Problem 1501

4. (a) The fundamental period of $p$ is the smallest integer $d(p)$ such that $p \mid 10^{d(p)}-1$. Let $s$ be an arbitrary prime and set $N_{s}=10^{2 s}+10^{s}+1$. In that case $N_{s} \equiv 3 \pmod{9}$. Let $p_{s} \neq 37$ be a prime dividing $N_{s} / 3$. Clearly $p_{s} \neq 3$. We claim that such a prime exists and that $3 \mid d(p_{s})$. The prime $p_{s}$ exists, since otherwise $N_{s}$ could be written in the form $N_{s}=3 \cdot 37^{k} \equiv 3 \pmod{4}$, while on the other hand for $s > 1$ we have $N_{s} \equiv 1 \pmod{4}$. Now we prove $3 \mid d(p_{s})$. We have $p_{s} \mid N_{s} \mid 10^{3 s}-1$ and hence $d(p_{s}) \mid 3 s$. We cannot have $d(p_{s}) \mid s$, for otherwise $p_{s} \mid 10^{s}-1 \Rightarrow p_{s} \mid (10^{2 s} + 10^{s} + 1, 10^{s} - 1) = 3$; and we cannot have $d(p_{s}) \mid 3$, for otherwise $p_{s} \mid 10^{3}-1 = 999 = 3^{3} \cdot 37$, both of which contradict $p_{s} \neq 3, 37$. It follows that $d(p_{s}) = 3 s$. Hence for every prime $s$ there exists a prime $p_{s}$ such that $d(p_{s}) = 3 s$. It follows that the cardinality of $S$ is infinite. (b) Let $r = r(s)$ be the fundamental period of $p \in S$. Then $p \mid 10^{3 r} - 1$, $p \nmid 10^{r} - 1 \Rightarrow p \mid 10^{2 r} + 10^{r} + 1$. Let $x_{j} = \frac{10^{j-1}}{p}$ and $y_{j} = \{ x_{j} \} = 0.a_{j} a_{j+1} a_{j+2} \ldots$. Then $a_{j} < 10 y_{j}$, and hence
$$f(k, p) = a_{k} + a_{k+r} + a_{k+2 r} < 10 (y_{k} + y_{k+r} + y_{k+2 r}).$$
We note that $x_{k} + x_{k+s(p)} + x_{k+2 s(p)} = \frac{10^{k-1} N_{p}}{p}$ is an integer, from which it follows that $y_{k} + y_{k+s(p)} + y_{k+2 s(p)} \in \mathbb{N}$. Hence $y_{k} + y_{k+s(p)} + y_{k+2 s(p)} \leq 2$. It follows that $f(k, p) < 20$. We note that $f(2, 7) = 4 + 8 + 7 = 19$. Hence 19 is the greatest possible value of $f(k, p)$.