Source: Australian Mathematical Society Gazette Puzzle Corner 35

There are some weights on the two sides of a balance scale. The mass of each weight is an integer number of grams, but no two weights on the same side of the scale share the same mass. At the moment, the scale is perfectly balanced, with each side weighing a total of W grams. Suppose W is less than the number of weights on the left multiplied by the number of weights on the right. Is it always true that we can remove some of the weights from each side and still keep the two sides balanced?

Rem: Ignore the trivial solution of removing all weights from both sides.

  • $\begingroup$ Are you interchangeably using Mass and weight or is it part of the trick $\endgroup$
    – skv
    Nov 7, 2014 at 9:19
  • $\begingroup$ @skv, it's not part of the trick. Mass and weight are interchangeable in the context of this puzzle. $\endgroup$
    – Kenshin
    Nov 7, 2014 at 9:21
  • $\begingroup$ That seemed really easy to disprove at first but every time I try I find something that doesn't quite work. :) $\endgroup$
    – Chris
    Nov 7, 2014 at 9:37
  • $\begingroup$ I removed my answer because was based on a wrong assumption $\endgroup$ Nov 7, 2014 at 9:57
  • 1
    $\begingroup$ Yes, that why my question is also a proof by contradiction. I forgot, that I made an additional assumption in search of a counter-example. $\endgroup$
    – Gere
    Nov 7, 2014 at 10:30

1 Answer 1


Yes, because there must a mass shared by two weights on different sides.

Suppose for contradiction that there is no such mass. Then, the weights are all distinct. Let $a$ and $b$ be the number of weights on the two sides. At minimum, the total weight of the two sides is $(a+b)(a+b+1)/2$ when the masses are $1,2,\dots,a+b$.

Then, we have

$$2w \geq(a+b)(a+b+1)/2 > (a+b)^2/2 \geq 2ab $$

where the last inequality follows from being equivalent to $(a-b)^2/2 \geq 0$. But this contradicts the statement that $w<ab$.


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