Comment by Peritract

The algorithm they're using must be "Until you hit the limit, take the highest denomination coin that fits beneath the limit. If you can't hit the limit, fall back one step."

That fits your definition of "use as many coins as possible of a given large denomination before dropping back to the next-lower denomination" but will find 10-10-10-1-1-1-1-1-1-1 and stop before it even tries 10-9-anything.