(Canada, 1988). Consider a set of $r$ natural numbers $S=\left\{a_{1}, \ldots, a_{r}\right\}$. If $A$ is a non-empty subset of $S$, we denote $p_{A}$ as the product of its elements, and $m(S)$ as the arithmetic mean of the $p_{A}$ for all non-empty subsets $A$ of $S$.

We assume that $m(S)=13$ and $m\left(S \cup\left\{a_{r+1}\right\}\right)=49$. Calculate $a_{r+1}$ as well as the values of the elements of $S$.
. The sum of $p_{A}$ for all subsets of $S$, including the empty set, is equal to the product of the $1+a_{i}$ (setting $p_{\varnothing}=1$). We thus define

$$
\pi(S):=\left(2^{r}-1\right) m(S)+1=\left(1+a_{1}\right) \cdots\left(1+a_{r}\right)
$$

and we denote $S^{\prime}=S \cup\left\{a_{r+1}\right\}$.

We have $\pi(S)=2^{r} \times 13-12$ and $\pi\left(S^{\prime}\right)=2^{r} \times 98-48$. Furthermore, $\pi(S)\left(1+a_{r+1}\right)=\pi\left(S^{\prime}\right)$. The formula

$$
\left(2^{r} \times 98-48\right)+\left(2^{r} \times 6-48\right)=8 \times\left(2^{r} \times 13-12\right)
$$

shows that $\pi(S)$ must divide $2^{r} \times 6-48$. If $r>3$, this is impossible, because we would have $0<2^{r} \times 6-48<2^{r} \times 13-12$.

We calculate:

| $r$ | $2^{r} \times 13-12$ | $2^{r} \times 6-48$ |
| :---: | :---: | :---: |
| 1 | 14 | -36 |
| 2 | 40 | -24 |
| 3 | 92 | 0 |

Thus $r$ is 3, $\pi(S)=92$, $\pi\left(S^{\prime}\right)=736$, and $a_{4}=7$. Then

$$
\pi(S)=\left(1+a_{1}\right)\left(1+a_{2}\right)\left(1+a_{3}\right)=2^{2} \times 23
$$

has divisors $1,2,4,23,46,92$, leading to the possibilities $S=\{0,1,45\}$ and $S=\{0,3,22\}$.