\section*{Task 1 - 111011}

A room is to be equipped with 32 light bulbs such that a total power of 1800 W is achieved. There are sufficient numbers of light bulbs available in \(40 \mathrm{~W}, 60 \mathrm{~W}\) and \(75 \mathrm{~W}\), but no others.

List all possible ways to achieve this configuration.
}

Let the number of 40, 60, and 75-watt light bulbs be denoted by \(x, y, z\) respectively. Then an equipment meets the conditions of the problem if \(x, y, z\) are natural numbers such that

\[
x+y+z=32 \quad(1) \quad ; \quad 40 x+60 y+75 z=1800
\]

Assuming \(x, y, z\) are three such natural numbers, it follows from (1) and (2) that

\[
4 y+7 z=104 \quad(3) \quad \text { so } \quad z=\frac{104-4 y}{7} \leq \frac{104}{7}=14+\frac{6}{7}
\]

which implies \(0 \leq z \leq 14\) (4). From (3), we further have

\[
y=26-\frac{7 z}{4}
\]

Since \(y\) is a natural number, \(7 z\) and thus \(z\) must be divisible by 4.

Hence, from (4), the only possible values for \(z\) are \(0,4,8,12\). From (5), the corresponding values for \(y\) are 26, 19, 12, 5, and from (1), the corresponding values for \(x\) are 6, 9, 12, 15. Therefore, the only possible equipment configurations that meet the conditions of the problem are:

\[
\begin{array}{ccc|ccc|ccc|ccc}
x & y & z & x & y & z & x & y & z & x & y & z \\
\hline 6 & 26 & 0 & 9 & 19 & 4 & 12 & 12 & 8 & 15 & 5 & 12
\end{array}
\]

By substitution, it can be shown that in all four cases, equations (1) and (2) are satisfied.

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