Fomin $D$ d.

Consider a set of weights, each weighing an integer number of grams, and the total weight of all weights is 200 grams. Such a set is called correct if any body with a weight expressed as an integer number of grams from 1 to 200 can be balanced by some number of weights from the set, and in a unique way (the body is placed on one pan of the scales, the weights - on the other; two ways of balancing, differing only in the replacement of some weights with others of the same weight, are considered the same).

a) Provide an example of a correct set where not all weights are 1 gram each.

b) How many different correct sets exist?

(Two sets are different if some weight participates in these sets a different number of times.)
The correct set must correspond to the factorization of the number 201 (see the solution to problem $\underline{98056}$), and it only factors into two factors: $201=3 \cdot 67$.

## Answer

a) Two weights of 67 grams each and 66 weights of 1 gram each, or 66 weights of 3 grams each and two of 1 gram each.

b) 3 sets.

## Author: Fomin D:

Consider a set of weights, each weighing an integer number of grams, with the total weight of all weights being 500 grams. Such a set is called correct if any object with a weight expressed as an integer number of grams from 1 to 500 can be balanced by some number of weights from the set, and in a unique way (the object is placed on one pan of the scales, the weights on the other; two ways of balancing, differing only in the replacement of some weights with others of the same weight, are considered the same).

a) Provide an example of a correct set where not all weights are 1 gram each.

b) How many different correct sets exist?

(Two sets are different if a certain weight participates in these sets a different number of times.)

## Solution

Let the largest weight of a weight in some correct set be $M$ (grams). This means that any smaller weight can be balanced by smaller weights. Let the weight of all smaller weights be $m$. Clearly, $m \geq M-1$. But if $m \geq M$, then we have two ways to balance the weight $M+r$, where $r$ is the remainder of the division of $m$ by $M$. Therefore, $m=M-1$.

Let there be $k$ weights of the maximum weight. Then the total weight of all weights $k M+m=500$, which means 501 is divisible by $M$. Determining $M$, we can determine the weight of the second heaviest weight. Repeating the previous reasoning shows that it must be a divisor of $M$. But 501 has exactly two divisors other than 1 and 501: 3 and 167, each of which is a prime number. Therefore, there are exactly two correct sets, not counting the trivial set of 500 weights of 1 gram each: two weights of 167 grams each and 166 weights of 1 gram each; 166 weights of 3 grams each and two weights of 1 gram each.

## Answer

a) Two weights of 167 grams each and 166 weights of 1 gram each, or 166 weights of 3 grams each and two of 1 gram each.

b) 3 sets