Find a solution to the following system of simultaneous equations. All letters stand for integers (positive, zero, or negative), and different letters are different integers.

The preferred solution will be the one in which the difference between the largest and smallest of the values is the least.

C+E+R+O = 0

U+N+O = 1

D+O+S = 2

T+R+E+S = 3

C+U+A+T+R+O = 4

C+I+N+C+O = 5

S+E+I+S = 6

S+I+E+T+E = 7

O+C+H+O = 8

N+U+E+V+E = 9

D+I+E+Z = 10


1 Answer 1


This system of equations is underdetermined, and it has 3 degrees of freedom. Denoting these with $x$, $y$ and $z$, a possible substitution of the original letters is:

At this point I don't have any better idea than quasi-randomly trying $x$, $y$ and $z$ values to see if these expressions indeed produce distinct values, and if they do, calculate the span of them.

To provide a working example, $x=0$, $y=-1$, $z=1$ seems to produce distinct numbers in the range of $[-16; 27]$, so that produces a span of 43. I doubt this is minimal, feel free to use this input for further finetuning.

also, confirmed with programming: $24$ is the minimal span if $\max(|x|, |y|, |z|) \le 120$

Update 2
Now it can easily be proven, that $24$ is the minimal span in general.
If we want the span to be $24$ at most, any letter pairs should have a difference of at most $24$, so $|3x+2|\le24$, so we get $|x|\le8$.
But also $\text{S}-\text{I}$ and $\text{S}-\text{C}$ has to be at most $24$ (in its absolute value), which then gives $|y|\le32$ and $|z|\le32$, and since every tuplet in this range has been checked, we are done.

So the value of the letters can be any of the following two sets to achieve the minimal span of $24$:

     A   C   D   E   H   I   N   O   R   S   T   U   V   Z
    10  -9   2  -2  11  14   6   3   8  -3   0  -8  15  -4
    -2  12   3   5  10 -11  -1  -7 -10   6   2   9  -9  13
  • $\begingroup$ @JonMarkPerry, but if x=-8, |3x+2|=22 $\endgroup$
    – elias
    Mar 19, 2019 at 9:16
  • $\begingroup$ For my proof, the bound |x|<=8 is enough, no need to be more strict. I could not use it anyway in the next step, when trying to bound y and z. $\endgroup$
    – elias
    Mar 19, 2019 at 9:22
  • $\begingroup$ I'm not stating that $|3x+2|\le24$ is equivalent with $|x|\le8$. I'm just saying the latter is a consequence of the former. $\endgroup$
    – elias
    Mar 19, 2019 at 9:27
  • $\begingroup$ So you claim that $|3x+2|\le24$ doesn't cause $|x|\le8$? Could you provide a counterexample? I mean an $x$ that fulfills the first, but does not fulfill the second? $\endgroup$
    – elias
    Mar 19, 2019 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.