Question

Lost weight. The masses of weights in a set are 1 g, 2 g, 3 g, …, 101 g. The 19-g weight was lost. Is it possible to divide the remaining 100 weights into two groups with the same number of pieces and the same total weight?

Answer 1

I would say yes. The sum of the masses would be
\[\sum_{n=1}^{101} n = 5151g \space \text{less the 19 g mass gives } \space 5132 g \]
Half of this is 2566 g, and so the problem becomes is it possible to arrange exactly half of the remaining 100 masses in such a way that
\[\sum_{i = 1}^{50} m_i = 2566 g\]
If we sum the first 50 masses, i.e. 1 gram through 50 grams (for the moment let us pretend the 19 g mass is still with us) we would find
\[\sum_{n=1}^{50} = 1275 g\]
much too low. Note however that by summing masses 2-51, we add 50g, by summing masses 3-52 we add 100 g, etc. Therefore, notice that
\[2566 - 1275 = 1291\text{, and } \frac{1291}{50} = 25.82. \]
Let us therefore shift upward by 25 masses, i.e. sum masses 26 through 75. This yields
\[\sum_{n=26}^{75} = 2525 g\]
this differs from our desired mass by 41 grams, so let us shift the heavier 41 grams up by one. Thus, we have
\[\sum_{n=26}^{34} n + \sum_{n=36}^{76}n = 2566 g \]
Therefore, we have found an arrangement of 50 masses (half the total number) that weighs 2566 g (half the total mass).