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

tertre  Mar 30, 2020
#4
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Sorry; try the numbers close to this:  30 and 45.

geno3141  Mar 30, 2020
#5
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Thank you! I somehow forgot to add 18 to both the numbers, 12 and 27.

tertre  Mar 30, 2020