Will Orrick
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 1d comment Confusion About “Stars and Bars” Method Your method is fine; the original question is really asking for triangular numbers, and the formula you get is the usual one. Is there maybe a bug in your Mathematica code? If you used loops to enumerate solutions, did you perhaps hard-code an upper limit somewhere? Apr 26 revised Known classes of Hadamard matrices edited tags Apr 26 comment Known classes of Hadamard matrices Reference: Hadamard matrix. Apr 26 answered Known classes of Hadamard matrices Apr 24 revised Puzzle About Cubes (from the book thinking mathematically) added 5547 characters in body Apr 15 revised Puzzle About Cubes (from the book thinking mathematically) added 4065 characters in body Apr 12 comment Puzzle About Cubes (from the book thinking mathematically) @pjs36: I've posted an answer that contains one possible generalization to the $3\times 3\times 3$ cube. Indeed, there is one cube that is hidden from view. Nevertheless, one can find solutions in which all nine colors appear on every face. Higher $n$ may also be possible. Apr 12 answered Puzzle About Cubes (from the book thinking mathematically) Apr 12 comment Puzzle About Cubes (from the book thinking mathematically) In the arrangement f1:red, f2:green,f3:blue,f4:white. b1:green b2:red,b3:white b4:blue, the top of the cube, consisting of f1, f2, b1, b2 has two reds and two greens. The bottom has two whites and two blues. Apr 12 comment Puzzle About Cubes (from the book thinking mathematically) ... colors 1, 4, 7, 9, 6, 3, which means that colors 2, 5, 8 need to be added (in the middle layer). But that means that two of the colors 2, 5, 8 will also touch the top face. This is a problem because colors 2 and 8 are already present on the top face. The same issue also occurs on the bottom face. One can, in fact, show without much difficulty that six different colors must be used on the center cubes of the six faces of the $3\times3\times3$ cube. I will post an answer with the proof. Apr 12 comment Puzzle About Cubes (from the book thinking mathematically) In the $3\times3\times3$ case you can arbitrarily color the front face $$\begin{array}{ccc}1&2&3\\ 4&5&6\\ 7&8&9\end{array}.$$ With diagonally opposite cubes being the same color, the back face would be colored $$\begin{array}{ccc}9&8&7\\ 6&5&4\\ 3&2&1\end{array}.$$ Here the back face is being viewed from the front, as if the layers in front of it were transparent. Now both the left side face and the right side face already contain... Apr 11 comment Puzzle About Cubes (from the book thinking mathematically) I'm also uncertain about the generalization. Are you assuming that diagonally opposite cubes are of the same color, so that in the $3\times3\times3$ cube, the center of the front face is the same color as the center of the back face, the middle cube of the top front edge is the same color as the middle cube of the bottom back edge, and so on, and top front left corner cube is the same color as the bottom back right corner cube, and so on? If so, then, after applying this idea to the coloring of the front and back faces, you run into conflicts when you try to color the middle layer of cubes. Apr 11 comment Puzzle About Cubes (from the book thinking mathematically) I'm not sure the $\sqrt{n}\times\sqrt{n}\times\sqrt{n}$ generalization has been completely thought through. Just one possible issue: when $\sqrt{n}>2$, there will be interior cubes that are completely hidden from view, which may therefore be permuted arbitrarily. There are other problems as well. Apr 11 comment Puzzle About Cubes (from the book thinking mathematically) Another way to say it: one of the two possible arrangements of the left side face in your analysis is not going to work when you later go to color the top and bottom faces, but the other one will. Apr 11 comment Puzzle About Cubes (from the book thinking mathematically) As in @browngreen's solution, once the colors on the front face are fixed, the arrangement of colors on the other five faces are uniquely determined. So you need to omit the two factors of $2!$. One way to think about it: if the colors on the front face are 1234, read in clockwise order from the top left, then, since the top face already uses 1 and 2, we need to add 3 and 4; since the right side face already uses 2 and 3, we need to add 1 and 4. But the top face and the right side face share an edge, so the missing color on that edge can only be 4. Apr 11 comment Puzzle About Cubes (from the book thinking mathematically) The wording of the problem seems to suggest that the two red cubes are not distinguished from each other, and, in general, the two cubes of each color are not distinguished from each other. If one does choose to distinguish then, however, wouldn't fixing the identity of the cube in one of the corners reduce the count by $8$ rather than $4$? Apr 11 revised Puzzle About Cubes (from the book thinking mathematically) added 13 characters in body Apr 5 revised How many ways to choose \$a