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You have $3$ boxes $A, B$ and $C$. Boxes $A$ and $B$ are empty to start with and box $C$ has $27$ balls. In "$i$-th" move you make, you must transfer exactly "$i$" balls from one box to another (any transfer to two different boxes are two distinct steps). You CANNOT transfer balls between boxes $A$ and $B$.

Your task is to get to equal number of balls in each box. Can this be done? If yes, how many minimum steps will it take?

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1 Answer 1

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9
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A solution is

possible in 7 steps. 

   A B C
   0 0 27
1: 1 0 26
2: 3 0 24
3: 3 3 21
4: 7 3 17
5: 2 3 22
6: 2 9 16
7: 9 9 9

Proof:

Proof of minimality: At the end each box has 9 balls. Clearly the last move is at most 9, so it has to be done in 9 moves or fewer. Every move adds or removes from box C. Since C must be reduced by 18, an even number, the total number of balls transferred must also be even. The total number of balls transferred by $n$ moves is $T_n=\frac{n(n+1)}2$, so this must be even and at least $18$. This leaves only $n=7$ and $n=8$ as possibilities. I don't think there is an $8$ move solution.

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  • $\begingroup$ Yes. As the sum of numbers from 1 to 5 is 15 (< 18), it is clear that the boundary is between 6 and 9. As you rightly mentioned, from odd even of sum of transferred numbers, it is also clear that it has to be 7 or 8. I also do not think there is a solution for 8 but even if there was, it is not the minimum. For i = 7, there are 2 solutions. One that you have mentioned where you first get to 7 in a box, then to 2 and then to 9. The other way is to first put 2 in a box and then put 7 in that box later. (0,0,27),(0,1,26),(2,1,24),(2,4,21),(2,8,17),(2,3,22),(2,9,16),(9,9,9). $\endgroup$
    – Math Lover
    Aug 7, 2020 at 13:22
  • $\begingroup$ @MathLover: Yes, the second solution simply puts the back-and-forth transfers of +1+4-5 in the other box. $\endgroup$
    – Jaap Scherphuis
    Aug 7, 2020 at 13:25
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    $\begingroup$ I confirm that there is no solution for 8. There are 168 solutions for 7. $\endgroup$
    – RobPratt
    Aug 7, 2020 at 14:21
  • $\begingroup$ @RobPratt That can't be right. What are some of those solutions you found? $\endgroup$
    – Jaap Scherphuis
    Aug 7, 2020 at 14:45
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    $\begingroup$ Sorry, I had not properly constrained C=27 initially. There are actually only 6 solutions: (0,0,27),(0,1,26),(2,1,24),(2,4,21),(2,8,17),(2,3,22),(2,9,16),(9,9,9); (0,0,27),(0,1,26),(0,3,24),(3,3,21),(7,3,17),(2,3,22),(2,9,16),(9,9,9); (0,0,27),(0,1,26),(0,3,24),(3,3,21),(3,7,17),(3,2,22),(9,2,16),(9,9,9); (0,0,27),(1,0,26),(3,0,24),(3,3,21),(7,3,17),(2,3,22),(2,9,16),(9,9,9); (0,0,27),(1,0,26),(3,0,24),(3,3,21),(3,7,17),(3,2,22),(9,2,16),(9,9,9); (0,0,27),(1,0,26),(1,2,24),(4,2,21),(8,2,17),(3,2,22),(9,2,16),(9,9,9) $\endgroup$
    – RobPratt
    Aug 7, 2020 at 15:07

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