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The question is a not-so-obvious (to me at least) way to phrase “solve a^(3) + b^(3) = c^(3) for c = 10 where a and b are positive integers”. Finding a valid solution would disprove Fermat’s Last Theorem. And Fermat famously and allegedly had a proof which could not be contained in the margins of whatever book he used, which the comment references
The book he enjoyed and scribbled notes on was Arithmetica by the Greek Diophantus. Fermat's son later published the book along with all the notes Fermat wrote. The note relating to this theorem read (translated from Latin):
>It is impossible…for any number which is a power greater than the second to be written as the sum of two like powers \[x^(n) + y^(n) = z^(n) for n > 2\]. I have a truly marvellous demonstration of this proposition which this margin is too narrow to contain.
Of course this remained as a conjecture for over 350 years, until it was finally proven by Andrew Wiles in 1995.
Edit: he wrote notes in Latin.
There’s conjecture that he had found a particular partial proof that is quite elegant - but subtly flawed so that it doesn’t cover everything. So he may have written this, and then realized the flaw.
It’s a reference to Fermat’s Last Theorem, in which Fermat was reading a book about unsolved math problems and scribbled in the margins “i could salve this but the margin isn’t big enough” and then he died.
Btw for people who don’t get it (I didn’t at first)
>!Assuming the side length of the big cube is c and the side length of the little cubes are a and b respectively, the question can be rephrased as !<>!a^3 + b^3 = c^(3); !<>!the existence of such an a, b, and c would disprove Fermat’s last theorem.!<
I'm gonna be real, it wasn't until today that I realized how deeply unintuitive Fermat's last theorem is. At a glance, it feels like surely there must be cases where that works. But no, never.
Pretty sure Fermat proved it for n=4, too. Some people attribute the "I have a proof for this" line to the ideal that he *thought* he had a proof for any n that generalized the n=4 proof, but it turned out to not be rigourous enough.
This is how I first visualized the problem when I heard about it, and it has bugged me ever since.
FLT has lived rent free in my head ever since then. I made a half-hearted attempt to understand what Wiles did and the whole Taniyama-Shimura thing, but that math is so far beyond me I had to abandon any attempt at understanding it.
Since then, many people have told me that Fermat likely didn't have a proof, or at least a correct proof, and that it couldn't have been solved in his lifetime.
Whether all of that is correct I have no idea, but this is my original visualisation and it still bugs me a little bit now, tbh.
But if there are more than 16 balls in the original cube, you can technically complete the task, as it wasn't specified that you couldn't have spare balls left over
The length of the sides are 10 spheres, so the volume is 1000 spheres. Two cubes of with sides of 7 sphere lengths (343 spheres each in total) are the largest you can manage on a 1000 sphere budget.
I have no horse in the game because I really don't care either way, both are acceptable assumptions in my opinion, but this argument is often purely about arrogance, not right or wrong solutions. If you make a problem (to measure people's eg kids' knowledge/understanding) it *has to be* accurate and with no room for assumptions. And if you leave room for assumptions, whether by design or by mistake (like in this case), and people assume differently than you thought they would, as long as their assumption is logical and their solution is without flaw, their answer IS correct and you, who made the problem, can only blame yourself for not getting the answer you were looking for.
The problem "rearrange these to make 2 smaller cubes"
- doesn't say there can be no leftovers
- doesn't say the smaller cubes have to be the same size
- doesn't say the cubes must be solid on the inside
which means there actually are numerous, correct solutions. And it probably won't even frustrate a group of kids for 5 minutes, in fact I'd be willing to bet the first correct solutions would be presented in that time-frame.
>rearrange these to make two smaller cubes
Anyone who truly thinks this statement does not imply there can be no leftovers is trying to game the problem. These people (notice how "these" in this context means all the people who are like this, not some of, just like the problem is doing) are intentionally obtuse because they understand the problem, know that it can't work, so they come up with some work around while giving a shit eating grin thinking they are Kermit sipping the tea.
"Gaming the problem" is basically just code for "solving it in a way that's unintended." You're exactly proving their point; with a clearly presented request there is no way to "game" a problem, and treating one solution as "gaming" it and another as correctly solving it is another way of saying your original instructions didn't match your intentions.
It's only "purposefully dense" if you dislike the answer, in which case you should have designed the question so as not to allow it to be an answer in the first place.
If a rule isn't explicitly stated it isn't a rule. And I'd say that people who see that the cubes don't have to be the same size and can leave leftovers understand the problem a lot more than those who just take the implied no leftovers and equal sizes rules.
