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Topic: Painted Cubes (Read 435 times) |
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ThudnBlunder
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Painted Cubes
« on: Jan 2nd, 2004, 5:01am » |
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Some unit cubes are stuck together to form a larger cube.
Some of the faces of the larger cube are then painted at random.
When the cube is taken apart it is found that 217 of the unit cubes have paint on them.
How many unit cubes are there?
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| « Last Edit: Jan 4th, 2004, 2:31am by ThudnBlunder » |
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LZJ
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Re: Painted Cubes
« Reply #1 on: Jan 2nd, 2004, 7:58am » |
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729 cubes, 3 sides painted
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ThudnBlunder
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Re: Painted Cubes
« Reply #2 on: Jan 2nd, 2004, 2:47pm » |
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Which 3 sides?
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TenaliRaman
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Re: Painted Cubes
« Reply #3 on: Jan 2nd, 2004, 3:30pm » |
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Adjacent faces i.e each face adjacent to every other face.
3*(n-1)2+3*(n-1)-216=0
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Sir Col
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Re: Painted Cubes
« Reply #4 on: Jan 3rd, 2004, 6:01pm » |
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An unknown number of faces of a cube, n3, are painted. Find the set of values, C, representing the number of cubes that will have paint on them.
For 23, C={0, 4, 6, 7, 8}.
For 33, C={0, 9, 15, 18, 19, 21, 23, 24, 25, 26}
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TenaliRaman
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Re: Painted Cubes
« Reply #5 on: Jan 3rd, 2004, 9:54pm » |
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for n3,
C={0, n2, 2n2, 2n2-n, 3n2-2n, 3n2-3n+1, 4n2-5n+2, 5n2-8n+4, 6n2-12n+8}
what!! did i miss something?? please fill in them will ya! 
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| « Last Edit: Jan 3rd, 2004, 9:56pm by TenaliRaman » |
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LZJ
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Re: Painted Cubes
« Reply #6 on: Jan 3rd, 2004, 11:05pm » |
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You missed out 4n2 - 4n 
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ThudnBlunder
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Re: Painted Cubes
« Reply #7 on: Jan 4th, 2004, 2:51am » |
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For each of the 10 configurations, how many ways are there to paint them?
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TenaliRaman
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Re: Painted Cubes
« Reply #8 on: Jan 4th, 2004, 4:51am » |
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on Jan 3rd, 2004, 11:05pm, LZJ wrote:| You missed out [hidden] |
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Thx m8!!!
on Jan 4th, 2004, 2:51am, THUDandBLUNDER wrote:| For each of the 10 configurations, how many ways are there to paint them? |
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0 face : C(6,0)
1 face : C(6,1)
2 face : C(6,1)*C(1,1)
2 face : C(6,1)*C(4,1)
3 face : C(6,1)*C(4,1)*C(2,1)
3 face : C(6,1)*C(4,1)*C(2,1)
4 face : C(6,1)*C(4,1)*C(2,1)*C(1,1)
4 face : C(6,1)*C(4,1)*C(2,1)*(C(2,1)+C(3,1))
5 face : C(6,5)
6 face : C(6,6)
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i am just finished with my exams, so chances that i have seriously goofed up in the above solution is more than 50%.
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ThudnBlunder
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Re: Painted Cubes
« Reply #9 on: Jan 4th, 2004, 5:35am » |
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TenaliRaman, congratulations on finishing your exams.  
Shouldn't the answers for 2 faces = the answers for 4 faces?
It's probably easier to do it by inspection (of a die).
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| « Last Edit: Jan 5th, 2004, 8:09am by ThudnBlunder » |
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TenaliRaman
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Re: Painted Cubes
« Reply #10 on: Jan 4th, 2004, 9:39am » |
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on Jan 4th, 2004, 5:35am, THUDandBLUNDER wrote:TenaliRaman, congratulations on finishing your exams.
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thanks!!!
but i would also like to have some luck!!! 
Quote:
Shouldn't the answers for 2 faces = the answers for 4 faces?
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yeah i guess so!! 
Quote:
It's probably easier to do it by inspection (of a die).
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that's it!! i don't have a die!! 
though i will recheck on my calcs..
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TenaliRaman
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Re: Painted Cubes
« Reply #11 on: Jan 8th, 2004, 10:56pm » |
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My answers to 0,1,3,5,6 are correct it "seems"!
The answers to 2 and 4 are same and should be equal to the answer i gave for 2 it "seems"!
But it all "seems" doubty!!
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ThudnBlunder
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Re: Painted Cubes
« Reply #12 on: Jan 9th, 2004, 12:01am » |
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TenaliRaman, your answers may be correct. But they are not clearly correct.
That is, it is not clear which of your answers is which.
Let's call the numbers 2a, 2o, 3c, 3u, 4a, and 4c, for reasons which I hope are obvious.
Anyway, I think your numbers are too high.
For example, consider the case 2o (two opposite faces).
By inspection, the answer is clearly 3; but it "seems" you have an answer of 6 (or 24).
For 2a (two adjacent faces) I get 12, but you "seem" to get 24 (or 6).
"Seems" like you are forgetting to divide by 2.
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| « Last Edit: Jan 9th, 2004, 2:16am by ThudnBlunder » |
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TenaliRaman
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Re: Painted Cubes
« Reply #13 on: Jan 9th, 2004, 9:48am » |
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Quote:| "Seems" like you are forgetting to divide by 2. |
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oops!right!! my configurations get repeated twice!!
For the 2 opposite sides, one side can be chosen out 6 in C(6,1) and the opposite side can be chosen only in 1 way that C(1,1).Hence total number of ways it can be done is C(6,1)*C(1,1). However as it is obvious the total number of ways repeats twice.
For the 2 adjacent sides, one side can be chosen out 6 in C(6,1) and the adjacent side can be chosen in 4 ways that C(4,1).Hence total number of ways it can be done is C(6,1)*C(4,1). However as it is obvious again total number of ways repeats twice.
That's how i got my remaining answers as well. So for the 3 side case, i think we might have to divide my answer by 3!=6 i.e.
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