenge with this type of problem - the equation looks intimidating, but the key is strategic testing with small positive integers.

The Strategic Approach

Since  \frac{4}{3} \approx 1.33  is greater than  1 , at least one of  x ,  y , or  z  must be small (likely  1  or  2 ). Here's how to systematically test:

Testing Option II: Can  x + y + z = 7 ?

Try  x = 2 ,  y = 2 ,  z = 3 :
 \frac{1}{2} + \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{3}{6} + \frac{2}{6} = \frac{8}{6} = \frac{4}{3}  ✓

This works! So  7  is possible.

Testing Option III: Can  x + y + z = 13 ?

Since we need a larger sum, try  x = 1  (which contributes  \frac{1}{1} = 1  to our sum):
Remaining:  \frac{1}{y} + \frac{1}{z} = \frac{4}{3} - 1 = \frac{1}{3}

Try  y = 6 ,  z = 6 :
 \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}  ✓

So  x = 1 ,  y = 6 ,  z = 6  gives  x + y + z = 13 . This works!

Testing Option I: Can  x + y + z = 5 ?

Possible combinations:  (1,1,3) ,  (1,2,2) 
 
  (1,1,3) :  \frac{1}{1} + \frac{1}{1} + \frac{1}{3} = \frac{7}{3} 
eq \frac{4}{3}  ✗ 
  (1,2,2) :  \frac{1}{1} + \frac{1}{2} + \frac{1}{2} = 2 
eq \frac{4}{3}  ✗  
Neither works, so  5  is not possible.

Answer: (E) II and III only

Key Strategy for Similar Problems

When solving  \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = k  where you need to find specific sums:
 
 Compare  k  to  1  to determine if you need small values (if  k > 1 , at least one variable must be  1  or  2 ) 
 Start with the smallest integers and test systematically 
 If using  x = 1 , calculate what remains:  \frac{1}{y} + \frac{1}{z} = k - 1  
 Look for symmetric solutions (like  y = z ) to simplify calculations

I hope this helps! You can practice similar questions  here  (you'll find a lot of OG questions) - select  Algebra - Equations.

adewale223 · Answer

1/x + 1/y + 1/z = 4/3

yz + xz + xy/ xyz = 4/3

Which means the numerator must be a multiple of 4 and the denominator a multiple of 3.

Since x, y and z are positive integers, start from 3

xyz = 3

Possible combinations 1, 1, 3. Plugging the values for the numerator gives 7 which is not a multiple of 4

xyz = 12

Possible combinations 2,2, 3, this works.

xyz = 36

Possible combinations 6, 6, 1, this works

Answer E.

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

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

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