A store sells shirts, water bottles, and chocolate bars. Shirts cost $\$ 10$ each, water bottles cost $\$ 5$ each, and chocolate bars cost $\$ 1$ each. On one particular day, the store sold $x$ shirts, $y$ water bottles, and $z$ chocolate bars. The total revenue from these sales was $\$ 120$. If $x, y$ and $z$ are integers with $x>0, y>0$ and $z>0$, how many possibilities are there for the ordered triple $(x, y, z)$ ?
The store sold $x$ shirts for $\$ 10$ each, $y$ water bottles for $\$ 5$ each, and $z$ chocolate bars for $\$ 1$ each.

Since the total revenue was $\$ 120$, then $10 x+5 y+z=120$.

Since $z=120-10 x-5 y$ and each term on the right side is a multiple of 5 , then $z$ is a multiple of 5 .

Set $z=5 t$ for some integer $t>0$.

This gives $10 x+5 y+5 t=120$. Dividing by 5 , we obtain $2 x+y+t=24$.

Since $x>0$ and $x$ is an integer, then $x \geq 1$.

Since $y>0$ and $t>0$, then $y+t \geq 2$ (since $y$ and $t$ are integers).

This means that $2 x=24-y-t \leq 22$ and so $x \leq 11$.

If $x=1$, then $y+t=22$. There are 21 pairs $(y, t)$ that satisfy this equation, namely the pairs $(y, t)=(1,21),(2,20),(3,19), \ldots,(20,2),(21,1)$.

If $x=2$, then $y+t=20$. There are 19 pairs $(y, t)$ that satisfy this equation, namely the pairs $(y, t)=(1,19),(2,18),(3,17), \ldots,(18,2),(19,1)$.

For each value of $x$ with $1 \leq x \leq 11$, we obtain $y+t=24-2 x$.

Since $y \geq 1$, then $t \leq 23-2 x$.

Since $t \geq 1$, then $y \leq 23-2 x$.

In other words, $1 \leq y \leq 23-2 x$ and $1 \leq t \leq 23-2 x$.

Furthermore, picking any integer $y$ satisfying $1 \leq y \leq 23-2 x$ gives a positive value of $t$, and so there are $23-2 x$ pairs $(y, t)$ that are solutions.

Therefore, as $x$ ranges from 1 to 11, there are

$$
21+19+17+15+13+11+9+7+5+3+1
$$

pairs $(y, t)$, which means that there are this number of triples $(x, y, z)$.

This sum can be re-written as

$$
21+(19+1)+(17+3)+(15+5)+(13+7)+(11+9)
$$

or $21+5 \cdot 20$, which means that the number of triples is 121 .

ANSWER: 121