Probability of another 3 integers with same sum and product as the first 3 integers

Asked Mar 17 '15 at 08:41

Active Mar 18 '15 at 16:44

Viewed 147 times

Let us suppose $3$ integers are selected at random from a large range, say $$-1000\leq x\leq y\leq z\leq 1000$$

Now, we define the sum and product: $$\begin{align*}s&=x+y+z \\p&=xyz\end{align*}$$

($s$ and $p$ will not be equal in most cases, sorry for the confusion)

What is the probability that there exists another solution for $(x,y,z)$ that satisfies above 3 equations? (reordering of x, y and z not allowed)

My friend gave me this question, and I have no idea where to start. If we limit ourselves to positive integers, is there a unique solution, or not?

edited Mar 18 '15 at 16:44

Jimmy R.

asked Mar 17 '15 at 08:41

1

If reordering is not allowed then you could write the first constraint as $-1000\le x \le y \le z \le 1000$. – Jimmy R. Mar 17 '15 at 08:55
Among positive integers $1+5+8=2+2+10$ and $1\times5\times8 =2 \times 2 \times 10$ and there are more examples – Henry Mar 17 '15 at 08:58
@Henry Unfortunately not quite what the question is asking. It's asking for $xyz=x+y+z$. – Mar 17 '15 at 08:58
2

@Emrakul You may be correct but I read the question as looking for duplicate pairs of $s$ and $p$ but not requiring $s=p$ – Henry Mar 17 '15 at 09:03
@Henry The title says "same sum and product", which is why I concluded that, though I agree it is mildly unclear. – Mar 17 '15 at 09:17
@Emrakul Sorry for not stating it clearing, but Henry is right. – ghosts_in_the_code Mar 18 '15 at 16:40
@Stef Yes, I could, but does it benefit in any way? – ghosts_in_the_code Mar 18 '15 at 16:40
@ghosts_in_the_code No, it is only a matter of a more compact presentation of the constraints. – Jimmy R. Mar 18 '15 at 16:42
@Stef Okay, I'll do it. – ghosts_in_the_code Mar 18 '15 at 16:43

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