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# Given positive integers $x$ and $y$ such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{18}$, what is the smallest possible value

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Given positive integers $x$ and $y$ such that $x\neq y$ and $\frac{1}{x} + \frac{1}{y} = \frac{1}{18}$, what is the smallest possible value for $x + y$?

May 17, 2021

#1
-1

You get the minimum when x = 27 and y = 54, so x + y = 81.

May 17, 2021
#2
0

how did you get the minimum

ch1ck3n  May 17, 2021
#3
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1/ x   +   1/y    =  1/18

x and y  must  be  >  18

Let   z = 18

Let   x =   z +  a

Let y =  z  +  b

So  we  have  that

1 / ( z + a)   +  1  /(z + b)  =  1 /  z   simplify

[ z + b + z + a ]  / [ (z + a) (z + b) ]    =  1  / z    cross-multiply

z  [2z  + a + b ]   =     (z + a) ( z + b)

2z^2  + az + bz   =  z^2 + az  + bz  +  ab

z^2  =   ab

This implies  that

18^2   =  ab

324  =  ab

Factors of 324   are   1 | 2 | 3 | 4 | 6 | 9 | 12 | 18 | 27 | 36 | 54 | 81 | 108 | 162 | 324 (15 divisors)

x  +  y   is minimized   when     a  = 12   and  b = 17

So....x = z + a  =  18  + 12  =  30

And  y  = z + b  =   18  + 27  =   45

So  x  + y =   30  +  45    =  75   May 17, 2021
#4
+1

thank you for the explanation!

ch1ck3n  May 17, 2021