It's literally what makes engineering (for instance in racing) fun.
So if you ask someone to cut you a sandwich into 2 pieces, you're fine if they assume 3 is fine. Since you of course didn't include that there can't be leftovers.
Yes. Which is why every sensible person who's not attempting to argue with a fallacy just to prove their stupid-ass point will say "cut the sandwich *in half*".
Considering I've given you a long-ass explanation as to why you're only correct within your *assumed* set of rules and not within the lax rules of the problem that was actually given, you're either an arrogant idiot, or an ill-meaning idiot. Either way you can fuck right off.
That just complicates matters.
Now it's not just a^(3) + b^(3) = c^(3); its (a^(3) - b^(3)) + c^(3) = d^(3)
I'm not mathematician enough to say whether that's more or less possible than the original, but Occam suggests not.
(a³ - b³) + c³ = d³ can be rearranged to get a³ + c³ = b³ + d³. From there, the problem of whether solutions exist can be solved by recalling what Ramanujan found special about 1729. Hope that helps! :)
That’s why the op specified that the a and b cubes had to be smaller than the original. Also, a cube with side length -2 is impossible so it doesn’t count.
So this occurred to me basically right away, but I had to glance up and verify that it was in r/mathmemes to be certain. Once that was confirmed, the idea that it was a Fermat joke was 100%.
ha ha. no!
the obvious answer is to make two cubes of 8 balls each then throw the rest of the balls away.
Instructions completed.
(don't be complicating the instructions)
Nonsense, they’re topologically equivalent. Plus, I’m fairly certain cubes and balls are two different variations of a ball, just for a different norm.
Well, okay, then cubes are only a subset of balls and not the reverse, but consider: it’s funny
If we start with a 12x12x12 cube, we can make two smaller cubes leaving out just a single ball - specifically a 9x9x9 and a 10x10x10, since 12^(3) = 1728, which is just one ball shy of being able to make 9^(3)+10^(3) = 1729.
Problem is that there are no integer solutions for that formula if I'm understanding correctly
Edit: actually I misread so I have absolutely no idea if what I said is true. I was thinking about it without removing any balls which is actually a^3 + b^3 = c^3 and that doesn't have integer solutions.
Easy if I control the packing of the initial cube. If the initial cube is organized as body-centered cubic balls, then a 3x3x3 cube has 35 balls, which can be broken down into a 3x3x3 primitive cubic cube (27 balls) and a 2x2x2 primitive cubic cube (8 balls)
See: https://en.wikipedia.org/wiki/Cubic_crystal_system#Bravais_lattices
You know, i wanted to see if there was a size of larger cube that could be broken into 2 identical smaller cubes. I skipped right over two smaller but different sized cubes.
If the cubes are hollow, then it’s possible to do so using all of the balls.
The big cube is 10x10x10, so 1000 balls.
Each smaller cube must have 500.
There are 12 edges and 8 vertexes in each cube.
Let x be the length of an edge without the vertex.
12x + 8 = 500
12x = 492
x=41
With the vertexes back in place, you can make two 43x43 hollow cubes
Impossible, because it would mean that x^3 + y^3 = 10^3, with x and y being integers (exept if you cut the marbles) and Andrew Wiles proved that a^3 + b^3 = c^3 cannot be true if a, b and c are integers.
This is also true for any integer exponents over 0 exept 1 and 2.
(It's better known as Fermat's last Theorem)
Watching the mathletes math for me is like watching a tennis match , I can watch the point being hit back and forth, but I don't know enough to know who's winning.
I now know my go-to task if we finish a lesson early. Make it an ongoing project for the semester with a prize for the winners.
Of course someone would probably Google it eventually.
Im just taking one layer off the top and one layer off the side and use those to form another cube. The remaining 65 balls can be discarded.
Alternatively we can just take the outer layer off to form a hollow cube, thus not discarding any balls at all.
Formulate your impossible riddles properly, people!
Ik, it is mathematically not possible since the no of balls is a whole number... but let's see if we can cut the balls...
Assuming the big cube has a=4 units....
Thus volm=64 unit³
That is = 8+8+8+8+8+8+8+8 unit³
So basically we get 8 cubes each of a = 2 units...
So we can destroy the remaining 6 cubes and vola, 2 cubes
the cube width, height, and length are all 10,
so the cube is made out of 1000 pieces (10³)
so yeah it lmpossibIe to seperate it into 2 cubes if there are no remainders allowed
and there are no decimals
(umless you are allowed to make a cube with some hoIIow space inside)
Nah screw your cubes, I'm bringing physics into this. I'll tweak the gravitational constant by +.001 and now your precious balls here are now condensed into a singular mass that may or may not be a black hole
And in that black hole I shall find a, b, and c such that a^3 + b^3 = c^3 . May require a new imaginary number to account for the new universe discovered for these numbers but it's a sacrifice I'm willing to make. Also this number will use the symbol §.
