MHB Divide a big cube into 49 small cubes

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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.
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Please divide a big cube into 49 small cubes
 
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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]

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$\chi$ $\sigma$
 
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
 

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See A014544 - OEIS for information about the number of sub-cubes into which a cube can be dissected.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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