1988 AIME Problem 8

Attempt Problem 8 of the 1988 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AIME solutions, or check the answer key.

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8.

The function f,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),(x+y)f(x,y)=yf(x,x+y).\begin{gathered}f(x,x)=x,\\f(x,y)=f(y,x),\\(x+y)f(x,y)=yf(x,x+y).\end{gathered}
Calculate f(14,52).f(14,52).
Answer: 364
Concepts:functional equationgreatest common divisorleast common multiple
Small hint:
Apply the third property to (x,y−x)(x,y-x) when y>xy\gt x
Big hint:
Compare the resulting subtraction rule with the same rule for lcm⁡(x,y)\operatorname{lcm}(x,y)
Solution:
For y>x,y\gt x, the third property applied to (x,y−x)(x,y-x) gives f(x,y)=yy−xf(x,y−x).f(x,y)=\frac{y}{y-x}f(x,y-x). The least common multiple obeys the identical relation, because gcd⁡(x,y)=gcd⁡(x,y−x).\gcd(x,y)=\gcd(x,y-x). Hence the quotient f(x,y)lcm⁡(x,y)\frac{f(x,y)}{\operatorname{lcm}(x,y)} is unchanged by each subtraction step of the Euclidean algorithm. On the diagonal it equals f(g,g)g=1,\frac{f(g,g)}{g}=1, so f(x,y)=lcm⁡(x,y).f(x,y)=\operatorname{lcm}(x,y). Therefore f(14,52)=14⋅52gcd⁡(14,52)=364.\begin{aligned}f(14,52)&=\frac{14\cdot52}{\gcd(14,52)}\\&=364.\end{aligned}
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