Dividing a cube

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MartinMuller
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Joined:Sun Jan 24, 2021 10:56 pm
Dividing a cube

Unread post by MartinMuller » Sun Jan 24, 2021 11:02 pm

A cube with an edge length of 10 is divided into two cuboids with integer edge length by a flat cut.
Afterwards, one of these cuboids is again being divided into two smaller cuboids with integer edge lengths by a second flat cut.
What´s the smallest possible volume of the biggest of the three cuboids? (The result needs to be PROVEN!)

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Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm
Location:Dhaka, Bangladesh

Re: Dividing a cube

Unread post by Mehrab4226 » Sun Jan 31, 2021 10:06 pm

Ans:
$350$ Cubic Units
Proof:
I don't think I need to prove the following statement,

$V \geq \frac{10^3}{3}$[Where V is the required volume] which is quite obvious.
But if you still want to prove it can be easily done with contradiction.

Now the edges of the cuboid must be integers, thus,
$V \geq 334$[Where $V$ is an integer]

Again since the cubes were cut 2 times, so one of the edges of the cuboid must be 10 units making $V$ a multiple of 10,
$\therefore V \geq 340$[where $V$ is a multiple of 10]

Now $340$ can't be a valid cuboid as its factors are $10,2,17$ and all the lengths must be less than or equal to $10$,
$\therefore V \geq 350$

350 can be the volume of a valid cuboid. So we are done! $\square$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

idealisticcormorant
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Joined:Wed Jun 15, 2022 12:53 pm

Re: Dividing a cube

Unread post by idealisticcormorant » Wed Jun 15, 2022 12:54 pm

A flat cut divides a cube with ten edge lengths into two cuboids with integer edge lengths.
After that, a second flat cut divides one of these cuboids into two smaller cuboids with integer edge lengthsoctordle

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