Problem 11.2. Vera has a set of weights of different masses, each weighing an integer number of grams. It is known that the lightest weight in the set weighs 71 times less than all the other weights combined. It is also known that the two lightest weights in the set together weigh 34 times less than all the other weights combined. What is the smallest number of grams the lightest weight can weigh?
Answer: 35.

Solution. All weights in the solution are expressed in grams.

Let the lightest weight be $m$, then the other weights are $71 m$, and the total weight is $72 \mathrm{~m}$. Let also the two lightest weights together weigh $n$, then the other weights weigh $34 n$, and the total weight is $35 n$, which is divisible by 35. Therefore, $72 m$ is divisible by 35, i.e., $m$ is divisible by 35, since 72 and 35 are coprime.

Thus, $m \geqslant 35$. Now let's provide an example of a possible set of weights where $m=35$. For the set of weights with different weights $35,37,48,2400$, all conditions are met: $35 \cdot 71=$ $2485=37+48+2400,(35+37) \cdot 34=2448=48+2400$.