## Task 2 - 300832

Given are three different types of balls; there are 100 pieces of each type:

Type $A$: balls with a mass of $0.3 \mathrm{~g}$ each,

Type $B$: balls with a mass of $1.5 \mathrm{~g}$ each,

Type $C$: balls with a mass of $7.0 \mathrm{~g}$ each.

Investigate whether it is possible to select exactly 100 balls from these such that their total mass is exactly $100 \mathrm{~g}$! If this is the case, determine all such possibilities for the three quantities of balls to be selected from types $A, B$, and $C$!
For every selection of balls $A$ and $B$, these together have a mass $m$, which, measured in tenths of a gram, is a number divisible by 3. To achieve one of the possibilities sought in the problem, one must thus choose a number $z$ of balls $C$ such that the mass of these balls $C$ complements such a mass $m$ to $100 \mathrm{~g}$.

The smallest number $z$ for which this is true is $z=1$, as is evident from $100 g - 1 \cdot 7 g = 93.0 g$.

Further such numbers $z$ of balls $C$ arise only when the number 1 is increased by multiples of 3. Thus, the numbers

$$
z=4,7,10, \ldots
$$

with the remaining masses for balls $A$ and $B$

$$
m=93 g, 72 g, 51 g, 30 g, \ldots
$$

Every number $z=13+n$ with $n \geq 0$ would result in a mass $m$ of at most $100 - (13+n) \cdot 7 < 9 - 6.9 \cdot n$ grams. Even if one were to use only balls $A$ to make up this mass, this would result in no more than $30 - 23 n$ balls $A$, thus a total of no more than

$$
(30-23 n) + 0 + (13+n) = 43 - 22 n
$$

balls; using balls $B$ as well would result in even fewer.

Since this does not reach a total of 100, only the first four initial values mentioned in (1) and (2) remain.

Of these, the 1st, 3rd, and 4th values are eliminated; for to achieve these values with $y$ balls $B$ and consequently (100 - z - y) balls $A$, it would have to be

$$
\begin{array}{lll}
(99-y) \cdot 0.3 + y \cdot 1.5 = 93, & y \cdot 1.2 = 63.3 & \text{or} \\
(93-y) \cdot 0.3 + y \cdot 1.5 = 51, & y \cdot 1.2 = 23.1 & \text{or} \\
(90-y) \cdot 0.3 + y \cdot 1.5 = 30; & y \cdot 1.2 = 3 &
\end{array}
$$

Since this is not possible with integer $y$, only the 2nd value remains in (1) and (2). For this value, the requirements of the problem are exactly met if

$$
(96-y) \cdot 0.3 + y \cdot 1.5 = 72, \quad y \cdot 1.2 = 43.2, \quad y = 36, \quad 96 - y = 60
$$

Thus, it is proven:

It is possible to meet the requirements of the problem, and they are exactly met with 60 balls $A$, 36 balls $B$, and 4 balls $C$.