Exercise 8. A list of numbers is said to be pretty if it consists of strictly positive integers such that the sum of these integers is equal to their product. Determine the smallest number of ones that a pretty list of 100 numbers can contain.
Solution to Exercise 8 The list $1,1, \ldots, 1,2,2,3,3,3$ is pretty and contains 95 numbers equal to 1. Let's show that we cannot do better.

We can assume that the first numbers are $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{r}$ with $a_{1} \geqslant 2$ and that the others are equal to 1. We want to show that if $a_{1} \cdots a_{r}=(100-r)+a_{1}+\cdots+a_{r}$ then $r \leqslant 5$.

![](https://cdn.mathpix.com/cropped/2024_06_04_ff4f5d9cabdcb92d54e8g-5.jpg?height=66&width=1694&top_left_y=1390&top_left_x=237)
unit, while the right-hand side decreases by one unit. We deduce that if we replace all $a_{j}$ by 2, then the left-hand side is less than or equal to the right-hand side. Consequently, $2^{r} \leqslant 100+r$. Since $r \leqslant 100$, we have $2^{r} \leqslant 200$, so $r \leqslant 7$. We then have $2^{r} \leqslant 107$, so $r \leqslant 6$. It remains to show that $r=6$ is impossible.

If $a_{i} \geqslant 3$ for all $i$, then by the same reasoning as above, we have $3^{r} \leqslant 100+2 r \leqslant$ 112, so $r \leqslant 4$. Contradiction.

We deduce that $a_{1}=2$, so $2 a_{2} \cdots a_{6}=96+a_{2}+\cdots+a_{6}$. We show similarly that $a_{2}=2$, then $a_{3}=2$, then $a_{4}=2$, so $16 a_{5} a_{6}=102+a_{5}+a_{6}$.

If $a_{5}=2$ then $32 a_{6}=104+a_{6}$, so $a_{6}=\frac{104}{31}$ which is not an integer.

If $a_{5} \geqslant 3$ then the same reasoning as above gives $16 \times 9 \leqslant 102+3+3$, which is false.