# Cube and Numbers

my question is the following. On an cube are numbers. The numbers are v, l, r, o, u and h. The twelve absolute amounts of the differences of these numbers are the numbers from 1 to 12. The differences are from the two sides which are next to each other. So the differences are

|v - l|; |v - r|; |v - o|; |v - u|; |u - l|; |u - r|; |u - h|; |h - l|; |h - r|; |h - o|; |o - l|; |o - r|

Now I have to find out one example of this distribution. And then I have to show that the absolute amount of the difference from two opposite sides of the cube with such a distribution is not larger than 17.

My problem is that I don't have an idea to analyse this distribution, so I just can try a long time. Thank you.

• (The letters refer to the German words vorn, links, rechts, oben, unten, hinten.) In order to answer the second question one has to have an overview over all solutions. Note that you may assume $u=0$. – Christian Blatter Sep 21 '14 at 12:33
• What's "twelf"? – barak manos Sep 21 '14 at 12:59
• It should be twelve. – N. F. Taussig Sep 21 '14 at 13:00

An example is: $v = 0,h=17,l=5,r=6,o=7,u=9$

For the second part of the question: suppose there are two opposite sides of the cube with a difference of 18. Consider the 4 paths from one of these sides to the other. Each of these paths have 2 differences that have to add up to 18 or have a difference of 18. We know the difference cant be 18 because that would imply that one of the differences of the path is larger than 12, and that is not allowed. So we need 4 times 2 differences that add op to 18 the possibilities are: $0+18,1+17,2+16,3+15,...,9+9$

But differences larger than 13 are not allowed, and each difference can only occur once, so the only possibilities are $10+8,11+7,12+6$

There are only 3 possibilities but we need 4, so it is impossible to have 2 opposing sides with a difference of 18. You can see that the same holds for any number larger than 18 too.

• Thank you very much for your absolute easily understandable answer. But I have got a question. Why is the pair 9+9 not allowed? Thank you one times again. – groh Sep 21 '14 at 14:14
• $9+9$ is not allowed because then there would be 2 differences that are equal to 9. This means that one of the numbers from 1 to 12 is not a difference. (because there are only 12 differences). I hope this makes it a bit more clear? – Ward Beullens Sep 21 '14 at 14:17
• ohh yes your right, I had realized it a few seconds before your command but I thank you very much. – groh Sep 21 '14 at 14:18
• You mean comment :D Here is a command if you want one: "Accept my answer!" – Ward Beullens Sep 21 '14 at 14:20
• ;D "comment" ;D – groh Sep 21 '14 at 14:22