6 different positive numbers; sum = product

Question

As the title says, I want to know if there exists a set of 6 DIFFERENT POSITIVE numbers such that their sum equals their product. (a+b+c+d+e+f=abcdef)

Answer 1

If you choose $a, b, c, d, e$ randomly and solve for $f$, you will almost surely get something different from your first five numbers, and there's at least a good chance it will be positive.

For example, for $a=1$, $b=2$, $c=3$, $d=4$, $e=5$ (not very random, but works well enough) we get $$ 15+f = 120 f$$ which we can solve to get $f=\frac{15}{119}$.

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This can be extended to show that there is a sequence of distinct positive rational numbers $(a_i)_{i\in\mathbb N}$ such that $$ \sum_{i=1}^n a_i = \prod_{i=1}^n a_i $$ for all $n\ge 1$.

Answer 2

Let $C=\sqrt[\Large5]{\dfrac{1+2+\dotsb+6}{1\times2\times\dotsb\times6}}$. That is, let $C=\sqrt[\Large5]{\dfrac{21}{720}}\approx0.49315$. Then the following works: \begin{align} a&=\phantom1C\\ b&=2C\\ c&=3C\\ d&=4C\\ e&=5C\\ f&=6C \end{align}