Let the distance of cat running 1 step be a, and the distance of mouse running 1 step be b.

Then 5a=9b

The cat runs 1 step, and the mouse runs 3/2 steps.

The shortened distance between them = a-(3/2) b = a-(5/6) a = (1/6) a, which is positive, the cat can catch up with the mouse.

10a divided by (1/6)a=60.

So you can catch up in 60 steps.