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Given positive integers x and y such that x not equal to y and 1/x + 1/y = 1/18, what is the smallest possible value for x + y?

 Mar 29, 2020
 #1
avatar+4569 
0

1/x + 1/y = 1/18

 

\(\frac{x+y}{xy}=\frac{1}{18}\).

\(18x+18y=xy\)

 

\(xy-18x-18y=0\)

 

Note that the expression is very close to \((x-18)(y-18)\) is if I add 324 to both sides.

 

\((x-18)(y-18)=324\).

 

324=2^2*3^4

 

12+18=30

 

27+18=45

 

30 and 45 give the largest result, thus x+y=30+45=75.

 Mar 30, 2020
edited by tertre  Mar 30, 2020
 #2
avatar+20768 
0

Since this is a symmetric equation (you can interchange x and y without changing the equation),

the minimum value will occur when x = y.

Solving 1/x + 1/y  =  1/18   --->   1/x + 1/x  =  1/18   --->   2/x  =  1/18   --->   x  =  36   --->   x + y  =  72.

 Mar 30, 2020
 #3
avatar+4569 
0

@geno3141, "x and y such that x not equal to y" in the problem.

tertre  Mar 30, 2020
 #4
avatar+20768 
0

Sorry; try the numbers close to this:  30 and 45.

geno3141  Mar 30, 2020
 #5
avatar+4569 
0

Thank you! I somehow forgot to add 18 to both the numbers, 12 and 27.

tertre  Mar 30, 2020

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