2.1. The seller has weights of $1, 2, 4, 8, 16, 32$ grams (one of each) and a balance scale. On the first pan, a candy weighing 25 grams and some three weights were placed, and on the second pan, the remaining three weights, with the scales coming into balance. Indicate the weights of all three weights on the second pan.
Answer: $4,8,32$ (All answers)

## Solution.

1st method. Let's find the total weight of all the weights and candies to understand how many grams we need to balance. $1+2+4+8+16+32+25=88.88: 2=44$ - on each pan. 44 can be obtained in only one way: $32+8+4$, which is the answer.

2nd method. Notice that in any set of weights, the heaviest one weighs more than the sum of the rest. Therefore, the 32 g weight must definitely be on the second pan. Since $25<32$, the 8 g weight must also definitely be on the second pan. Similarly, continuing further, we get that the 4 g weight should also lie on the second pan (its weight is currently less, $25+16>32+8$), while the remaining weights 1,2 are placed on the first, balancing the pans.