Algebra question

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If 27 cubes are glued together to form a 3 x 3 x 3 block, one cube will be completely hidden. In a 4 x 4 x 4 block, eight of the sixty-four cubes are completely hidden. What are the dimensions of the smallest block in which more than half of the cubes are hidden.

Is the solution?

((n/(n-1))^3) >= 0.5

I think answer is 5x5x5
 
If you assume the total number of cubes is (n^3)/2 and the number of hidden cubes is (n-2)^3

(Note: these two assumptions work for the two examples, but who knows if the pattern continues)

At any rate, then you are trying to find when:

(n-2)^3 > (N^3)/2

Rather than solve this inequality, you might try just plugging in some numbers until you get your answer.

Dave
 
Hello, americo74!

If 27 cubes are glued together to form a 3 x 3 x 3 block, one cube will be completely hidden.
In a 4 x 4 x 4 block, eight of the sixty-four cubes are completely hidden.
What are the dimensions of the smallest block in which more than half of the cubes are hidden?

Is the solution: \(\displaystyle \left(\frac{n}{n-1}\right)^3\:> \:0.5\) . . . . close, but wrong
Click to expand...
The entire cube has \(\displaystyle n^3\) unit cubes.

The hidden cubes have: \(\displaystyle (n - 2)^3\) unit cubes.
 
Code: 
n    n^3   (n-2)^3
3     27       1
4     64       8
5    125      27
...
9    729     343
10  1000     512 : AHEM!
Make it equal:
x^3 = 2(x - 2)^3; leads to
x^3 - 12x^2 + 24x - 16 = 0
try 9: x^3 - 12x^2 + 24x - 16 = -43
try 10: x^3 - 12x^2 + 24x - 16 = 24
So clearly has to be 10.....
 
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