At the AoPS office, mice vary inversely with cats, that is, $\text{mice}=\frac{k}{\text{cats}}$, for some value of $k$. When there are $3r-19$ cats, there are $2r+1$ mice, and when there are $6r+2$ mice, there are $4r+3$ cats. Find $k$.

Answer 1

From the first statement, we know that \[2r+1=\frac{k}{3r-19}.\]From the second statement, we know that \[4r+3=\frac{k}{6r+2}.\]Since $\dfrac{k}{3r-19}=\dfrac{k}{6r+2}$, \begin{align*} 2r+1&=4r+3 \\ -2r&=2 \\ r&=-1.\end{align*}The values $r=-1$ satisfy both original statements, so \[2(-1)+1=4(-1)+3=-2+1=-1.\]At $r=-1$, we can calculate \[2r+1=2(-1)+1=-1.\]Thus, $k=(2(-1)+1)(3(-1)-19)=-1(-4-19)=23(-1)=\boxed{-23}$, which is our answer.