Thread: How many basketballs will fit into a room? (Dimensions and amounts included)

1) Cubes is not a good plan. That's definitely a lower bound. Put four balls together, their centers making a lovely tetrahedron, and see how much space is between them. This might be a good start on increasing the number. There are other problems at the walls, however.

2) You'll love this. If you number the balls, you might get a different outcome. I explain:

a1) Throw in 10 balls. b1) Find #1 and remove it.
a2) Throw in 10 balls. b2) Find #2 and remove it.
a3) Throw in 10 balls. b3) Find #3 and remove it.
...
an) Throw in 10 balls. bn) Find #n and remove it.
...

Continue this process. This leads to no balls at all. Every ball eventually will be removed. :-)

You seem to be "missing the point" in question#1; WHY are you calculating volumes?
Let's take a simple case: you have a 7by7by7 square box and a 4by4by4 square box:
how many 4by4by4 boxes can you fit in the 7by7by7 box? ONLY 1, right?
But 7by7by7 box volume = 343 and 4by4by4 box volume = 64 ; 64 * 5 = 320:
so that "seems" to mean that 5 boxes can be fit in...but ONLY 1 can...GET IT?

Could you elaborate on the throwing the balls please, I'm not grasping the concept.
You put 10 balls into the room, and take out the ball labelled #1?
Why?

Who cares why? Follow the logic.