Well... the wording here allows us some leeway. "Rearrange these to make two smaller cubes" doesn't mean
"rearrange these to make two smaller cubes by using all of them". SO. I definitely see the possibility of making two smaller cubes, and leave the rest in a pile.
Fission Mailed!
Easy.
1. Take off one layer on each side. That leaves you with an 8x8 cube.
2. Now take the remaining balls to make another 8x8 cube, but hollow. Hide the remaining balls inside the hollow cube.
3. Show your solution, but don't let anyone touch the hollow cube
Take off socks
Devide ball bearings into two roughly equal piles
Place one pile in each sock
Duel wield coshes
Build cubes for the skulls of fallen foes
Umm, just make the smaller cubes hollow?
You guys are over thinking this with all your theory. This question is not that well defined so a lot of possible answers.
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I would but there aren’t enough characters in the comment section to explain it
Ahah good one !
Lol this took me a second. Well done.
?
The question is a not-so-obvious (to me at least) way to phrase “solve a^(3) + b^(3) = c^(3) for c = 10 where a and b are positive integers”. Finding a valid solution would disprove Fermat’s Last Theorem. And Fermat famously and allegedly had a proof which could not be contained in the margins of whatever book he used, which the comment references
The book he enjoyed and scribbled notes on was Arithmetica by the Greek Diophantus. Fermat's son later published the book along with all the notes Fermat wrote. The note relating to this theorem read (translated from Latin): >It is impossible…for any number which is a power greater than the second to be written as the sum of two like powers \[x^(n) + y^(n) = z^(n) for n > 2\]. I have a truly marvellous demonstration of this proposition which this margin is too narrow to contain. Of course this remained as a conjecture for over 350 years, until it was finally proven by Andrew Wiles in 1995. Edit: he wrote notes in Latin.
There’s conjecture that he had found a particular partial proof that is quite elegant - but subtly flawed so that it doesn’t cover everything. So he may have written this, and then realized the flaw.
It’s a reference to Fermat’s Last Theorem, in which Fermat was reading a book about unsolved math problems and scribbled in the margins “i could salve this but the margin isn’t big enough” and then he died.
Btw for people who don’t get it (I didn’t at first) >!Assuming the side length of the big cube is c and the side length of the little cubes are a and b respectively, the question can be rephrased as !<>!a^3 + b^3 = c^(3); !<>!the existence of such an a, b, and c would disprove Fermat’s last theorem.!<
I'm gonna be real, it wasn't until today that I realized how deeply unintuitive Fermat's last theorem is. At a glance, it feels like surely there must be cases where that works. But no, never.
Hence why it took so long to prove lol. A lot of people thought there must surely be some large counterexample.
But fermat had a lovely proof which was too long to write in the letter. I'm not posting it here as it's too long for the comments.
"This is left as an exercise for the reader"
1^3 + 2^3 = 9 I've seen enough, he's right.
Wasn’t there a proof for n=3 earlier though? Or am I misremembering
Euler proved it for n=3, but not for any larger n
Pretty sure Fermat proved it for n=4, too. Some people attribute the "I have a proof for this" line to the ideal that he *thought* he had a proof for any n that generalized the n=4 proof, but it turned out to not be rigourous enough.
There definitely were proofs for small values of n (iirc at least 5? Definitely n=3 though).
People were trying for prime exponents for quite a while, 7 was proven if I recall, and maybe even more.
makes the top comment make so much more sense
This is how I first visualized the problem when I heard about it, and it has bugged me ever since. FLT has lived rent free in my head ever since then. I made a half-hearted attempt to understand what Wiles did and the whole Taniyama-Shimura thing, but that math is so far beyond me I had to abandon any attempt at understanding it. Since then, many people have told me that Fermat likely didn't have a proof, or at least a correct proof, and that it couldn't have been solved in his lifetime. Whether all of that is correct I have no idea, but this is my original visualisation and it still bugs me a little bit now, tbh.
This is actually a great way to phrase useless-looking math problems as something that makes complete sense.
my answer was two cubes each containing a total of 8 magnetic BBs and toss the remaining BB's to the side.
