Eight zeros are written on a blackboard...

Question

Eight numbers, all of them zero, are written on a blackboard. Each move, 4 of the 8 numbers are randomly chosen, say $a,b,c$ and $d$ and replaced with $a+3,b+3,c+2$ and $d+1$ respectively.

Find all positive integers $n$ for which it is possible after some moves that there are eight consecutive numbers on the blackboard, the smallest of which is $n$.

My working: Each move, the sum of the numbers increase by 9. Sum of 8 consec. numbers are multiple of 4. Hence the sum of 8 consecutive numbers after some moves must be a multiple of lcm$(4,9)=36$

Answer 1

As Gribouillis said the sum is $8n+28$. This is multiple of $4$, but not of $8$. Then, as $9m=8n+28$, using the same argument you used, the possible values are $36+72k$.

Now, it is easy to obtain the solution $99999999$. For example, if we consider the first 4 digits: $0000\rightarrow3321\rightarrow4554\rightarrow7866\rightarrow9999$. (You obtain $99999999$ repeating the same in the other digits).

Then, if you have a solution for $k=0$ (hence, for $12345678$) you will have a solution for $k=1$ (adding up $99999999$). And, by induction, for any $k\geq 0$.

Hence, we have reduced the problem to solve $12345678$. You can try to solve it or see the spoiler below:

\begin{align} 00000000 \ (+ 12330000) \\12330000 \ (+00010233)\\12340233\ (+00002133) \\ 12342366 \ (+00003312) \\ 12345678 \ (+00003312) \end{align}

Pd: This is my first post here, hope everything is clear because my MathJax skills are not the best :)