# 1988 AIME Problems/Problem 8

## Problem

The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: $$f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).$$ Calculate $f(14,52)$.

## Solution 1 (Algebra)

Let $z = x+y$. By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\ f(x,z) &= \frac{z}{z-x} \cdot f(x,z-x). \end{align*} Using the properties of $f,$ we have \begin{align*} f(14,52) &= \frac{52}{38} \cdot f(14,38) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot f(14,24) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(14,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(10,14)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(10,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(4,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(4,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(2,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot f(2,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot 2\\ &=\boxed{364}. \end{align*} ~MRENTHUSIASM (credit given to AoPS)

## Solution 2 (Algebra)

Since all of the function's properties contain a recursive definition except for the first one, we know that $f(x,x) = x$ in order to obtain an integer answer. So, we have to transform $f(14,52)$ to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one.

Note that $$f(14,52) = f(14,14 + 38) = \frac{52}{38}\cdot f(14,38).$$

Repeating the process several times, \begin{align*} f(14,52) & = f(14,14 + 38) \\ & = \frac{52}{38}\cdot f(14,38) \\ & = \frac{52}{38}\cdot \frac{38}{24}\cdot f(14,14 + 24) \\ & = \frac{52}{24}\cdot f(14,24) \\ & = \frac{52}{10}\cdot f(10,14) \\ & = \frac{52}{10}\cdot \frac{14}{4}\cdot f(10,4) \\ & = \frac{91}{5}\cdot f(4,10) \\ & = \frac{91}{3}\cdot f(4,6) \\ & = 91\cdot f(2,4) \\ & = 91\cdot 2 \cdot f(2,2) \\ & = \boxed{364}. \end{align*}

## Solution 3 (Number Theory)

Notice that $f(x,y) = \mathrm{lcm}(x,y)$ satisfies all three properties:

For the first two properties, it is clear that $\mathrm{lcm}(x,x) = x$ and $\mathrm{lcm}(x,y) = \mathrm{lcm}(y,x)$.

For the third property, using the identities $\gcd(x,y) \cdot \mathrm{lcm}(x,y) = x\cdot y$ and $\gcd(x,x+y) = \gcd(x,y)$ gives \begin{align*} y \cdot \mathrm{lcm}(x,x+y) &= \dfrac{y \cdot x(x+y)}{\gcd(x,x+y)} \\ &= \dfrac{(x+y) \cdot xy}{\gcd(x,y)} \\ &= (x+y) \cdot \mathrm{lcm}(x,y). \end{align*} Hence, $f(x,y) = \mathrm{lcm}(x,y)$ is a solution to the functional equation.

Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value of $f(14,52)$.

Therefore, we have \begin{align*} f(14,52) &= \mathrm{lcm}(14,52) \\ &= \mathrm{lcm}(2 \cdot 7,2^2 \cdot 13) \\ &= 2^2 \cdot 7 \cdot 13 \\ &= \boxed{364}. \end{align*}

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 