Problem:
Let f be a function defined on the set of positive rational numbers with the property that f(aâ‹…b)= f(a)+f(b) for all positive rational numbers a and b. Furthermore, suppose that f also has the property that f(p)=p for every prime number p. For which of the following numbers x is f(x)<0?
Answer Choices:
A. 3217​
B. 1611​
C. 97​
D. 67​
E. 1125​
Solution:
If n is a positive integer whose prime factorization is n=p1​p2​⋯pk​, then f(n)=f(p1​)+f(p2​)+⋯+f(pk​)= p1​+p2​+⋯+pk​. Because f(1)=f(1⋅1)=f(1)+f(1), it follows that f(1)=0. If r is a positive rational number, then 0=f(1)=f(r⋅r1​)=f(r)+f(r1​), which implies that f(r1​)=−f(r) for all positive rational numbers r. Hence
f(3217​)=17−5⋅2=7>0
f(1611​)=11−4⋅2=3>0
f(97​)=7−2⋅3=1>0
f(67​)=7−2−3=2>0, and
f(1125​)=2⋅5−11=−1<0
Of the choices, only x=(E)1125​​ has the property that f(x)<0.
The problems on this page are the property of the MAA's American Mathematics Competitions