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 $\delta :D_1⟶D_2$ defined by $\delta (d)=\frac{n}{d}$ for all $d\in D_1$. If $d\equiv 3\mathrm{mod}10$ and $\delta (d)\equiv 3\mathrm{mod}10$ for some $d\in D_1$, then $n=d\cdot \delta (d)\equiv 9\mathrm{mod}10$ 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 $p\le 50$.

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 $\delta :D_1⟶D_2$ defined for $d\in D_1$ by $\delta (d)=\frac{d}{p}$ if $p$ divides $d$ and by $\delta (d)=pd$ 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 $d_1\equiv d_2\equiv 3\mathrm{mod}10$, we deduce that $p^2\equiv 1\mathrm{mod}10$, which contradicts the fact that $p^2\equiv -1\mathrm{mod}10$ since $p\equiv 3,7\mathrm{mod}10$. Hence $\delta$ is injective and therefore the cardinality of $D_1$ is less than or equal to the cardinality of $D_2$. Thus $p\le 50$.

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