Take 13 integers. Prove that if any 12 of them can be partitioned into two sets of six each with equal sums, then all the integers are the same.

Question

Giancarlo Callahan · Accepted Answer

To answer the more general question,Given any positive integers       k   and       n    &amp;gt;    1  , if there are       k    n    +    1   real numbers             r      i      , such that for any       i  , all except             r      i       can be split into       n   groups of       k   each, such that the sum of the reals in any group is the same as any other, then             r      i        =          r      j      .This is true.Instead of reals, start with integers.Notice that if       S   is the total sum, then       S    =          r      i            mod            n   and so             r      i        =          r      j            mod            n  .wlog assume             r      1       is the smallest.Then we have a new set of weights                              r          i                −                  r          1                    n       which has lower       max    {          r      i        }   (note all have become non-negative so we can assume they were non-negative to begin with) and hence we end up with all zeroes. Since this is reversible, the original must have been equal.Thus if the weights were integers, they all have to be the same. This easily extends to rationals weights.Now since             R       is an infinite dimensional vector space over             Q      , we are done.

Gretchen Allison · Accepted Answer

They must all be even or all be odd. If not, you could find a set of 12 with odd sum and could not divide it evenly. Now subtract 1 if they are odd, then divide by 2, and make the same argument, so they are the same mod 4. Repeat.

Take 13 integers. Prove that if any 12 of them can be partitioned into two sets of six each with equal sums, then all the integers are the same.

Answered question

Answer & Explanation

New Questions in Precalculus