1. **State the problem:** We want to find the number of ways to make 80 cents using only quarters (25 cents), dimes (10 cents), and nickels (5 cents). 2. **Set variables:** Let $q$ be the number of quarters, $d$ the number of dimes, and $n$ the number of nickels. 3. **Write the equation:** The total value must be 80 cents: $$25q + 10d + 5n = 80$$ 4. **Simplify the equation by dividing all terms by 5:** $$\cancel{5} \times 5q + \cancel{5} \times 2d + \cancel{5} \times n = \cancel{5} \times 16$$ which simplifies to $$5q + 2d + n = 16$$ 5. **Express $n$ in terms of $q$ and $d$:** $$n = 16 - 5q - 2d$$ 6. **Constraints:** Since $n$, $q$, and $d$ represent counts of coins, they must be non-negative integers: $$q \geq 0, \quad d \geq 0, \quad n \geq 0$$ 7. **Find all integer solutions:** For each integer $q$ such that $0 \leq q \leq 3$ (because $5q \leq 16$), find all $d$ such that $n \geq 0$: - For $q=0$: $n = 16 - 2d \geq 0 \Rightarrow d \leq 8$ - For $q=1$: $n = 16 - 5 - 2d = 11 - 2d \geq 0 \Rightarrow d \leq 5$ - For $q=2$: $n = 16 - 10 - 2d = 6 - 2d \geq 0 \Rightarrow d \leq 3$ - For $q=3$: $n = 16 - 15 - 2d = 1 - 2d \geq 0 \Rightarrow d \leq 0$ 8. **Count the number of solutions:** - $q=0$: $d=0$ to $8$ gives 9 solutions - $q=1$: $d=0$ to $5$ gives 6 solutions - $q=2$: $d=0$ to $3$ gives 4 solutions - $q=3$: $d=0$ to $0$ gives 1 solution 9. **Total number of ways:** $$9 + 6 + 4 + 1 = 20$$ **Final answer:** There are **20** ways to make 80 cents using quarters, dimes, and nickels.