20050807, 08:47  #1 
Jun 2005
Near Beetlegeuse
604_{8} Posts 
Counting Cubes
Not one of mine, this original problem by Susan Denham originally appeared in New Scientist.
I took a large wooden cube and painted two of its faces red, two of its faces white, and its two remaining faces blue. Then I cut the cube into lots of identically sized smaller cubes and gave them to my niece to play with. First she counted all the small cubes out loud, “One, two, three…”. Then she counted the number of small cubes that had both at least one red face and at least one white face. Then she counted the number of small cubes that had at least one red face and at least one blue face but no white faces. These last two answers were different but one was the reverse of the other. How many of the small cubes have at least one white face, but no red faces and no blue faces? 
20050807, 17:50  #2 
Jun 2003
1,579 Posts 
What do you mean by reverse?
Citrix 
20050807, 18:39  #3 
Jun 2005
Near Beetlegeuse
2^{2}·97 Posts 
345 is the reverse of 543

20050807, 19:19  #4 
Jun 2003
1,579 Posts 
ok!
Citrix Last fiddled with by Citrix on 20050807 at 19:22 
20050813, 05:49  #5 
Jun 2005
Near Beetlegeuse
388_{10} Posts 
Well, it's been a week with no response, so here is my solution.
Solution. The large cube is cut into x^3 smaller cubes, so that x represents the number of smaller cubes along one side of the large cube. A cube has 12 edges and 8 corners. I shall consider these two separately. Small cubes having more than one colour will be from the edges of the cube. If we say RW is the number of edges where a Red face meets a White face, and C_1 is the number of corners where these two colours meet, then my nieces first count was RW(x2) + C_1. If we say RB is the number of edges where a Red face meets a Blue face, and C_2 is the number of corners where these two colours meet but which do not have a White face, then my nieces second count was RB(x2) + C_2. Since we know that the first count was the reverse of the second, we know that their difference is a multiple of 9, which gives us:  (RW(x2) + C_1) – (RB(x2) + C_2)  = 9y Now, if we assume that the opposite faces are painted the same colour (so that no two adjacent faces are the same colour), this gives us  (4(x2) + 8) – (4(x2) + 0)  = 9y. Since for any value of x this gives 8 = 9y we know that the faces are not painted in this order. Numbering the faces as though the cube were a dice, 1 opposite 6, 3 opposite 4, etc, and trying various combinations of different coloured faces, we eventually come to 1 = W, 2 = W, 3 = B, 4 = B, 5 = W, and 6 = W. (Obviously, other arrangements will give the same result, but they are all geometric transpositions of the same arrangement). Putting the numbers from this combination into our equation we get  (2(x2) + 4) – (4(x2) + 2)  = 9y, which has the solution x = 12, y = 2. Therefore, there were 12^3 small cubes for my niece to count. So the two White faces of the cube each have 12^2 small cubes. Since these two faces are adjacent, to find the number of small cubes that have only White faces we need 2*(x2)^2 + x2. So the answer is 2 * 100 + 10 = 210 small cubes with only White faces. Last fiddled with by Wacky on 20050813 at 11:08 Reason: A nice puzzle. I "spoilerized" the answer in hopes that future readers will try it themselves. 
20050902, 14:16  #6 
Aug 2005
2 Posts 
is there another case for this question?
Hi, Numbers, thanks for this interesting question, and also the solution is so smart. I actually tried it myslef before looking at the solution, and I am wondering what if the cube is painted like this? 1 & 2 are red, 3 & 4 are white, and 5 & 6 are blue (still we imagine it's a dice). This gives us RW: (4(x2) + 6) = 4x2, and RB: 2(x2) = 2x 4. and then we know (4x2)  (2x 4) = 9y. Solve it we have 2x+2 = 9y, so x = 8, when y = 2. But according to this, RW is 30 and RB is 12. It obiously doesn't work. I am puzzled. I didn't know the charactristic of reversed #s eariler though (i.e.xyz  zyx= 9y), maybe I used it wrong?
And a related question is how did you decide x = 12, y = 2 in your solution with only one equation? or any x&y as long as they make equation equate? Thanks 
20050903, 00:26  #7 
Jun 2005
Near Beetlegeuse
2^{2}×97 Posts 
Armstrong,
Thank you for your interest and welcome to the forum. You quite correctly point out that (4x – 2)  (2x – 2) gives 2x + 2 = 9y, which has the solution x = 8, y = 2. However, that is only a necessary case, but is I’m afraid not sufficient. If you evaluate 4(x2) + 6 when x = 8 you get 30. If you evaluate 2(x2) + 0 when x = 8 you get 12. And the difference between these two numbers is indeed 9 * 2. But 30 is not the reverse of 12. 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A counting problem  jasonp  Puzzles  1  20171224 19:38 
Fibonacci number as sum of cubes  jux  Miscellaneous Math  15  20150830 06:21 
Squares and Cubes:  mfgoode  Puzzles  24  20070806 16:20 
Prime cubes!  fivemack  Puzzles  4  20070704 00:18 
counting bricks  michael  Puzzles  8  20040114 16:27 