Problem:
Let f be the unique function defined on the positive integers such that
d∣n∑d⋅f(dn)=1
for all positive integers n, where the sum is taken over all positive divisors of n. What is f(2023)?
Answer Choices:
A. −1536
B. 96
C. 108
D. 116
E. 144
Solution:
The number 1 has 1 divisor, so 1⋅f(1)=1, and hence f(1)=1. If n is a prime number, say n=p, then n has 2 divisors, namely 1 and p, and so 1=1⋅f(p)+p⋅f(1)=f(p)+p, and hence f(p)=1−p for all primes p. If p is prime and n=p2, then n has 3 divisors, namely 1 , p, and p2, so
1=1⋅f(p2)+pf(p)+p2f(1)=f(p2)+p(1−p)+p2
which says that f(p2)=1−p. If p and q are distinct primes and n=pq, then n has 4 divisors, and
1=1⋅f(pq)+pf(q)+qf(p)+pqf(1)=f(pq)+p(1−q)+q(1−p)+pq
so f(pq)=(1−p)(1−q). If p and q are distinct primes, then pq2 has 6 divisors, and
1=1⋅f(pq2)+pf(q2)+qf(pq)+pqf(q)+q2f(p)+pq2f(1)=f(pq2)+p(1−q)+q(1−p)(1−q)+pq(1−q)+q2(1−p)+pq2
which gives f(pq2)=(1−p)(1−q). Because 2023=7⋅172, it follows that f(2023)= (−6)(−16)=(B)96.
The problems on this page are the property of the MAA's American Mathematics Competitions