1. (mathematics) There are scales with two pans, 4 weights of 2 kg each, 3 weights of 3 kg each, and two weights of 5 kg each. In how many different ways can a 12 kg load be balanced on the scales, if the weights are allowed to be placed on both pans?
Answer: 7 ways
![](https://cdn.mathpix.com/cropped/2024_05_06_7d3a82aa7ed091c0d22dg-1.jpg?height=1074&width=1284&top_left_y=796&top_left_x=154)

Solution: Let $x$ be the number of 2 kg weights used in weighing, $y$ be the number of 3 kg weights, and $z$ be the number of 5 kg weights. Then the equilibrium condition is given by

$$
\pm 2 x \pm 3 y \pm 5 z=12
$$

The plus sign before an unknown, for example $x$, means that the 2 kg weights are placed on the opposite pan from the 12 kg load. Otherwise, a minus sign is used. Keeping the same notation, we assume that $x, y$, and $z$ are integers that can take negative values, with $|x| \leq 4,|y| \leq 3,|z| \leq 2$. In this case, the equilibrium condition takes the form of the equation

$$
2 x+3 y+5 z=12
$$

Let $z=t \in[-2 ; 2]$. Then, solving the Diophantine equation,
$2 x+3 y=12-5 t \rightarrow\left\{\begin{array}{c}x=5 t-12-3 s \\ y=12-5 t+2 s, s \in Z\end{array}\right.$

Case 1. $t=-2$

$\left\{\begin{array}{c}x=-22-3 s \\ y=22+2 s, s \in Z\end{array} \rightarrow\left\{\begin{array}{c}|22+3 s| \leq 4 \rightarrow s=-8,-7,-6 \\ |22+2 s| \leq 3 \rightarrow s=-13,-12,-11,-10\end{array} \rightarrow s \in \varnothing\right.\right.$

Case 2. $t=-1$

$\left\{\begin{array}{c}x=-17-3 s \\ y=17+2 s, s \in Z\end{array} \rightarrow\left\{\begin{array}{c}|17+3 s| \leq 4 \rightarrow s=-5,-6,-7 \\ |17+2 s| \leq 3 \rightarrow s=-7,-8,-9,-10\end{array} \rightarrow s_{1}=-7 \rightarrow\left\{\begin{array}{c}x=4 \\ y=3 \\ z=-1\end{array}\right.\right.\right.$

Case 3. $t=0$

$\left\{\begin{array}{c}x=-12-3 s \\ y=12+2 s, s \in Z\end{array} \rightarrow\left\{\begin{array}{c}|12+3 s| \leq 4 \rightarrow s=-5,-4,-3 \\ |12+2 s| \leq 3 \rightarrow s=-7,-6,-5\end{array} \rightarrow s_{2}=-5 \rightarrow\left\{\begin{array}{l}x=3 \\ y=2 \\ z=0\end{array}\right.\right.\right.$

Case 4. $t=1$

$\left\{\begin{array}{c}x=-7-3 s \\ y=7+2 s, s \in Z\end{array} \rightarrow\left\{\begin{array}{c}|7+3 s| \leq 4 \rightarrow s=-1,-2,-3 \\ |7+2 s| \leq 3 \rightarrow s=-2,-3,-4,-5\end{array} \rightarrow s_{3}=-2 \rightarrow\left\{\begin{array}{l}x=-1 \\ y=3, s_{4}=-3 \\ z=1\end{array} \rightarrow\left\{\begin{array}{l}x=2 \\ y=1 \\ z=1\end{array}\right.\right.\right.\right.$

Case 5. $t=2$

$\left\{\begin{array}{c}x=-2-3 s \\ y=2+2 s, s \in Z\end{array} \rightarrow\left\{\begin{array}{l}|2+3 s| \leq 4 \rightarrow s=0,-1,-2 \\ |2+2 s| \leq 3 \rightarrow s=0,-1,-2\end{array} \rightarrow s_{5}=0 \rightarrow\left\{\begin{array}{l}x=-2 \\ y=2,2 \\ z=2\end{array},-1 \rightarrow\left\{\begin{array}{l}x=1 \\ y=0 \\ z=2\end{array}\right.\right.\right.\right.$

$s_{7}=-2 \rightarrow\left\{\begin{array}{c}x=4 \\ y=-2 \\ z=2\end{array}\right.$