Problem:
Let f1β(x)=32ββ3x+13β, and for nβ₯2, define fnβ(x)=f1β(fnβ1β(x)). The value of x that satisfies f1001β(x)=xβ3 can be expressed in the form nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Note that for xξ =32β or β31β,
βf1β(x)=32ββ3x+13β=9x+36xβ7βf2β(x)=f1β(f1β(x))=9β
(9x+36xβ7β)+36β
(9x+36xβ7β)β7β=9xβ6β3xβ7β, and f3β(x)=f1β(f2β(x))=9β
(9xβ6β3xβ7β)+36β
(9xβ6β3xβ7β)β7β=x.β
Thus it follows that f3kβ(x)=x for all positive integers k. Thus f1001β(x)= f2β(x)=9xβ6β3xβ7β=xβ3, which is equivalent to 9x2β30x+25=0, and the only solution to this quadratic is x=35β. Thus m+n=5+3=8β.
Challenge: Show that if p+q=1 and f1β(x)=pβx+q1β, then f3β(x)=x. The given problem highlights a special case of this fact in which p=32β and q=31β.
The problems on this page are the property of the MAA's American Mathematics Competitions