Property of any odd number of nonnegative integers:

Given $x_1 \leq \cdots \leq x_{2n + 1}$ with each $x_i \in \mathbb{Z}_{\geq 0}$, suppose that for any $x_i$ we remove, the remaining numbers can be assigned to disjoint multisets $A$ and $B$ such that $|A| = |B| (= n)$ and $\sum_{x \in A} A = \sum_{x \in B} B.$ Then all the $x_i$ must be equal.

To prove this, fix any $n \geq 0$ and note that all the $x_i$ must have the same parity. Suppose there were an assignment of values such that the $x_i$ weren't all equal; using the Well-Ordering Principle, consider such a non-trivial assignment with $\sum x_i$ minimal. If $x_1 = 0,$ divide each $x_i$ by $2$ to derive a contradiction; if $x_1 > 0,$ subtract $1$ from each $x_i$ to derive a contradiction. Thus, all the $x_i$ must be equal.

This property extends easily from the nonnegative integers to the rational numbers: if we had a non-trivial assignment of rational values, we could multiply through by their lcd to obtain a non-trivial assignment of integer values, and then subtract from each the smallest integer value to obtain a non-trivial assignment of nonnegative integers, thereby contradicting our earlier conclusion.

Question: Does this property also hold for any odd number of real (or complex) numbers?

Any connections between this question and other areas would, of course, be welcomed.

Question re-stated explicitly: Given $x_1, \ldots, x_{2n + 1}$ with each $x_i \in \mathbb{R}$ (or $\mathbb{C}$), suppose that for any $x_i$ we remove that the remaining numbers can be assigned to disjoint multisets $A$ and $B$ such that $|A| = |B| (= n)$ and $\sum_{x \in A} A = \sum_{x \in B} B.$ Is it true that all the $x_i$ must be equal?

Edit 1: An equivalent problem appears as #15.23 in "Problems and Theorems in Classical Set Theory" (solution on pp. 323-324) by Péter Komjáth and Vilmos Totik (2006).

Edit 2: The integer version of this problem was Putnam Problem B1 (1973); the question about extending to the reals is posed at that solution link as well. The integer version (for 2n+1 = 23) is also included as problem #3.4.31 in "The Art and Craft of Problem Solving" (2e, p. 107) by Paul Zeitz. An earlier extension to the positive reals (easily seen as equivalent to the reals) is AMM Problem 11002 (2003), though the solver whose answer is linked to proceeds in a manner quite different from that of Pierre or Péter.

Edit 3: In a 1992 article, Liong-shin Hahn notes the connection between B1/1973 and an article on abelian groups with no nontrivial element of odd order (Respective sources: L.-S. Hahn, "Alternate Solutions to Putnam Competition Problems," Mathematics Magazine, 65(2), 1992; G.A. Martin, "A class of Abelian groups arising from an analysis of a proof," Amer. Math. Monthly, 95, 1988).

Hahn then proceeds to solve the problem over R and C using the sort of approach used above in AMM 11002's solution. This appears to be an English version of Hahn's earlier write-up (1981), in which he claims to have solved this problem (known as Problem 4304 in Mathmedia). I couldn't find his first solution (see Edit 4); however, I did locate another citation for his having solved it, as well as the original problem. All were written in Chinese (Hahn was born in Taiwan; see his AMS Citation for Public Service), so I will type out the original problem and provide an English translation below.

Traditional Chinese:

4304 (編輯部提供) 有17個球,假定無論任取一個後,剩下的16個都可以分為重重相等的兩堆,每堆8個。試證此17個球的重量必然相同。

Simplified Chinese:

4304 (编辑部提供) 有17个球,假定无论任取一个后,剩下的16个都可以分为重重相等的两堆,每堆8个。试证此17个球的重量必然相同。

English Translation:

4304 (Proposed by the editors) Given 17 balls, suppose that no matter which one is removed, the remaining 16 can be separated into two piles of equal weight, with 8 balls in each pile. Prove that these 17 balls must all be equal in weight.

Edit 4: I managed to obtain a copy of Hahn's original solution (1981) to this problem (over $\mathbb{R}_{+}$, which easily extends to $\mathbb{C}$) by contacting an administrator at Academia Sinica, Taipei, Taiwan. His approach proceeds using only basic linear algebra with an application of Cramer's Rule as the coup de grâce. The solution in Chinese can be viewed here; I will not endeavor to translate it unless it is somehow essential to someone's research. Nonetheless, it's interesting to see a full solution that preceded the AMM solution by 22 years. Though this is nothing like Weil's popularization of the Shimura-Taniyama conjecture from a Japanese journal, it does make one wonder what other gems have been hidden away in Asian journals of mathematics.

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    $\begingroup$ The integer version was a Putnam problem many years ago. $\endgroup$ Aug 24, 2012 at 23:25
  • $\begingroup$ Thanks. I've tracked down the Putnam problem, and then done some further sleuthing with regard to the problem's history. $\endgroup$ Aug 25, 2012 at 6:46
  • $\begingroup$ (See also: A Problem That Bears Repeating: 11002 in the AMM.) $\endgroup$ Jan 16, 2015 at 21:50
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    $\begingroup$ The condition that $x_1 \le \dotsb \le x_{2n + 1}$ is never used (and can't be, because the stated property has nothing to do with the order of the elements), which is a good thing, because it doesn't make sense once the $x$'s are complex numbers. $\endgroup$
    – LSpice
    Feb 23, 2020 at 23:02
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    $\begingroup$ @LSpice Good point! I changed the (re)statement of the problem. $\endgroup$ Feb 24, 2020 at 0:11

1 Answer 1


Yes, the result can be deduced from the rational case. Let $x_1, \ldots, x_{2n+1}\in \mathbb C$ be given. Fix a basis of the $\mathbb Q$-vector space spaned by the $x_i$'s and write the coordinates of $x_i$ in this basis as $(x_i^1, \ldots, x_i^r)$. Now the same property holds for each family $x_1^k,\ldots,x_{2n+1}^k$. As those are now rationals, we conclude that they are all equal. Hence the result follows.

  • $\begingroup$ That's a nice argument! $\endgroup$
    – user9072
    Aug 24, 2012 at 17:52
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    $\begingroup$ This problem, with the above solution, can be found in my book: Problems and solutions in classical set theory with Totik, in the chapter on Hamel bases as problem 15.23. $\endgroup$ Aug 24, 2012 at 19:10
  • $\begingroup$ @Pierre: Thanks! @Péter: I've tracked down a copy and will be sure to take a look through it. (I also love that you included solutions!) $\endgroup$ Aug 24, 2012 at 19:23
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    $\begingroup$ One can prove along similar lines that the result holds for all abelian groups with no element of odd order, by reducing it to $\mathbb Z$ and $\mathbb Z/2\mathbb Z$. $\endgroup$ May 6, 2014 at 14:19
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    $\begingroup$ @EmilJeřábek See jstor.org/stable/2322479 = Martin, G. A. (1988). A class of abelian groups arising from an analysis of a proof. American Mathematical Monthly, 433-436. $\endgroup$ May 7, 2014 at 2:58

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