Selection tests for the International Mathematical Olympiad, 2013

Let $n$ be a positive integer. Consider $D_{1}$, the set of all divisors of $n$ with unit digit $3$, and $D_{2}$ the set of all the other divisors of $n$.

If the unit digit of $n$ is different from $9$, consider the map $\delta: D_{1} \longrightarrow D_{2}$ defined by $\delta(d) = \frac{n}{d}$ for all $d \in D_{1}$. If $d \equiv 3 \bmod 10$ and $\delta(d) \equiv 3 \bmod 10$ for some $d \in D_{1}$, then $n = d \cdot \delta(d) \equiv 9 \bmod 10$ which is a contradiction. Then the map $\delta$ is well defined. Clearly, the map $\delta$ is an injective map. Therefore the cardinality of $D_{1}$ is less than or equal to the cardinality of $D_{2}$. Hence $p \leq 50$.

If the unit digit of $n$ is equal to $9$, the integer $n$ is neither divisible by $2$ nor by $5$. This means that all prime divisors of $n$ have unit digit equal to $1, 3, 5$, or $9$.

If all prime divisors of $n$ have unit digits equal to $1$ or $9$, all divisors of $n$ have unit digits equal to $1$ or $9$. Therefore $p = 0$.

If the integer $n$ has a prime divisor $p$ with unit digit $3$ or $7$, consider the map $\delta: D_{1} \longrightarrow D_{2}$ defined for $d \in D_{1}$ by $\delta(d) = \frac{d}{p}$ if $p$ divides $d$ and by $\delta(d) = p d$ otherwise. We can see easily in both cases, that the unit digit of $\delta(d)$ is equal to $1$ or $9$. Hence $\delta$ is well defined.

Assume that there exist $d_{1}, d_{2} \in D_{1}$ such that $\delta\left(d_{1}\right) = \delta\left(d_{2}\right)$. It is clear that if both $d_{1}, d_{2}$ are divisible by $p$ or both $d_{1}, d_{2}$ are not divisible by $p$, we have $d_{1} = d_{2}$. Assume that $d_{1}$ is divisible by $p$ and $d_{2}$ is not divisible by $p$. We have $\frac{d_{1}}{p} = d_{2} p$, which is equivalent to $d_{1} = p^{2} d_{2}$. Because $d_{1} \equiv d_{2} \equiv 3 \bmod 10$, we deduce that $p^{2} \equiv 1 \bmod 10$, which contradicts the fact that $p^{2} \equiv -1 \bmod 10$ since $p \equiv 3, 7 \bmod 10$. Hence $\delta$ is injective and therefore the cardinality of $D_{1}$ is less than or equal to the cardinality of $D_{2}$. Thus $p \leq 50$.

Notice that for $n = 3$, its divisors are $1, 3$ and $p = 50$. This proves that the maximum possible value of $p$ is $50$.