How do I approach this problem? (Cubes within a larger cube...)

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1. Nov 26, 2016

Thiru07

1. The problem statement, all variables and given/known data
A cube of 8cm x 8cm x 8cm is divided into smaller cubes of 1cm x 1cm x 1cm and all the smaller cubes are numbered and arranged to form the larger cube. The smaller cubes are numbered such that the number on the cube represents the smallest volume enclosed by extending the sides of the cube to the outer surface of the largest cube and each cube bears the same number on each surface.
Find the sum of the numbers on the cubes along the two body diagonals of the largest cube.

2. Relevant equations

Volume of a cube = side*side*side
3. The attempt at a solution
I'm not able to visualise what's given in the problem.

I need a little push here.

2. Nov 26, 2016

Staff: Mentor

This is the only way (in two dimensions) I can think of it, since the question confuses me, too.

3. Nov 26, 2016

Thiru07

I went through it's solution , but I hardly understood anything.

4. Nov 26, 2016

Staff: Mentor

So the difference to my interpretation is, that I required an expansion in all directions, and the book only in one direction, the shortest to the outer boundary.

5. Nov 26, 2016

Thiru07

How did you get all those perfect squares of odd numbers?
What do you mean by
?

6. Nov 26, 2016

Staff: Mentor

Then let me choose the version where expansion isn't required to be in all directions.

Here the square (3,4) has a distance of 3 squares to hit the boundary first, here to the top. So a minimum of $3^2=9$ grey squares can be covered without leaving the outer square. The blue one is bigger (25), but as far as I understood it, the smallest one is meant.

7. Nov 26, 2016

Thiru07

Cubes(1cm x 1cm x 1cm) numbered 4,9 and 16 are present inside the largest cube(8cm x 8cm x 8cm) and not on the surface as it appears above in the 2D image right?
If that's the case, then these cubes numbered 4,9 and 16 are invisible right?
Only cubes numbered 1 are visible i.e all the six surfaces have only one numbered cubes .
am I right?

8. Nov 26, 2016

Staff: Mentor

The problem statement doesn't say anything about visibility. In fact, it asks for the numbers of the body diagonal of the large (8,8,8) cube, which aren't visible except of the two at the ends.

9. Nov 26, 2016

Thiru07

I'm getting 4*(1+4+9+16) = 120 as the sum of the numbers on the cubes along the two body diagonals of the largest cube.
What am I doing wrong?

10. Nov 26, 2016

Staff: Mentor

That's the two-dimensional answer. In three dimensions, the squares turn into cubes and the powers turn from $2$ into $3$.

11. Nov 26, 2016

Thiru07

Thank you fresh_42

12. Nov 26, 2016

Thiru07

I have 3 more questions from the same problem.
1) Find the number of cubes bearing the numbers which are multiple of three.
2) Find the sum of numbers on all the smaller cubes on the surface of the larger cube.
3) Find the number of cubes bearing the number 8 on them.

I have following questions.
We are required to extend the sides of the cube to the furthest outer surface or closest?
If we are required to extend the sides of the cube to the closest outer surface , then all the cubes on all the surfaces will bear number 1 right?

13. Nov 27, 2016

haruspex

I don't think there's any different extending involved in these three questions.

Edit: I left out the crucial word "different"

Last edited: Nov 27, 2016
14. Nov 27, 2016

Thiru07

Number of cubes bearing the numbers which are multiple of three is 296
What I'm getting is (4*8)+(4*8)+(4*8) = 96
I missed a lot of cubes. But I don't know which cubes I skipped.
are 1,4,9 and 16 the only numbers smaller cubes can bear?

15. Nov 27, 2016

haruspex

My understanding of the extension process is that for the given small cube you find the smallest cube which contains it and has an exterior face. Thus, all the surface cubes are numbered 1. If you peel those off, the next layer are all numbered 8, and so on.
The numbers 4 and 9 only arise in the 2D version that fresh_42 used as illustration.
What multiples of three will there be?

16. Nov 27, 2016

Thiru07

Okay, if I peel off twice , I will get a 4x4x4 cube with number 27(multiple of 3) on all 6 faces.
Why do I think that the total number of cubes along all six faces of this 4x4x4 cube is the answer which is 48 and not 296?
Why I'm wrong?

17. Nov 27, 2016

haruspex

4x4x4 after two? You start with 8x8x8. After taking off one layer you have 7x7x7.... You have to go through all the layers, looking for all multiples of 3.

18. Nov 27, 2016

Thiru07

Previously, I thought peeling off as peeling off all the six surfaces, which reduces the 8x8x8 cube to 6x6x6 cube.
When you say peeling off a layer you mean peeling off just one surface right? i.e After peeling off , you get a 8x8x7 cube. Please correct me if I'm wrong.

19. Nov 27, 2016

haruspex

Whoops - my blunder.

Yes, sorry, after two you have only 4x4x4 left. So I calculate there are 56 with numbers divisible by 3.

Edit: 296 is the answer to a different question: how many have the number 1?

20. Nov 27, 2016

Thiru07

How did you get 56?
I got only 48 from counting all these 6 surface cubes which are completely covered with number 9.

By the way , Attached is the given solution which I did not understand.
Can you please tell if it makes any sense?

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