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# Question, help quick

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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
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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....   Jul 15, 2019
edited by CPhill  Jul 15, 2019
edited by CPhill  Jul 15, 2019
#3
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Thanks!!! Guest Jul 15, 2019
#4
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Sorry, the answer was incorrect, here's the solution:

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

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Jul 15, 2019
#5
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