Problem:
Let f be a function defined on the set of positive rational numbers with the property that f(a. b)=f(a)+f(b) for all positive rational numbers a and b. 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=p1p2⋯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=1125 has the property that f(x)<0.
The problems on this page are the property of the MAA's American Mathematics Competitions