+0

# Halp

0
48
3

$$\text{}$$$$\text{Given positive integers x and y such that x\neq y and \frac{1}{x} + \frac{1}{y} = \frac{1}{12}, what is the smallest possible value for x + y?}$$

Dec 31, 2019

#1
0

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.

Dec 31, 2019
#2
+106516
+1

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   !!!!

Dec 31, 2019
edited by CPhill  Dec 31, 2019
#3
+106516
+1

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

Dec 31, 2019