Difference between revisions of "2024 DMC Mock 10 Problems/Problem 14"

Latest revision as of 12:15, 22 December 2024

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 $2+3+6=\boxed{11}$ .

@@ Line 2: / Line 2: @@
 <cmath> \frac{1}{a} + \frac{1}{b} + \frac{1}{c} < \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1</cmath>
-and if <math>a=1</math> then the expression is greater than one. The problem then reduces to <math>\frac{1}{b}+\frac{1}{c} = \frac{1}{2}.
+and if <math>a=1</math> then the expression is greater than one. The problem then reduces to <math>\frac{1}{b}+\frac{1}{c} = \frac{1}{2}</math>.
-Using a similar process, we find </math>b=3<math> and </math>c=6<math>, so the only solution is </math>(a,b,c) = (2,3,6)<math>. Therefore the answer is </math>2+3+6=\boxed{11}$.
+Using a similar process, we find <math>b=3</math> and <math>c=6</math>, so the only solution is <math>(a,b,c) = (2,3,6)</math>. Therefore the answer is <math>2+3+6=\boxed{11}</math>.