MHB Divide a big cube into 49 small cubes

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Cube
Click For Summary
To divide a large cube into 49 smaller cubes, first divide the large cube with a side length of 1 into 343 smaller cubes, each with a side length of 1/7. From these, 336 cubes can be combined to create 42 cubes with a side length of 2/7. This results in a total of 49 cubes, consisting of 42 larger cubes and 7 smaller ones. The discussion also references the OEIS sequence A014544 for additional insights on cube dissection. The method effectively demonstrates a systematic approach to achieving the desired division.
Albert1
Messages
1,221
Reaction score
0
Please divide a big cube into 49 small cubes
 
Mathematics news on Phys.org
Albert said:
Please divide a big cube into 49 small cubes

[sp]You can be assumed to follow a procedure of this type. Il the side of big cube is 1, then You first divide it into 343 cubes of side $\frac{1}{7}$. Then You use 336 of them to form 42 cubes of side $\frac{2}{7}$, so that at the end 42+7=49 cubes remain...[/sp]

Kind regards

$\chi$ $\sigma$
 
chisigma said:
[sp]You can be assumed to follow a procedure of this type. Il the side of big cube is 1, then You first divide it into 343 cubes of side $\frac{1}{7}$. Then You use 336 of them to form 42 cubes of side $\frac{2}{7}$, so that at the end 42+7=49 cubes remain...[/sp]

Kind regards

$\chi$ $\sigma$
very good ! here is my solution
if the big cake of side 6
divide the cake into three lays:
the first lay of side 2 (9 cakes)
the second lay of side 1 (36 cakes)
the third lay of side 3 (4 cakes)
9+36+4=49
 

Attachments

  • divide cake.jpg
    divide cake.jpg
    14 KB · Views: 90
Last edited:
See A014544 - OEIS for information about the number of sub-cubes into which a cube can be dissected.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
12K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K