AIME 1988 · 第 8 题

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 $f(x,z):$

$$\begin{aligned} zf(x,z-x) &= (z-x)f(x,z) \\ f(x,z) &= \frac{z}{z-x} \cdot f(x,z-x). \end{aligned}$$ Using the properties of $f,$ we have

$$\begin{aligned} f(14,52) &= \frac{52}{38} \cdot f(14,38) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot f(14,24) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(14,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(10,14)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(10,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(4,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(4,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(2,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot f(2,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot 2\\ &=\boxed{364}. \end{aligned}$$ ~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,

$$\begin{aligned} f(14,52) & = f(14,14 + 38) \\ & = \frac{52}{38}\cdot f(14,38) \\ & = \frac{52}{38}\cdot \frac{38}{24}\cdot f(14,14 + 24) \\ & = \frac{52}{24}\cdot f(14,24) \\ & = \frac{52}{10}\cdot f(10,14) \\ & = \frac{52}{10}\cdot \frac{14}{4}\cdot f(10,4) \\ & = \frac{91}{5}\cdot f(4,10) \\ & = \frac{91}{3}\cdot f(4,6) \\ & = 91\cdot f(2,4) \\ & = 91\cdot 2 \cdot f(2,2) \\ & = \boxed{364}. \end{aligned}$$

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

$$\begin{aligned} y \cdot \mathrm{lcm}(x,x+y) &= \dfrac{y \cdot x(x+y)}{\gcd(x,x+y)} \\ &= \dfrac{(x+y) \cdot xy}{\gcd(x,y)} \\ &= (x+y) \cdot \mathrm{lcm}(x,y). \end{aligned}$$ 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

$$\begin{aligned} f(14,52) &= \mathrm{lcm}(14,52) \\ &= \mathrm{lcm}(2 \cdot 7,2^2 \cdot 13) \\ &= 2^2 \cdot 7 \cdot 13 \\ &= \boxed{364}. \end{aligned}$$

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Proof that $f(x,y)=\mathrm{lcm}(x,y)$ is the only function instead of just using the fact that this is an AIME problem (Proof credited to AMSP's 101 Algebra Problems):

FTSOC, suppose that there is another function $g(x,y)$ also satisfying the given conditions. Let $S$ be the set of all pairs of positive integers $(x,y)$ such that $f(x,y) \ne g(x,y)$, and let $(m,n)$ be such a pair with minimal sum $m+n$. It is clear that $m \ne n$, otherwise $f(m,n) = f(m,m) = m = g(m,m) = g(m,n)$. By symmetry, we can also assume that $n>m$. Note that

$$nf(m,n-m)=[m+(n-m)]f(m,n-m)$$ $$=(n-m)f(m,m+(n-m))=(n-m)f(m,n)$$ or

$$f(m,n-m=\dfrac{n-m}{n} \cdot f(m,n).$$ Likewise,

$$g(m,n-m) = \dfrac{n-m}{n} \cdot g(m,n).$$ Since $f(m,n) \ne g(m,n)$, $f(m,n-m) \ne g(m,n-m)$. Thus $(m,n-m) \in S$. But $(m,n-m)$ has a smaller sum then $(m,n)$, a contradiction. Therefore, $f(x,y) = \mathrm{lcm}(x,y)$ is the only solution.

~Yiyj1