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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

0
323
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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 $x + y$?

Jul 13, 2021

### 2+0 Answers

#1
+208
0

$$\frac{1}{x}+\frac{1}{y}=\frac{1}{18} \\x+y=\frac{xy}{18} \\18x+18y=xy \\xy-18x-18y=0 \\xy-18x-18y+324=324 \\(x-18)(y-18)=324$$

factors of 324: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 81, 108, 162, and 324

can you do it yourself now?

JP

Jul 13, 2021
#2
+26319
+1

Given positive integers x and y such that $$x \ne y$$ and

$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{18}$$,
what is the smallest possible value for $$x+y$$?

$$\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{18} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{18} \\\\ \mathbf{xy} &=& \mathbf{18*(x+y)} \\ \hline \end{array}$$

$$\large{AM\ge GM}$$

$$\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 18*(x+y) \\ (x+y)^2 &\ge& 4*18*(x+y) \\ x+y &\ge& 4*18 \\ \mathbf{x+y } &\ge& \mathbf{72} \\ \hline \end{array}$$

$$\text{The smallest possible value for x+y is 72}$$
Source: https://www.quora.com/Given-positive-integers-x-and-y-x-does-not-equal-y-and-frac-1-x-frac-1-y-frac-1-12-what-is-the-smallest-possible-value-for-x-y

In general:

$$\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{n} \\\\ \mathbf{x+y } &\ge& \mathbf{4n} \\ \hline \end{array}$$

$$\text{The smallest possible value for x+y is 4n}$$

Jul 14, 2021