Problem 1079

Obviously, if the two pans can stay in equilibrium, the sum of the masses must be even (namely twice the sum of the masses of the weights placed on one of the pans of the balance).

Conversely, we prove by "strong" induction after $n$ that, if the sum of the masses is even, then one can place the weights conveniently. Assume the statement to be true for any $k < n$ weights whose total mass is even. We distinguish two cases:

• if $m_{n-1} = m_n$, then $m_1 + m_2 + \dots + m_{n-2}$ is even, hence, according to the inductive hypothesis for $k = n-2$, we can place these weights on the two pans of the balance such that the balance stays in equilibrium. Now we simply add the two weights of mass $m_n$, one in each pan, and the balance remains in equilibrium.

• If $m_n \neq m_{n-1}$, instead of the two weights we produce a new one of mass $m = |m_n - m_{n-1}|$. Then $1 \le m \le n-1$ and the sum of the masses of these $n-1$ weights,
$$m_1 + m_2 + \dots + m_{n-2} + m = m_1 + m_2 + \dots + m_n - 2 \cdot \min\{m_{n-1}, m_n\},$$
is even. From the inductive hypothesis for $k = n-1$ it follows that we can place these $n-1$ weights on the pans of the balance such that the balance stays in equilibrium. Now, instead of the weight of mass $|m_n - m_{n-1}|$ we put the weight $\max\{m_{n-1}, m_n\}$, and on the other pan we put the weight of mass $\min\{m_{n-1}, m_n\}$; the balance stays in equilibrium.

The statement is thus proven.

Obviously, if the two pans can stay in equilibrium, the sum of the masses must be even (namely twice the sum of the masses of the weights placed in one of the pans of the balance).

Conversely, assume that the total mass is even. We place the weight of mass $m_n$ on one of the pans. The modulus of the difference of the total weights of the two pans is $m_n \le n$. We place successively the weights of masses $m_{n-1}, m_{n-2}, \dots, m_1$ on the pan whose total mass is smaller. (In case the two pans are in equilibrium, we place the weight on any of two pans.) After placing the weight of mass $m_k$, the difference between the total masses on the two pans is at most $k$. Proceeding this way, we insure that in the end, after placing the weight of mass $m_1 = 1$, the modulus of the difference between the two pans is at most 1. But the total mass being even, it follows that the difference is 0, which is what we wanted.