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Rubik's Cube

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kamalakarthi Senior | Next Rank: 100 Posts Default Avatar
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Rubik's Cube

Post Thu Aug 31, 2017 5:17 pm
Hi Experts,

This attached problem is from One of the Manhattan CATs. While I approached this problem geometrically and wasted lot of time, I also thought this could be done thinking visually.

Is there any geometric method that I can use to solve the problem ?

ttps://postimg.org/image/bc53ludbp/" target="_blank">

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Post Tue Sep 05, 2017 7:53 am
Hey kamalakarthi,

Great question. This requires some visualization.

Our initial large cube has dimensions n x n x n.

ttps://postimg.org/image/48qbxo979/" target="_blank">

All of the small cubes that DO face out (on the outside of the large cube) will have red sides. All of the cubes that DO NOT face out (on the inside of the large cube) will NOT have red sides.

We can visualize removing all of the small cubes with red sides (the ones that face out) like so. First we need to remove all of the cubes facing out on the left and right:

ttps://postimg.org/image/xis8x06fp/" target="_blank">

The original width of the cube was n. We just removed 2 cubes from the width (one on the left and one on the right). This gives us n-2 small cubes that DO NOT have red sides as our width. Then we can remove all of the cubes facing out on the top and bottom:

ttps://postimg.org/image/6g8cqtimt/" target="_blank">

Again, the original height of the cube was n, and we just removed 2 cubes from the height, giving us n-2 small cubes that DO NOT have red sides as our height. Finally, we can remove all of the cubes facing out on the front and back:

ttps://postimg.org/image/ilxk17xcl/" target="_blank">

This gives us n-2 small cubes that DO NOT have red sides as our length. Putting this all together, we have (n-2) x (n-2) x (n-2) small cubes that DO NOT have red sides, or (n-2)^3. We can then subtract from our original number of cubes (n^3) to solve for the number small cubes that DO have red sides:

n^3 - (n-2)^3
n^3 - (n^3 - 6n^2 + 12n - 8)
6n^2 + 12n - 8

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Matt@VeritasPrep GMAT Instructor
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Post Wed Sep 20, 2017 4:19 pm
kamalakarthi wrote:
Matt,

Thanks for your reply. Say if I have a cube 8 as a side, How can I know the cube that will fit inside this cube will have side 6 upon cutting the sides of a bigger cube. Is it a generic knowledge that any cube to fit in it should be N-2 size.

I can understand from the drawing that if we have 64 same sized cubes, then the inner most 4 cubes would not have the red coloured picture.
When you want to remove all painted cube from the larger cube, you need to remove a painted face from both sides. So if I had eight cubes before, once I remove one from both sides (the furthest on the left and the furthest on the ride), I'll be left with six cubes, or 8 - 2.

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Post Tue Sep 05, 2017 7:53 am
Hey kamalakarthi,

Great question. This requires some visualization.

Our initial large cube has dimensions n x n x n.

ttps://postimg.org/image/48qbxo979/" target="_blank">

All of the small cubes that DO face out (on the outside of the large cube) will have red sides. All of the cubes that DO NOT face out (on the inside of the large cube) will NOT have red sides.

We can visualize removing all of the small cubes with red sides (the ones that face out) like so. First we need to remove all of the cubes facing out on the left and right:

ttps://postimg.org/image/xis8x06fp/" target="_blank">

The original width of the cube was n. We just removed 2 cubes from the width (one on the left and one on the right). This gives us n-2 small cubes that DO NOT have red sides as our width. Then we can remove all of the cubes facing out on the top and bottom:

ttps://postimg.org/image/6g8cqtimt/" target="_blank">

Again, the original height of the cube was n, and we just removed 2 cubes from the height, giving us n-2 small cubes that DO NOT have red sides as our height. Finally, we can remove all of the cubes facing out on the front and back:

ttps://postimg.org/image/ilxk17xcl/" target="_blank">

This gives us n-2 small cubes that DO NOT have red sides as our length. Putting this all together, we have (n-2) x (n-2) x (n-2) small cubes that DO NOT have red sides, or (n-2)^3. We can then subtract from our original number of cubes (n^3) to solve for the number small cubes that DO have red sides:

n^3 - (n-2)^3
n^3 - (n^3 - 6n^2 + 12n - 8)
6n^2 + 12n - 8

_________________


Erika John - Content Manager/Lead Instructor
http://gmat.prepscholar.com/gmat/s/

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Matt@VeritasPrep GMAT Instructor
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Post Wed Sep 20, 2017 4:19 pm
kamalakarthi wrote:
Matt,

Thanks for your reply. Say if I have a cube 8 as a side, How can I know the cube that will fit inside this cube will have side 6 upon cutting the sides of a bigger cube. Is it a generic knowledge that any cube to fit in it should be N-2 size.

I can understand from the drawing that if we have 64 same sized cubes, then the inner most 4 cubes would not have the red coloured picture.
When you want to remove all painted cube from the larger cube, you need to remove a painted face from both sides. So if I had eight cubes before, once I remove one from both sides (the furthest on the left and the furthest on the ride), I'll be left with six cubes, or 8 - 2.

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kamalakarthi Senior | Next Rank: 100 Posts Default Avatar
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Post Fri Sep 01, 2017 4:12 pm
Matt,

Thanks for your reply. Say if I have a cube 8 as a side, How can I know the cube that will fit inside this cube will have side 6 upon cutting the sides of a bigger cube. Is it a generic knowledge that any cube to fit in it should be N-2 size.

I can understand from the drawing that if we have 64 same sized cubes, then the inner most 4 cubes would not have the red coloured picture.

Erika,

"If we take all of the smaller cubes facing out on the larger cube off, we are left with (n-2)^3 smaller cubes". How do I know this.

My approach to this question was , I considered 9 as a side so volume of this bigger cube is 9*9*9

and the small cube is of size 3*3*3 so the number of cubes that would fit in a bigger cube is 9*9*9/3*3*3 so I got 27 cubes. By this time, I spent around 3 mins so I made a guess.

What am I missing., Can you please help.

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