2024 DMC Mock 10 Problems/Problem 14
Revision as of 12:14, 22 December 2024 by Pateywatey (talk | contribs) (Created page with "First, <math>a=2</math>, since if <math>a\geq 3</math>, then <cmath> \frac{1}{a} + \frac{1}{b} + \frac{1}{c} < \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1</cmath> and if <ma...")
First, 
, since if 
, then
![\[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} < \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\]](http://latex.artofproblemsolving.com/7/f/2/7f29d749a412ce643b5b8db588681c0ab7100491.png)
and if 
then the expression is greater than one. The problem then reduces to $\frac{1}{b}+\frac{1}{c} = \frac{1}{2}.
Using a similar process, we find$ (Error compiling LaTeX. Unknown error_msg)b=3
c=6
(a,b,c) = (2,3,6)
2+3+6=\boxed{11}$.
