Team Selection Test

We will show that the set $S=\{k\cdot 5^{k-1}:k\in {\mathbb{Z}}^{+}\}$ is an appropriate example.

We first prove that $\mathbb{P}$ is $S$-proper for $N=4$. Let $p\ne 5$ be prime and $0\le n<p$. Then by Chinese remainder theorem there exists a positive integer $k$ satisfying the conditions $k\equiv n(\mathrm{mod}p)$ and $k\equiv 1(\mathrm{mod}p-1)$. Therefore, by Fermat's theorem we obtain that $k\cdot 5^{k-1}\equiv n(\mathrm{mod}p)$. For $p=5$; as $1,10\in S$, $N=4$ works.

Now suppose that ${\mathbb{Z}}^{+}$ is $S$-proper. Then there exists a positive integer $N$ satisfying the given conditions. Therefore we can get any residue in $(\text{mod }{5}^N)$ as sum of at most $N$ elements of $S$. Since $k\cdot 5^{k-1}\equiv 0(\mathrm{mod}{5}^N)$ for all $k>N$, we have that for each integer $0\le n<5^N$, there exist non-negative integers $x_1,x_2,\dots ,x_N$ such that $$x_1\cdot 1+x_2\cdot 2\cdot 5+\cdots +x_i\cdot i\cdot 5^{i-1}+\cdots +x_N\cdot N\cdot 5^{N-1}\equiv n(\mathrm{mod}{5}^N)$$ and $x_1+x_2+\cdots +x_N\le N$. Observe that for each $n$, we get a different $N$-tuple. On the other hand, the number of $N$-tuples of non-negative integers satisfying $x_1+x_2+\cdots +x_N\le N$ is $\left(\genfrac{}{}{0ex}{}{2N}{N}\right)<5^N$ and hence we get a contradiction.