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In the equation 1/j +1/k = 1/3, both j & k are positive integers. What is the sum of all possible values for k?

 Nov 10, 2019
 #1
avatar+142 
+1

WHAT I HAVE SO FAR:

 

J+K           1

___     =  _____

JK              3

 

Right?

What can I do from here?

 Nov 10, 2019
 #2
avatar+142 
+1

After that, I got 

 

3(j+k)=jk

3j+3k=jk

       0=-3j+jk-3k

 

What from here?

MagicKitten  Nov 10, 2019
 #3
avatar+109518 
+1

I have not worked out how to do it formally and I won't guarentee that what i think is correct.

I don't think what you have done is helpful because the answers for j and k must be integers

 

I have only found 1 solution set for j and k, well 2 really because k and j are interchangable.

Can you find one answer, just by mucking around with the numbers?

 Nov 10, 2019
 #7
avatar+142 
0

Ooof...

 

Okay, I will try mucking around!

MagicKitten  Nov 10, 2019
 #4
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+1

Not sure if it helps but,

Solving for k

k= 3j/j-3  , Notice J can't be equal to 3 

If J and K are positive integers, 

Is there a restricted domain?

Also J can't be equal to=(1,2,3,5,7,8,9,10,11,13,14,15,16,17,18,19,20,21,22, etc..) The pattern contuines so, k can be equal to 4, 6 ,12only out of the first 20 numbers.. I don't think the equation works for more... Idk. 

so the sum of possible values of k is 4+6+12=22 

 Nov 10, 2019
edited by Guest  Nov 10, 2019
 #5
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+1

P.S for these type of questions, the best strategy is just trying numbers and see if you notice a pattern as this question doesn't have a specific pattern but only 3 out of the first 30 numbers worked, Then they only can work i guess.(There is usually a restricted domain for these questions like x

Guest Nov 10, 2019
 #6
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+1

Uhh, lol I just noticed i got the sum of the J values not K but as melody said, J and K are interchangable so either way it works.

For example: J=4, then K=12

J=6, K=6 so K=J here.

J=12 , K=4 

So we can interchange between both.

Guest Nov 10, 2019
 #8
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+1

https://www.desmos.com/calculator 

Type your question here but let k=x and j=y or vice versa and notice that k=j at 6 

also when k or j = 4 the other = 12 

Also notice that it will approach 3 but can't equal to 3 as it will be undefined. 

 Nov 10, 2019
 #9
avatar+2854 
+1

Here is my attempt:

 

Good job on finding \( 0=-3j+jk-3k\) guys, we can work from here?

 

We factor:

 

\(0=j(-3+k)-3k\)

Simplifies to:

\(0=j(k-3)-3k\)

 

We add 9 to both sides, this allows us to factor better:

\(9=j(k-3)-3k+9\)

 

We factor once again

\(9=j(k-3)-3(k-3)\)

 

We then simplify:

\(9=(j-3)(k-3)\)

 

Now that we have a binomial expression, we can list the factors of 9 to guess solutions.

 

1 and 9

 

3 and 3.

 

So for 1 and 9, we set them equal

\(j-3=1\)

\(k-3=9\)

 

Solving this, we get J=4 and k=12, since they can be reversed, the values of K and be 4 or 12.

 

Now we solve 3 and 3

\(j-3=3\)

\(k-3=3\)

 

Solving this, we get J=6 and K=6, since they can be reversed, the values of K can only be 6.

 

 

So the possible values for K can be 4, 6, and 12.

 

The sum of that is \(\boxed{22}\)

 

 

 

 

A factor of 9 is also -3 and -3. But do you know why it doesn't work?

.
 Nov 10, 2019
edited by CalculatorUser  Nov 10, 2019
edited by CalculatorUser  Nov 10, 2019
 #10
avatar+142 
0

Undefined?

MagicKitten  Nov 10, 2019
 #11
avatar+2854 
0

both j & k are positive integers

 

The question says that j and k have to be positive.

 

This is why it doesn't work, becuase when you add three to both sides, j and k become 0.

 

0 is neither positive or negative. So 0 is not a solution

CalculatorUser  Nov 10, 2019
 #12
avatar+142 
0

OhHHhhhhHHHHHhhhhhhhhhHhhhhh

MagicKitten  Nov 10, 2019

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