A function $f : \text{N}^+ \to \text{N}^+$, defined on the set of positive integers $\text{N}^+$, satisfies the following properties: $$f(n) = f(n/2) \quad \text{if } n \text{ is even}$$ $$f(n) = f(n+5) \quad \text{if } n \text{ is odd}$$ Let $R = \{i \mid \exists j : f(j) = i\}$ be the set of distinct values that $f$ takes. The maximum possible size of $R$ is ________ .

We can analyze the equivalence classes induced on $\mathbb{N}^+$ by the transitions:

1. If $n$ is even, $n \to n/2$.

2. If $n$ is odd, $n \to n+5 \to (n+5)/2$.

Let us analyze these transformations modulo 5:

• If $n$ is a multiple of 5 ($n \equiv 0 \pmod 5$):

  • If $n$ is even, $n/2$ is also a multiple of 5.

  • If $n$ is odd, $(n+5)/2$ is also a multiple of 5.

  All positive multiples of 5 eventually reduce to the cycle $5 \to 10 \to 5$, meaning $f(n) = f(5)$ for every $n$ divisible by 5.

• If $n$ is not a multiple of 5 ($n

ot\equiv 0 \pmod 5$): - Tracing small non-multiples of 5:

This forms a single cycle containing $\{1, 2, 3, 4, 6, 8\}$. - Any other non-multiple of 5 will eventually transition into this cycle (e.g., $7 \to 12 \to 6$, $9 \to 14 \to 7 \to 6$, $11 \to 16 \to 8$, etc.).

Thus, the domain $\mathbb{N}^+$ is partitioned into exactly two connected components:

1. Numbers that are multiples of 5, which all map to the same value $f(5)$.

2. Numbers that are not multiples of 5, which all map to the same value $f(1)$.

Hence, $f$ can take at most 2 distinct values, so the maximum size of $R$ is 2.