jmc

We use the distance formula: $$\sqrt{(8-0{)}^2+(0-15{)}^2}=\sqrt{64+225}=\boxed{17}.$$- OR -

We note that the points $(0,15)$, $(8,0)$, and $(0,0)$ form a right triangle with legs of length 8 and 15. This is a Pythagorean triple, so the hypotenuse must have length $\boxed{17}$.