# What is the result of the equation?

Considering from $a$ until $z$:

$x = (x -a) \cdot (x -b) \cdot (x -c) \dots (x - z)$

What I want is the value of $x$

It´s something easy, but I could not find this here and I think it´s a nice puzzle.

• How can we solve this without a clue what a, b, c, d, e, f, g, h, I, j, k, l, m, n, o, p, q, r, s, t, u, v, w, y, or z is? – warspyking Oct 28 '14 at 3:33
• @warspyking well, look at the answers :P – Mathias711 Oct 28 '14 at 4:16
• @hvd The sample shown in the question includes c, though, making it unable to be the Estonian alphabet ;) – Niet the Dark Absol Oct 28 '14 at 11:44
• @hvd As the language used in this site is English, I believe you can assume I am using the English alphabet. – Doguita Oct 28 '14 at 12:03
• There is a catch: (x-i) is a complex number! – Florian F Oct 28 '14 at 18:10

if $x = \textrm{anything}\cdot(x-x)$ then $x$ has to be $0$

Without knowing the answer, if you start with an assumed value for each of the variables, regardless of the value you assume for any of the variables when you evaluate $x-x$ this would become $0$.

So regardless of other variables, if $x = \textrm{anything}\cdot(x-x)$ then $x$ has to be $0$

• I'm not sure why you would be assuming values for X. What you say after applies for all values of x including zero so in fact you are just noting that the right hand side equals zero and thus the left hand side is too. You sound like you are using proof by contradiction which is just not necessary here. – Chris Oct 28 '14 at 12:38
• Its not necessary to prove, but would it be wrong to use that? – skv Oct 28 '14 at 14:02
• I think my issue with the phrase "If you assume any other value for x" is that the following logic applies if you assume x is 0 as well which means you're not really needing to make an assumption. I'm certainly not saying that you are wrong, just that the explanation seems to be a bit overcomplicated. To put it another way if the equation was "x = (x-x)" you wouldn't say "if you assume x is non-zero", you'd just evaluate the right and be done with it. I should disclaim that I am a mathematician by education so I may have a different view on this to others. :) – Chris Oct 28 '14 at 14:08
• I wrote it that way, because I did it that way :) I tried to assume x is 1 to be precise and assume values when this idea stuck me, so its more an approach to arriving at the answer (a generic one) than, the answer needing it, not sure if many others feel the way you do because the other simpler answer apparently has received many more upvotes than mine, which incidentally is the accepted answer, I will edit it. – skv Oct 28 '14 at 14:13
• That makes a lot of sense. I think this is because of my mathematician's mindset since I read yours more as a formal proof rather than "I did this to get to the answer". – Chris Oct 28 '14 at 14:15

The third from last factor is $x-x$, so the right side is $0$, so $x=0$

• That´s correct but skv was 2 seconds faster. – Doguita Oct 28 '14 at 10:43
• I don't think you're supposed to accept whoever's faster. You accept whatever answer is better, unless the rules for accepting are different for puzzling.sx for some reason. – Matthew Oct 28 '14 at 20:52
• @Matthew Both answers are good to me. So how choose only one? I picked up the one who wrote first. – Doguita Oct 28 '14 at 21:20
• @Matthew, Doguita, you are certainly encouraged to accept the best answer. In fact, you are encouraged to wait a bit before accepting an answer to see if a better one comes in-in theory if an answer is accepted there is no need to supply another one. If the answers are equally useful, first posted is one way to break the tie. That said, I don't get excited about reputation points, so if between us we have made a good answer I am happy. – Ross Millikan Oct 29 '14 at 0:11
• @Matthew as RossMillikan mentioned, reputation points are not the thing I am worried about, but since you mentioned "You accept whatever answer is better" is there any reason to think that my answer is worse – skv Oct 29 '14 at 8:12