1. **Problem statement:** We need to find all possible combinations of the number of saftevand cups ($x$), kakao cups ($y$), and pieces of chokolade ($z$) that sum up to 90 kr. 2. **Given prices:** - Saftevand: 6 kr. each - Kakao: 15 kr. each - Chokolade: 8 kr. each 3. **Equation:** $$6x + 15y + 8z = 90$$ where $x,y,z \geq 0$ and are integers. 4. **Approach:** We will find all non-negative integer solutions $(x,y,z)$ to the equation. 5. **Step-by-step:** - For each possible $y$ (kakao cups), calculate the remaining amount: $$90 - 15y$$ - Then solve for $x$ and $z$ in: $$6x + 8z = 90 - 15y$$ - Since $x,z \geq 0$, iterate over possible $z$ values and check if $x$ is an integer. 6. **Calculations:** - $y$ can be from 0 to 6 because $15 \times 6 = 90$. For each $y$: - Compute $R = 90 - 15y$ - For $z$ from 0 to $\lfloor R/8 \rfloor$: - Check if $x = \frac{R - 8z}{6}$ is a non-negative integer. 7. **Solutions found:** - $y=0$: $6x + 8z = 90$ - $z=0$: $x=15$ - $z=3$: $x=9$ - $z=6$: $x=3$ - $y=1$: $6x + 8z = 75$ - $z=3$: $x=7$ - $z=6$: $x=2$ - $y=2$: $6x + 8z = 60$ - $z=0$: $x=10$ - $z=3$: $x=4$ - $y=3$: $6x + 8z = 45$ - $z=3$: $x=1$ - $y=4$: $6x + 8z = 30$ - $z=0$: $x=5$ - $y=5$: $6x + 8z = 15$ - $z=0$: $x=2$ - $y=6$: $6x + 8z = 0$ - $z=0$: $x=0$ 8. **Summary of all combinations $(x,y,z)$:** - (15,0,0), (9,0,3), (3,0,6), (7,1,3), (2,1,6), (10,2,0), (4,2,3), (1,3,3), (5,4,0), (2,5,0), (0,6,0) These are all the possible ways the children could have spent exactly 90 kr. on saftevand, kakao, and chokolade, possibly not buying all three items.