mathisfun7 wrote: Numbers $0$ and $2013$ are written at two opposite vertices of a cube. Some real numbers are to be written at the remaining $6$ vertices of the cube. On each edge of the cube the difference between the numbers at its endpoints is written. When is the sum of squares of the numbers written on the edges minimal? Let $x$ be a vertex and $y,z,t$ the three adjacent vertices. The sum of squares of edges is a quadatic in $x$ : $3x^2-2x(y+z+t)+...$ and is minimum when $x=\frac{y+z+t}3$ If vertices in one face are $0,a,b,c$ and on opposite face are $d,e,2013,f$, we get : $3a-b-e=0$ $3b-a-c=2013$ $3c-b-f=0$ $3d-e-f=0$ $3e-a-d=2013$ $3f-d-c=2013$ And so $a=c=d=\frac 252013$ and $b=e=f=\frac 352013$ and sum os squares of edges is $\frac 652013^2$

You can think of a potential difference applied to the cube across the main diagonal (with each edge being an identical resistor). Then the current distribution minimizes the power dissipated, which is $\sum \frac{V^2}{R}$. The potentials at different vertices of the cube are easy to find (a cube in this manner can be thought of as three resistances in parallel, in series with six resistances in parallel, in series with three resistances in parallel giving that the potential difference between points that are one edge away from grounded vertex are at $\frac{2v}{5}$, two edges away are at $\frac{3v}{5}$ where $v$ is the potential difference across the main diagonal. $2013$ in this case), and from there we get the result, conveniently minimized already. Generalizations (to higher powers of cubes, different arrangements, etc) are also rather doable this way. The answer is just $V^2/R_{eq}$ where $R_{eq}$ is the equivalent resistance.