Interesting Olympiad style problem.

Question

Problem: Seven vertices of a cube are marked by $0$ and one by $1$. You may repeatedly select an edge and increase the numbers at its both ends by $1$. Your goal is to reach $8$ equal numbers.

Solution: My solution does not match with the author's, so I don't know whether I am correct or not.

Let $S$ be the sum of numbers on all vertices. Then initially $S\equiv 1\pmod{2}.$ Whenever we add $1$ to the number on two vertices $S$ remains invariant. At the end if we have $8$ numbers equal then $S\equiv 0\pmod {2}.$ Which is a contradiction and so we can never reach our goal. Is this argument correct or have I missed something?

Except you meant to say "$S$ mod $2$ remains invariant" rather than "$S$ remains invariant". Good job. — DanielWainfleet, Dec 30 '17 at 06:17

score 2 · Accepted Answer · answered Jan 05 '18 at 12:37

Yes, your answer is correct. Basically, the sum S=1 will keep increasing by 2 each time we choose an edge of the cube. It will go to 3,5,7... As you can see it is always an odd number, got that? And we want to reach eight equal numbers, let's say equal to 'k'. The total sum would be 8k=2*4k an even number. That is impossible, no matter how many steps you take. That's it

Interesting Olympiad style problem.

1 Answers1