Consider the bottom plane of the 7x7x7 cube.
The four corners of the 4x4x4 cube are at (0,0), (0,4), (4,4) and (4,0).
The corners of two 3x3x3 cubes are at (4,0), (4,3), (7,3), (7,0) and (0,4), (0,7), (3,7), (3,4) respectively.
The corners of three 2x2x2 cubes are at (3,5), (3,7), (5,7), (5,5) and (5,5), (5,7), (7,7), (7,5) and (5,3), (5,5), (7,5), (7,3) respectively.
The remaining three 3x3x3 cubes are stacked on top of the 4x4x4 cube and 3x3x3 cubes.
The remaining six 2x2x2 cubes are stacked on top of the three 2x2x2 cubes.
Hence, one 4x4x4 cubes = 64 units.
Five 3x3x3 cubes = 135 units.
Nine 2x2x2 cubes = 72 units.
The remaining are 72 units of 1x1x1 cubes.
Total 87 cubes make up 343 units.
Some adjustment:
Replace one layer of three 2x2x2 cubes with one 3x3x3 cube.
The total should be:
One 4x4x4 cube
Six 3x3x3 cubes
Six 2x2x2 cubes
Sixty nine 1x1x1 cubes
My solution looks like this: with the 7x7x7 cut into layers, and each number means the size of the cube it belongs to.
2222333
2222333
2222333
2222122
3331122
3332222
3332222
-------
2222333
2222333
2222333
2222122
3331122
3332222
3332222
-------
2222333
2222333
1111333
1111122
3331122
3331122
3331122
-------
2222122
2222122
1111111
4444122
4444122
4444122
4444122
-------
3331122
3331122
3331111
4444111
4444333
4444333
4444333
-------
3332222
3332222
3331122
4444122
4444333
4444333
4444333
-------
3332222
3332222
3331122
4444122
4444333
4444333
4444333
Obtained by first considering the same problem for squares; the minimum is 9 squares and one possible pattern for that is used for three of the sides, so this pattern is somewhat symmetric.
Maybe the answer is even better. However clearly only one 4x4x4 cube can exist in the cube, and using too many 3x3x3 cubes seems to greatly limit the number of remaining 2x2x2 cubes that fit, so that there are a lot of 1x1x1s.
回應
yipting
Re: 一道沒人答對的國際小學數學競賽試題
2008-11-04 00:333x3x3 5個
2x2x2 9個
1x1x1 72個
Total 87個
孫文先
Re: 一道沒人答對的國際小學數學競賽試題
2008-11-04 09:40yipting
Re: 一道沒人答對的國際小學數學競賽試題
2008-11-04 10:47The four corners of the 4x4x4 cube are at (0,0), (0,4), (4,4) and (4,0).
The corners of two 3x3x3 cubes are at (4,0), (4,3), (7,3), (7,0) and (0,4), (0,7), (3,7), (3,4) respectively.
The corners of three 2x2x2 cubes are at (3,5), (3,7), (5,7), (5,5) and (5,5), (5,7), (7,7), (7,5) and (5,3), (5,5), (7,5), (7,3) respectively.
The remaining three 3x3x3 cubes are stacked on top of the 4x4x4 cube and 3x3x3 cubes.
The remaining six 2x2x2 cubes are stacked on top of the three 2x2x2 cubes.
Hence, one 4x4x4 cubes = 64 units.
Five 3x3x3 cubes = 135 units.
Nine 2x2x2 cubes = 72 units.
The remaining are 72 units of 1x1x1 cubes.
Total 87 cubes make up 343 units.
yipting
Re: 一道沒人答對的國際小學數學競賽試題
2008-11-04 17:18Replace one layer of three 2x2x2 cubes with one 3x3x3 cube.
The total should be:
One 4x4x4 cube
Six 3x3x3 cubes
Six 2x2x2 cubes
Sixty nine 1x1x1 cubes
i.e. 82 cubes
孫文先
Re: 一道沒人答對的國際小學數學競賽試題
2008-11-04 18:19colfulus
Re: 一道沒人答對的國際小學數學競賽試題
2008-11-04 19:092222333
2222333
2222333
2222122
3331122
3332222
3332222
-------
2222333
2222333
2222333
2222122
3331122
3332222
3332222
-------
2222333
2222333
1111333
1111122
3331122
3331122
3331122
-------
2222122
2222122
1111111
4444122
4444122
4444122
4444122
-------
3331122
3331122
3331111
4444111
4444333
4444333
4444333
-------
3332222
3332222
3331122
4444122
4444333
4444333
4444333
-------
3332222
3332222
3331122
4444122
4444333
4444333
4444333
Obtained by first considering the same problem for squares; the minimum is 9 squares and one possible pattern for that is used for three of the sides, so this pattern is somewhat symmetric.
1x1x1 cubes: volume 51, 51 cubes
2x2x2 cubes: volume 120, 15 cubes
3x3x3 cubes: volume 108, 4 cubes
4x4x4 cubes: volume 64, 1 cube
Total: volume 343, 71 cubes.
Maybe the answer is even better. However clearly only one 4x4x4 cube can exist in the cube, and using too many 3x3x3 cubes seems to greatly limit the number of remaining 2x2x2 cubes that fit, so that there are a lot of 1x1x1s.
孫文先
Re: 一道沒人答對的國際小學數學競賽試題
2008-11-05 17:33根據您的答案,我們做了一個較詳細的解答
http://www.chiuchang.org.tw/download/iwymic/777.pdf