Play with numbers

Question

Find three numbers whose sum is equal to their products. (1,2,3 are those numbers) Now My question is how to find these numbers? (Is this information sufficient to apply some mathematics to it and get answer?)

Answer 1

So, we have 3 consecutive numbers, we call them $$n , n+1, n+2$$ We want their sum to be equal to their product $$n + (n+1) + (n+2) = n (n+1)(n+2)$$ $$3n+3 = n^3+3n^2+2n$$

$$n^3+3n^2-n-3=0$$ Solving the polynomial above

$$n^3+3n^2-n-3=(n-1)(n+1)(n+3)=0$$ The roots of the polynomial are $$x=-3,x=-1,x=1$$ Therefore we have three sequence whose sum equal to their product $$-3,-2,-1$$ $$-1,0,1$$ $$1,2,3$$

Answer 2

Here is a solution assuming natural numbers.

Let the numbers be $x \le y \le z$. Then $xyz = x+y+z\le 3z$ implies $xyz \le 3z$.

If $z=0$, then $x=y=z=0$.

If $z\ne 0$, then $xy \le 3$ and so there are only a small number of candidates for $(x,y)$.

Only $x=1$ and $y=2$ work with $xyz = x+y+z$, and give $z=3$.