, 2026

Suppose that when Serafine and Jamie stop playing catch, they
have a total of $c$ catches and $m$ misses.

Each catch increases the distance between them by $2\times 60\text{ cm}=120\text{ cm}$, and
so $c$ catches increases the distance
between them by $\$120c\text{}$
cm}$.

Each miss decreases the distance between them by $20\text{ cm}$, and so $m$ misses decreases the distance between
them by $20m\text{ cm}$.

They begin $1500\text{ cm}$ apart and
when they stop they are $\$1720\text{}$
cm}$apart, and so$1500+120c-20m=1720$.

Simplifying this equation, we get $120c-20m=1720-1500$ or $120c-20m=220$. Dividing each term by $20$, we get $6c-m=11$.

Since the question asks which number of throws is not possible, then we
let the number of throws be $n$, and
so $n=m+c$.

With $6c-m=11$ and $n=m+c$, we get the following equivalent
equations $$\begin{align*} 6c-m&=11\\ 6c&=11+m\\ 6c+c&=11+m+c\\ 7c&=11+n\end{align*}$$ and so $n=7c-11$.

If $n=21$, then $7c-11=21$ or $7c=32$, and so $c=\dfrac{32}{7}$. However, $\dfrac{32}{7}$ is not an integer and so
$21$ throws is not possible.

We can confirm that when $c$ is equal
to $4$, $12$, $8$, and $3$, then the number of throws, $n=7c-11$, is equal to $17$, $73$, $45$, and $10$, respectively.

In each case, $m=6c-11$ and so $m$ is also a positive integer.

Of the given possibilities, $21$
throws is not possible.