\(\text{} \)\(\text{Given positive integers $x$ and $y$ such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{12}$, what is the smallest possible value for $x + y$?}\)
I know that the question was asked here: https://web2.0calc.com/questions/given-positive-integers-x-and-y-such-that-x-not-y, but that is the wrong answer.
+128474 1/ x + 1/ y = 1 / 12
[ x + y ] / xy = 1 / 12
12 [ x + y ] = xy
12x + 12y = xy
12y - xy = - 12x
y ( 12 - x) = - 12x
y = 12x
_______
x - 12
We will have positive integers for x, y when
x y
13 156
14 84
15 60
16 48
18 36
20 30
24 24
28 21
The smallest sum of x , y when x ≠ y is 49
Thanks to EP for spotting my earlier error !!!!
+128474 Here's one more way :
Let x = 12 + a
Let y = 12 + b
1 / [ 12 + a] + 1 / [12 + b] = 1 /12
[ 24 + a + b ] / [ 144 + 12a + 12b +ab ] = 1/12
12* [ 24+ a + b ] = [ 144 + 12a + 12b + ab ]
288 + 12a + 12b = 144 + 12a + 12b + ab
144 = ab
So
a b
12 12
16 9
18 8
24 6
36 4
48 3
72 2
144 1
The smallest value of x + y =
12 + 16 , 12 + 9 =
28 , 21 =
49
