1988 AIME Problems/Problem 8

Problem

The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: \begin{eqnarray*} f(x,x) & = & x, \\ f(x,y) & = & f(y,x), \quad \text{and} \\ (x + y) f(x,y) & = & yf(x,x + y). \end{eqnarray*} Calculate $f(14,52)$.

Solution

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

\[f(14,52) = f(14,14 + 38) = \frac {52}{38}f(14,38)\]

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

Solution 2

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

Clearly, $\text{lcm}(x,x) = x$ and $\text{lcm}(x,y) = \text{lcm}(y,x)$.

Using the identities $\text{gcd}(x,y) \cdot \text{lcm}(x,y) = xy$ and $\text{gcd}(x,x+y) = \text{gcd}(x,y)$, we have:

$y \cdot \text{lcm}(x,x+y)$ $= \dfrac{y \cdot x(x+y)}{\text{gcd}(x,x+y)}$ $= \dfrac{(x+y) \cdot xy}{\text{gcd}(x,y)}$ $= (x+y) \cdot \text{lcm}(x,y)$.

Hence, $f(x,y) = \text{lcm}(x,y)$ 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, $f(14,52) = \text{lcm}(14,52) = \text{lcm}(2 \cdot 7,2^2 \cdot 13) = 2^2 \cdot 7 \cdot 13 = \boxed{364}$.

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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