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# How do you find positive integers when dealing with fractions

here's how the problem looks like:

1/x + 1/y = 1/12

HOW ON EARTH DO I DO THAT??

This is an integer equation. To solve integer equation, you can isolate either x or y.

Isolate x,

x = 12y / (y - 12) ......(1)

Since x and y have symmetry in the original equation, you only need to find half of the solutions, and then switch x and y to get another half solutions.

To get positive integer solution, y -12 > 0. So, you start searching y = 13.

y = 13, x = 156; and x = 13, y = 156;

y = 14, x = 84; and x = 84, y = 14;

y = 15, x = 60; and x = 15, y = 60;

y = 16, x = 48; and x = 16, y = 48;

y = 17, x has no integer solution

y = 18, x = 36; and x = 18, y = 36;

y = 19, x has no integer solution

y = 20, x = 30; and x = 20, y = 30;

y = 21, x = 28; and x = 21, y = 28;

y = 24, x = 24

First of all, observe that both x and y are greater than 12 and the smaller of x and y is at most 24.

To make it easier, let's solve for y first.

1/x + 1/y = 1/12

1/y = 1/12 - 1/x = (x - 12) / (12x)

y = 12x / (x - 12)

Now try all integers from 13 to 24 for x to see if y winds up being an integer. Also, swapping values of x and y in a solution gives another solution.

Results:

(13,156) , (14,84) , (15,60) , (16,48) , (18,36) ,

(20,30) , (21,28) , (24,24) , (28,21) , (30,20) ,

(36,18) , (48,16) , (60,15) , (84,14) , (156,13)