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), and (x+y)f(x,y)=yf(x,x+y)
Calculate f(14,52).
Solution:
The third property is most useful in the form f(x,z)=z−xzf(x,z−x), which is valid whenever z>x. To obtain this, set z=y+x, so that y=z−x. Now substitute for y in the original third property. Using this new form of the third property and the second given property of f repeatedly, we obtain
f(14,52)=3852f(14,38)=38522438f(14,24)=24521024f(14,10)=526f(10,14)=526414f(10,4)=591f(4,10)=591610f(4,6)=39126f(4,2)=91f(2,4)=9124f(2,2)=364
where the last equality is a consequence of the first given property of f.
The problems on this page are the property of the MAA's American Mathematics Competitions