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2024 DMC Mock 10 Problems/Problem 14

Revision as of 12:15, 22 December 2024 by Pateywatey (talk | contribs)
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First, $a=2$, since if $a\geq 3$, then \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} < \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\]

and if $a=1$ then the expression is greater than one. The problem then reduces to $\frac{1}{b}+\frac{1}{c} = \frac{1}{2}$.

Using a similar process, we find $b=3$ and $c=6$, so the only solution is $(a,b,c) = (2,3,6)$. Therefore the answer is $2+3+6=\boxed{11}$.

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