A price is a number $ \frac{x}{100}$ with a positive integer $x$. How many distinct prices can there be such that their product equals their sum, i.e., what is the largest $n$ such that positive integers $x_1 < \cdots < x_n$ exist with $x_1 \cdots x_n = 100^{n-1} \cdot(x_1 + \cdots + x_n)$?
As noted in OEIS sequence A382547, it is easy to see that $n \le 274$, but this is probably much too large an upper bound. Can anyone find a smaller bound?
Let $n$ be a positive integer, and let $x_1<\ldots<x_n$ be positive integers such that $$\prod_{k=1}^nx_k=100^{n-1}\sum_{k=1}^nx_k.\tag{1}$$ As noted in the comments under OEIS sequence A382547, for all $k$ you have $x_k\geq k$, and so $$\prod_{k=1}^nx_k\geq x_n\prod_{k=1}^{n-1}k=(n-1)!x_n.\tag{2}$$ Also for all $k$ you have $x_k\leq x_n$, and so $$100^{n-1}\sum_{k=1}^nx_k\leq100^{n-1}\sum_{k=1}^nx_n=100^{n-1}nx_n.\tag{3} $$ Substituting this into equation $(1)$, it follows that $$100^{n-1}nx_n\geq (n-1)!x_n,$$ and hence that $100^{n-1}n\geq(n-1)!$. This in turn implies that $n\leq274$.
We can expand the upper bound $(3)$ a bit to get $$100^{n-1}(n-1)x_{n-1}+100^{n-1}x_n\geq(n-1)!x_n,$$ from which it follows that $$\frac{x_{n-1}}{x_n}\geq\frac{(n-2)!}{100^{n-1}}-\frac{1}{n-1}.$$ If $n=274$ then this implies $x_{n-1}>0.4874x_n$. Then similar to $(2)$ we find $$\prod_{k=1}^nx_k>0.4874x_n^2\prod_{k=1}^{n-2}k=0.4874(n-2)!x_n^2,$$ and so as before $$100^{n-1}nx_n>0.4874(n-2)!x_n^2,$$ from which it follows that $$x_n<\frac{100^{n-1}n}{0.4874(n-2)!}<1145.$$ This gives us an upper bound on the number of factors of $5$ on the left hand side of $(1)$: We have $$v_5\left(\prod_{k=1}^nx_k\right)\leq v_5(x_n!)\leq v_5(1144!)=283.$$ But for $n=274$ we see on the right hand side of $(1)$ that $$v_5\left(100^{n-1}\sum_{k=1}^nx_k\right)=(n-1)v_5(100)+v_5\left(\sum_{k=1}^nx_k\right)\geq(n-1)v_5(100)=273\times2=546,$$ a contradiction. We conclude that $n\leq273$.
For positive integers $m$ and $v$, let $$b_5(m,v) := \min \{ a_m : a \in \mathbb N^m, 1 \le a_1 < \cdots < a_m, \ v_5(a_1 \cdot\ldots\cdot a_m) \ge v \}.$$ The values of $b_5(m,v)$ that we will use in the following can easily be determined by computer.
Let $n \ge 2$ be an integer and $x \in \mathbb N^n$ a solution, i.e. $1 \le x_1 < \cdots < x_n$ with $$x_1 \cdot\ldots\cdot x_n = 100^{n-1} (x_1 + \cdots + x_n),$$ which implies $x_n \ge b_5(n, 2n-2)$.
Following Servaes' idea, we estimate $$\begin{eqnarray*} (n-1)! \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1}(n-1)x_{n-1} + 100^{n-1}x_n \end{eqnarray*}$$ and get $$x_{n-1} \ge \alpha_n \, x_n \quad \mbox{with} \quad \alpha_n := \frac{(n-2)!}{100^{n-1}} - \frac1{n-1}.$$ In the following, let $n \ge 270$, which ensures $\alpha_n > 0$. Then we have: $$\begin{eqnarray*} (n-2)! \cdot x_{n-1} \cdot x_n &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} n x_n \end{eqnarray*}$$ $$(n-2)! \cdot \alpha_n \, x_n \le (n-2)! \cdot x_{n-1} \le 100^{n-1} n$$ $$x_n \le \frac{100^{n-1}n}{(n-2)! \cdot \alpha_n} =: \beta_n \tag{1}$$ Furthermore we have: $$\begin{eqnarray} (n-2)! \cdot x_{n-1} \cdot (x_{n-1} + 1) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1} (x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-1) x_{n-1} + 100^{n-1} \cdot \beta_n \end{eqnarray} \tag{2}$$ $$\begin{eqnarray} (n-3)! \cdot x_{n-2} (x_{n-2} + 1) (x_{n-2} + 2) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-2) x_{n-2} + 100^{n-1} \cdot 2 \beta_n \end{eqnarray} \tag{3}$$ $$\begin{eqnarray} (n-4)! \cdot x_{n-3} (x_{n-3} + 1) (x_{n-3} + 2) (x_{n-3} + 3) &\le& x_1 \cdot\ldots\cdot x_n\\ &=& 100^{n-1}(x_1 + \cdots + x_n)\\ &\le& 100^{n-1} (n-3) x_{n-3} + 100^{n-1} \cdot 3 \beta_n \end{eqnarray} \tag{4}$$ In the case of $n = 274$, $x_n \ge b_5(274, 546) = 5500$. But (1) gives $x_n \le 1144$, a contradiction.
In the case of $n = 273$, (1) reads $x_n \le 8549$, so $v_5(x_n) \le 5$ and $$544 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 5,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 539$, so $x_{n-1} \ge b_5(272,539) = 5375$. But (2) gives $x_{n-1} \le 1536$, a contradiction.
In the case of $n = 272$, (1) reads $x_n \le 64877$, so $v_5(x_n) \le 6$ and $$542 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-1}) + 6,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-1}) \ge 536$, so $x_{n-1} \ge b_5(271,536) = 5350$. But (2) gives $x_{n-1} \le 4293$, a contradiction.
In the case of $n = 271$, (1) reads $x_n \le 523728$, so $v_5(x_n) \le 8$, and (2) yields $x_{n-1} \le 12622$, so $v_5(x_{n-1}) \le 5$ and $$540 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-2}) + 13,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-2}) \ge 527$, so $x_{n-2} \ge b_5(269,527) = 5200$. But (3) gives $x_{n-2} \le 2680$, a contradiction.
In the case of $n = 270$, (1) reads $x_n \le 5395811$, so $v_5(x_n) \le 9$, (2) yields $x_{n-1} \le 43004$, so $v_5(x_{n-1}) \le 6$, and (3) yields $x_{n-2} \le 7188$, so $v_5(x_{n-2}) \le 5$ and $$538 = 2n-2 \le v_5(x_1 \cdot\ldots\cdot x_n) \le v_5(x_1 \cdot\ldots\cdot x_{n-3}) + 20,$$ thus $v_5(x_1 \cdot\ldots\cdot x_{n-3}) \ge 518$, so $x_{n-3} \ge b_5(267,518) = 5050$. But (4) gives $x_{n-3} \le 3396$, a contradiction.
Consequently, $n \le 269$.
