11.2. There are $n$ weights weighing $1,2 \ldots, n$ (g) and a two-pan balance. Can all the weights be arranged on the balance so that one pan has twice as many weights as the other, and the balance is in equilibrium: a) for $n=90$; b) for $n=99$?
Answer. a) cannot; b) can. Solution. a) The sum of the weights of all weights $\frac{90 \cdot 91}{2}=4095$ is an odd number, therefore, 90 weights cannot be divided into two piles with equal sums. b) Let's try to find 33 weights with consecutive weights, the total weight of which is equal to half the sum of the weights of all weights. Solving the equation $k+(k+1)+\cdots+(k+32)=\frac{99 \cdot 100}{4}$, i.e. $\frac{(2 k+32) \cdot 33}{2}=99 \cdot 25$, from which $k=59$. Therefore, on the first scale pan, you can place weights with weights $59,60, \ldots, 91$, and all the remaining on the second scale pan.