## Task 1 - 221231

All non-negative integer solutions of the system of equations

$$
\begin{aligned}
x_{1}+11 x_{2}+21 x_{3}+31 x_{4}+41 x_{5} & =55 \\
2 x_{1}+12 x_{2}+22 x_{3}+32 x_{4}+42 x_{5} & =60 \\
3 x_{1}+13 x_{2}+23 x_{3}+33 x_{4}+43 x_{5} & =65 \\
4 x_{1}+14 x_{2}+24 x_{3}+34 x_{4}+44 x_{5} & =70 \\
5 x_{1}+15 x_{2}+25 x_{3}+35 x_{4}+45 x_{5} & =75
\end{aligned}
$$

are to be determined.
The above system of equations is obviously equivalent to the simpler one

$$
\begin{aligned}
x_{1}+11 x_{2}+21 x_{3}+31 x_{4}+41 x_{5} & =55 \\
x_{1}+x_{2}+x_{3}+x_{4}+x_{5} & =5
\end{aligned}
$$

since (2) can be derived by simple subtraction from the first two equations, and conversely, all equations of the original system can be obtained by adding a suitable multiple of (2) to (1). Subtracting (2) from (1) and dividing by 10, we immediately obtain the additional equation

$$
x_{2}+2 x_{3}+3 x_{4}+4 x_{5}=5
$$

We now distinguish the following three mutually exclusive cases:

1. Case: $x_{5}=1$.

From (3), it immediately follows that

$$
x_{2}=1, x_{3}=x_{4}=0
$$

which, together with (2), yields $x_{1}=3$, which is indeed a solution to our system of equations.

2. Case: $x_{5}=0, x_{4}=1$

Here, (3) gives rise to the two subcases:

a) $x_{2}=0, x_{3}=1$, and $x_{1}=3$ from (2)

b) $x_{2}=2, x_{3}=0$, and $x_{1}=2$ from (2)

both of which also solve (1), and are thus solutions.

3. Case: $x_{4}=x_{5}=0$

Here, using (3) again, we get three subcases, namely

a) $x_{2}=1, x_{3}=2$, and $x_{1}=2$ from (2)

b) $x_{2}=3, x_{3}=1$, and $x_{1}=1$ from (2)

c) $x_{2}=5, x_{3}=0$, and $x_{1}=0$ from (2)

all of which also solve (1) and are thus solutions.

In summary, the following 6 solutions

$$
\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right) \in\{(0,5,0,0,0),(1,3,1,0,0),(2,1,2,0,0),(2,2,0,1,0),(3,0,1,1,0),(3,1,0,0,1)\}
$$

of the given system of equations have been obtained.