1. **State the problem:** Paul uses some number of trays A, B, and C to bake a total of 378 donuts. Tray A has 9 donuts, Tray B has 3 donuts, and Tray C has 6 donuts. We need to find how many of each tray was used to bake exactly 378 donuts. 2. **Set variables:** Let $x$ be the number of Tray A used, $y$ for Tray B, and $z$ for Tray C. 3. **Write the equation:** The total donuts baked is the sum of donuts from all trays: $$9x + 3y + 6z = 378$$ 4. **Simplify the equation:** Divide both sides by 3 to simplify: $$3x + y + 2z = 126$$ 5. **Find integer solutions:** Since $x$, $y$, and $z$ represent counts of trays, they are non-negative integers. We can express $y$ in terms of $x$ and $z$: $$y = 126 - 3x - 2z$$ 6. **Check for non-negative integer solutions:** $y \geq 0$, so $$126 - 3x - 2z \geq 0 \Rightarrow 3x + 2z \leq 126$$ 7. **Example solution:** For instance, choose $x = 0$, then $$2z \leq 126 \Rightarrow z \leq 63$$ If $z=0$, then $y=126$, so $(x,y,z) = (0,126,0)$ fits. If $x=10$ and $z=20$, then $$3(10)+2(20) = 30 + 40 = 70$$ and $$y = 126 - 70 = 56$$ Thus, $(x,y,z) = (10,56,20)$ is also a solution. 8. **Summary:** There are multiple integer solutions satisfying $$9x + 3y + 6z = 378$$ with $x,y,z \geq 0$. The number of each tray used depends on how Paul distributes trays. **Final answer:** The number of each tray $(x,y,z)$ used must satisfy $$9x + 3y + 6z = 378$$ where $x,y,z$ are non-negative integers. Example valid triplets include $(0,126,0)$ or $(10,56,20)$.