## Task B-1.6.

The hotel owners at the beginning of the tourist season bought new blankets, towels, and pillowcases for 4000 kn. They paid 120 kn for each blanket, 50 kn for each towel, and 25 kn for each pillowcase. If they bought a total of 100 items, how many blankets, how many towels, and how many pillowcases were bought? They bought at least one item of each type.
## First solution.

Let $x$ be the number of duvets bought, $y$ be the number of blankets bought, and $z$ be the number of pillows bought, $x, y, z \in \mathbb{N}$.

According to the conditions of the problem, we have:

$x+y+z=100 \quad$ and $\quad 120 x+50 y+25 z=4000$.

1 point

Express one unknown from the first equation and substitute it into the second:

$z=100-x-y$

$120 x+50 y+25(100-x-y)=4000$

$19 x+5 y=300$

From the last equation, express one unknown in terms of the other again:

$x=\frac{300-5 y}{19}=\frac{5(60-y)}{19}$

Since $x$, $y$, and $z$ must be natural numbers, we conclude that the numerator must be divisible by the denominator, that is, the expression $60 - y$ must be a multiple of 19, and $y<60$ must hold.

Then $60-y=19$ or $60-y=19 \cdot 2=38$ or $60-y=19 \cdot 3=57$.

For each of these possibilities, we get one of the three solutions to the problem.

From $y=41$ it follows that $x=5$ and $z=54$, that is, 5 duvets, 41 blankets, and 54 pillows were bought.

From $y=22$ it follows that $x=10$ and $z=68$, that is, 10 duvets, 22 blankets, and 68 pillows were bought.

From $y=3$ it follows that $x=15$ and $z=82$, that is, 15 duvets, 3 blankets, and 82 pillows were bought.