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April 18, 2014

April 18, 2014

Posted by **Lyon** on Wednesday, May 26, 2010 at 7:36pm.

An 11[times]11[times]11 wooden cube is formed by gluing together 11[cubed] unit cubes. what is the gratest number of unit cubes that cam be seen from a single point?

- Math -
**Reiny**, Wednesday, May 26, 2010 at 9:31pmLove this question!

I will take a viewpoint where I can see 3 of the faces, (can't do better than that)

I then set up a pattern

2by2by2

I see 2^2 cubes from the front row + 1 block along the edge between the other two faces + 2(1)^2 from the remaining top and side = 7

3by3by3

I see 3^2 cubes from the front row + 2 blocks along the edge between the other two faces + 2(2)^2 from the remaining top and side = 19

4by4by4

I see 4^2 cubes from the front row + 3 blocks along the edge between the other two faces + 2(3)^2 from the remaining top and side = 37

.

.

11by11by11

I see 11^2 cubes from the front row + 10 blocks along the edge between the other two faces + 2(10)^2 from the remaining top and side = 331

- Math -
**tchrwill**, Thursday, May 27, 2010 at 12:14pmHold a cube so that your line of sight is parallel to the diagonal from one vertex of the cube to the opposite xertex. This allows you to see three faces of the cube, and no more. Starting with a 1x1 cube, work your way up to 2x2, 3x3, 4x4, etc., and determine the number of individual cubes you see. Remember that the cube at the vertex nearest to your eye counts as one cube, not three cubes, since it is on each face. This results in 1 cube for a 1x1x1 cube, 3+3+3+1=10 cubes for a 2x2x2 cube, 8+8+8+1=25 for a 3x3x3 cube, 15+15+15+1= 46 for 1 4x4x4 cube, and so on. This results in the following data with n = the number of cubes to the side and N = the total number of cubes visible.

n...1...2...3...4...5...

N...1..10..25..46..73...

Diff.9...15..21..27

Diff...6...6...6

With the second differences being constant, the expression defining the total number of cubes visible for any "n" sided cube is of the form an^2 + bn + c.

Using the data derived so far, we can write

a(n^2) + b(n) +c =N

a(1)^2 + b(1) +c =1 or a + b + c = 1

a(2)^2 + b(2) +c =10 or 4a + 2b + c= 10

a(3)^2 + b(3) +c =25 or 9a + 3b + c= 25

Solving, a = 3, b = 0 and c = -2 yielding the general expression for the numbder of cubes visible in a cube of nxnxn sides

N = 3n^2 - 2

Therefore, an 11x11x11 cube has 361 visible cubes.

- Math -
**tchrwill**, Thursday, May 27, 2010 at 4:58pmClearly, this could have been solved quicker by recognizing that each visible surface shows 3n^2 cubes. since the apex cube counts as only one cube, the total seen becomes N = 3n^2 - 2.

- Math -
**tchrwill**, Saturday, May 29, 2010 at 9:12amCorrection

My previous statement

"Clearly, this could have been solved quicker by recognizing that each visible surface shows 3n^2 cubes. since the apex cube counts as only one cube, the total seen becomes N = 3n^2 - 2."

should have read

Clearly, this could have been solved quicker by recognizing that each visible surface shows n^2 cubes. Since the apex cube counts as only one cube, the total seen becomes N = 3n^2 - 2.

- Math -
**ABC**, Thursday, July 1, 2010 at 1:46amlim x infnitive (4x-1/2x+9)

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