tertre

Thank you! It's okay, my mind gets confused a lot....

Amazing, CPhill! Great job!

Thanks, EP! My hands hurt now.. 

EDIT: Wait, I'm getting 83... 

The number of painted faces is 9*6=54. However, the four edge faces painted on each face are also painted on the other face. Thus 4*6 of the painted faces are on cubes with two painted faces. These account for only 12 painted cubes, in addition to the 54-24=30 singly painted cubes. So there are 42 painted cubes, leaving 125-42=83 unpainted cubes.

@EP, I don't know how I'm getting 83.....Is this correct?

No problem, EP! 

(2) I think there are 96 ways...

Explanation: First, let us count the number of ways we can sum 9 with di erent combinations of 1, 2, and 4. We can have nine ones; seven ones and one two; ve ones and two twos; ve ones and one four; three ones and three twos; three ones, one two, and one four; one one and four twos; one one, two twos, and one four; or one one and two fours. Now, let us count the number of ways to rearrange each of these cases. Case One: Nine ones There is only 1 way to arrange these, since all of the ones are indistinguishable. Case Two: Seven ones and one two There are eight numbers total to sum. The two can be placed in any of the eight slots, after which the others are simply lled with ones. Hence, there are 8 ways to arrange these. Case Three: Five ones and two twos There are seven numbers total to sum. The rst two can be placed in any of the seven slots, and the second two can be placed in any of the remaining six. After this, we simply ll in the ones. However, because the twos are indistinguishable, we've overcounted by a factor of two. Thus, there are 7
6 2 = 21 ways to arrange these. Case Four: Five ones and one four There are six numbers to sum, and the four can be placed in any of the six slots, after which the other slots are lled with ones. So, there are 6 ways to arrange these. Case Five: Three ones and three twos There are six numbers to sum. Consider placing the twos and then lling in the ones. There are six slots for the rst two, ve for the second, and four for the third. However, given three distinguishable twos, there are six ways to arrange them. Thus, because the twos are indistinguishable, we have overcounted by a factor of six. Thus, there are 6
5
4 6 = 20 ways to arrange these. Case Six: Three ones, one two, and one four There are ve numbers to sum. The two can be placed in any of the ve slots, and then from there the four can be placed in any of the four slots and the other slots lled in with ones. Thus, there are 5
4 = 20 ways to arrange these. Case Seven: One one and four twos There are ve numbers to sum. The one can be placed in any of the ve slots, after which the twos are lled in. Thus, there are 5 ways to arrange these. Case Eight: One one, two twos, and one four There are four numbers to sum. The one can be placed in any of the four slots, the four placed after that in any of the remaining three, and the other two slots lled in with twos. So, there are 4
3 = 12 ways to arrange these. Case Nine: One one and two fours There are three numbers to sum. The one can be placed in any of the three slots, and then the other slots lled in with fours. Thus, there are 3 ways to arrange these. There are thus 1 + 8 + 21 + \(6 + 20 + 20 + 5 + 12 + 3 =\boxed{96} \)

Yes, that's called the "handshake" formula.

I don't think 95 is correct, EP.

You have a point, EP. 

Solution

For any given one of the 200 points, we can find exactly one square with that point as one of the vertices---that point, the point diametrically opposite it, and the endpoints of the diameter that is perpendicular to the diameter formed by connecting the first two points. Each square then accounts for 4 vertices, so there are \(200/4=\boxed{50}\) squares.