1. **State the problem:** We want to determine if exact change can be made for the amounts 23, 29, 19, and 33 using an infinite supply of coins with denominations 6, 10, and 15. 2. **Understand the problem:** We need to check if there exist non-negative integers $x$, $y$, and $z$ such that: $$6x + 10y + 15z = \text{amount}$$ for each amount. 3. **Check amount 23:** Try to find $x,y,z \geq 0$ such that: $$6x + 10y + 15z = 23$$ - If $z=0$, then $6x + 10y = 23$. - Possible $y$ values: 0,1,2... - For $y=0$, $6x=23$ no integer solution. - For $y=1$, $6x=13$ no integer solution. - For $y=2$, $6x=3$ no integer solution. - For $z=1$, $6x + 10y = 8$. - For $y=0$, $6x=8$ no integer solution. - For $y=1$, $6x=-2$ invalid. No solutions found. 4. **Check amount 29:** $$6x + 10y + 15z = 29$$ - $z=0$: $6x + 10y = 29$ - $y=0$: $6x=29$ no. - $y=1$: $6x=19$ no. - $y=2$: $6x=9$ no. - $z=1$: $6x + 10y = 14$ - $y=0$: $6x=14$ no. - $y=1$: $6x=4$ no. No solutions. 5. **Check amount 19:** $$6x + 10y + 15z = 19$$ - $z=0$: $6x + 10y = 19$ - $y=0$: $6x=19$ no. - $y=1$: $6x=9$ no. - $z=1$: $6x + 10y = 4$ - $y=0$: $6x=4$ no. No solutions. 6. **Check amount 33:** $$6x + 10y + 15z = 33$$ - $z=0$: $6x + 10y = 33$ - $y=0$: $6x=33$ no. - $y=1$: $6x=23$ no. - $y=2$: $6x=13$ no. - $y=3$: $6x=3$ no. - $z=1$: $6x + 10y = 18$ - $y=0$: $6x=18$ yes, $x=3$ So for 33, $x=3$, $y=0$, $z=1$ works. **Final answers:** - 23: No - 29: No - 19: No - 33: Yes Therefore, exact change can only be given for 33 using coins 6, 10, and 15.