\(\text{In the equation $\frac{1}{j}+\frac{1}{k}=\frac{1}{3},$ both $j$ and $K$ are positize integers. What is the sum of all possible values of $k$?}\)
Oops, I meant \(k\), not \(K\). And positive, not positize. Sorry.
+128474 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....
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 $4 + 6 + 12 = \boxed{22}$.}\)
