CEMC Cayley, 2015

Consider a stack of bills with a total value of $1000 that includes $x$ $20 bills and $y$ $50 bills.

The $20 bills are worth $\$20x$ and the $50 bills are worth $\$50y$, and so $20x+50y=1000$ or $2x+5y=100$.

Determining the number of possible stacks that the teller could have is equivalent to determining the numbers of pairs $(x,y)$ of integers with $x\geq1$ and $y \geq 1$ and $2x+5y=100$.

(We must have $x \geq 1$ and $y \geq 1$ because each stack includes at least one $20 bill and at least one $50 bill.)

Since $x \geq 1$, then $2x \geq 2$, so $5y = 100 -2x \leq 98$.

This means that $y \leq \frac{98}{5}=19.6$.

Since $y$ is an integer, then $y \leq 19$.

Also, since $5y = 100 -2x$, then the right side is the difference between two even integers, so $5y$ is itself even, which means that $y$ must be even.

Therefore, the possible values of $y$ are $2,4,6,8,10,12,14,16,18$.

Each of these values gives a pair $(x,y)$ that satisfies the equation $2x+5y=100$: $$(x,y)=(45,2),(40,4),(35,6),(30,8),(25,10),(20,12),(15,14),(10,16),(5,18)$$ Translating back to the original context, we see that the maximum number of stacks that the teller could have is 9.