Problem (Algebra):
Positive integers $x, y, z$ satisfy $x y+z=160$. Compute the smallest possible value of $x+y z$.

Solution:
First consider the problem with $x, y, z$ positive real numbers. If $x y+z=160$ and $z$ is constant, then $y=\frac{160-z}{x}$, yielding $x+y z=x+\frac{z(160-z)}{x}$. For $a, x>0$, the quantity $x+\frac{a}{x}$ is minimized when $x=\sqrt{a}$ (proof: use the Arithmetic-Geometric Mean Inequality $\frac{A+B}{2} \geq \sqrt{A B}$ with $A=x$ and $\left.B=\frac{a}{x}\right)$; in this case, $x+\frac{a}{x}=2 \sqrt{a}$. Thus $x+y z \geq 2 \sqrt{z(160-z)}$. Considered as a function of $z$, this lower bound is increasing for $z<80$.

These results suggest the following strategy: begin with small values of $z$, and find a factorization of $160-z$ such that $x$ is close to $\sqrt{z(160-z)}$. (Equivalently, such that $\frac{x}{y}$ is close to $z$.) The chart below contains the triples $(x, y, z)$ with the smallest values of $x+y z$, conditional upon $z$.

| $z$ | $(x, y, z)$ | $x+y z$ |
| :---: | :---: | :---: |
| 1 | $(53,3,1)$ | 56 |
| 2 | $(79,2,2)$ | 83 |
| 3 | $(157,1,3)$ | 160 |
| 4 | $(26,6,4)$ | 50 |
| 5 | $(31,5,5)$ | 56 |
| 6 | $(22,7,6)$ | 64 |

Because $x+y z \geq 2 \sqrt{z(160-z)}$, it follows that $x+y z \geq 64$ for $6 \leq z \leq 80$. And because $x+y z>80$ for $z \geq 80$, the minimal value of $x+y z$ is $\mathbf{5 0}$.

Final answer: 50