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\(\text{In the equation $\frac{1}{j}+\frac{1}{k}=\frac{1}{3},$ both $j$ and $K$ are positize integers. What is the sum of all possible values of $k$?}\)

 Jul 15, 2019
 #1
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Oops, I meant \(k\), not \(K\). And positive, not positize. Sorry.

 Jul 15, 2019
edited by Guest  Jul 15, 2019
edited by Guest  Jul 15, 2019
edited by Guest  Jul 15, 2019
edited by Guest  Jul 15, 2019
 #2
avatar+103999 
+2

Here's a trick I learned from one of our excellent  answerers  (heureka)

 

Let  3  = z

Let j = z + a

Let k =  z + b   .....so we have

 

  1                   1                     1

_____   +     ______   =   ______      get a common denominator on the left

z + a             z + b                z

 

[ z + b ] + [ z + a ]                   1

________________   =       ____

 z^2 + az+ bz + ab                  z

 

 

2z + b + a                                  1

_________________  =      _______        cross-multiply

z^2 + az + bz + ab                    z

 

 

2z^2 + az + bz  =  z^2  + az + bz +  ab

 

2z^2   = z^2  +  ab

 

z^2   = ab

 

Since z  = 3

 

Then

 

ab  =  3^2

 

ab  = 9

 

So....we are looking for all the pairs of positive integers  that multiply to 9

 

So we have

 

a      b

 

1      9

3      3

9      1

 

So  since  j  =  z + a       and   k =  z + b

 

Then we have

 

j        k

4      12

6      6

12    4

 

So.....the sum of all possible values for k   =  22

 

EDIT TO CORRECT AN ERROR....

 

 

cool cool cool

 Jul 15, 2019
edited by CPhill  Jul 15, 2019
edited by CPhill  Jul 15, 2019
 #3
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Thanks!!! laugh

Guest Jul 15, 2019
 #4
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Sorry, the answer was incorrect, here's the solution:

 

 

\(\text{Multiplying both sides of the equation by $3jk$ to clear the denominator gives $3k + 3j = jk$. Re-arranging and applying Simon's Favorite Factoring Trick, it follows tha t $$jk - 3j - 3k + 9 = (j-3)(k-3) = 9.$$ Thus, $j-3$ and $k-3$ are pairs of positive factors of $9$, so $(j-3,k-3) = (1,9),(3,3),(9,1)$. These give $k = 4,6,12$, and their sum is $4 + 6 + 12 = \boxed{22}$.}\)

.
 Jul 15, 2019
 #5
avatar+103999 
0

THX, Guest....correction made and my answer agrees with yours!!!

 

 

cool cool cool

CPhill  Jul 15, 2019

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