Carcass wrote:
If x and y are integers and \(3 < x < y < 9\), what is the minimum possible value of \(\frac{x + y}{xy}\) ?
Useful (and often tested) fraction property: \(\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}\)So we can rewrite and simplify our target expression as follows: \(\frac{x + y}{xy}=\frac{x}{xy}+\frac{y}{xy}=\frac{1}{y}+\frac{1}{x}\)
From here we must recognize that we can minimize the value of \(\frac{1}{y}+\frac{1}{x}\) by maximizing the values of \(x\) and \(y\)
Since \(x\) and \(y\) are
integers and \(3 < x < y < 9\), the maximum values \(x\) and \(y\) are \(x=7\) and \(y=8\)
This means the minimum value of \(\frac{1}{y}+\frac{1}{x}=\frac{1}{8}+\frac{1}{7}=\frac{7}{56}+\frac{8}{56}=\frac{15}{56}\)
Answer: \(\frac{15}{56}\)
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Brent Hanneson - founder of Greenlight Test Prep