HMMT-Nov 2010 team, Problem 9

A function $f(x, y, z)$ is linear in $x, y$, and $z$ such that $f(x, y, z)=\frac{1}{x y z}$ for $x, y, z \in\{3,4\}$. What is $f(5,5,5)$ ?

Solution: We use a similar method to the previous problem. Notice that $f(x, y, 5)=2 f(x, y, 4)-$ $f(x, y, 3)$. Let $f_{2}$ denote the function from the previous problem and $f_{3}$ the function from this problem. Since $3 f_{3}(x, y, 3)$ is linear in $x$ and $y$, and $3 f_{3}(x, y, 3)=\frac{1}{x y}$ for all $x, y \in\{3,4\}$, the previous problem implies that $3 f_{3}(5,5,3)=\frac{1}{36}=f_{2}(5,5)$. Similarly, $4 f_{3}(5,5,4)=f_{2}(5,5)$. Now we have \[ \begin{aligned} f_{3}(5,5,5) & =2 f_{3}(5,5,4)-f_{3}(5,5,3) \\ & =\frac{1}{2} f_{2}(5,5)-\frac{1}{3} f_{2}(5,5) \\ & =\frac{1}{6} f_{2}(5,5) \\ & =\frac{1}{6 \cdot 36} \\ & =\frac{1}{216} \end{aligned} \] $\fbox{\frac{1}{216}}$.

Answer: \frac{1}{216}