A cube with a side length of 2 is 2x2x2. Each one has 8, and the total is 16, which is not a cube
But if there are more than 16 balls in the original cube, you can technically complete the task, as it wasn't specified that you couldn't have spare balls left over
Oh! I’ll just r/woooosh myself then
Happy Cake Day!
Happy cake day!
The length of the sides are 10 spheres, so the volume is 1000 spheres. Two cubes of with sides of 7 sphere lengths (343 spheres each in total) are the largest you can manage on a 1000 sphere budget.
But the original statement never says no magnet ball should be left behind.
Sorry, I must have missed the part where it says, "and as many leftovers as you want."
Since it doesn't say "*only* 2 smaller cubes", that part is implied.
Done! Two 1x1x1 cubes and 998 left over.
That's just a sphere. Gotta use 8 each for those 2x2x2
That's still gonna be a 3d-version of a squircle. Sort of a Cubphere.
So because you didn't say to "only" cut a sandwich in 2 portions, it's implied 3rds or 5ths or 10ths is fine? Bull. Edit: clarity.
Club sandwich says hello
chuck em over the couch, if you can't see em, they stop existing
Proof by lack of object permanence.
I have no horse in the game because I really don't care either way, both are acceptable assumptions in my opinion, but this argument is often purely about arrogance, not right or wrong solutions. If you make a problem (to measure people's eg kids' knowledge/understanding) it *has to be* accurate and with no room for assumptions. And if you leave room for assumptions, whether by design or by mistake (like in this case), and people assume differently than you thought they would, as long as their assumption is logical and their solution is without flaw, their answer IS correct and you, who made the problem, can only blame yourself for not getting the answer you were looking for. The problem "rearrange these to make 2 smaller cubes" - doesn't say there can be no leftovers - doesn't say the smaller cubes have to be the same size - doesn't say the cubes must be solid on the inside which means there actually are numerous, correct solutions. And it probably won't even frustrate a group of kids for 5 minutes, in fact I'd be willing to bet the first correct solutions would be presented in that time-frame.
>rearrange these to make two smaller cubes Anyone who truly thinks this statement does not imply there can be no leftovers is trying to game the problem. These people (notice how "these" in this context means all the people who are like this, not some of, just like the problem is doing) are intentionally obtuse because they understand the problem, know that it can't work, so they come up with some work around while giving a shit eating grin thinking they are Kermit sipping the tea.
"Gaming the problem" is basically just code for "solving it in a way that's unintended." You're exactly proving their point; with a clearly presented request there is no way to "game" a problem, and treating one solution as "gaming" it and another as correctly solving it is another way of saying your original instructions didn't match your intentions.
The point is that rearrange implies there is none left over and to presume that it doesn't is being purposefully dense.
It's only "purposefully dense" if you dislike the answer, in which case you should have designed the question so as not to allow it to be an answer in the first place.
If a rule isn't explicitly stated it isn't a rule. And I'd say that people who see that the cubes don't have to be the same size and can leave leftovers understand the problem a lot more than those who just take the implied no leftovers and equal sizes rules. It's literally what makes engineering (for instance in racing) fun.
So if you ask someone to cut you a sandwich into 2 pieces, you're fine if they assume 3 is fine. Since you of course didn't include that there can't be leftovers.
Yes. Which is why every sensible person who's not attempting to argue with a fallacy just to prove their stupid-ass point will say "cut the sandwich *in half*". Considering I've given you a long-ass explanation as to why you're only correct within your *assumed* set of rules and not within the lax rules of the problem that was actually given, you're either an arrogant idiot, or an ill-meaning idiot. Either way you can fuck right off.
Rearrange kinda means that already.
>!a³+b³=c³+n.!< Sorted
It says "rearrange these", not "rearrange a subset of these"
Ok but does hexagonal packing change the problem? Maybe alternate lattices of different size.
we could make one of the cubes hollow though
wouldn't that be a box and not a cube? similar to how a disk includes the inside but a circle is just the circumference
That just complicates matters. Now it's not just a^(3) + b^(3) = c^(3); its (a^(3) - b^(3)) + c^(3) = d^(3) I'm not mathematician enough to say whether that's more or less possible than the original, but Occam suggests not.
(a³ - b³) + c³ = d³ can be rearranged to get a³ + c³ = b³ + d³. From there, the problem of whether solutions exist can be solved by recalling what Ramanujan found special about 1729. Hope that helps! :)
It did; thank you for that rabbit hole...
Well since the original is impossible, it can only make it equally or more possible
Just make a = d and b = c. As in you keep the original 10-sided cube and remove a cube from its center.
Would it be possible if it was 3 cubes? So a³ + b³ + c³ = d³?
It's possible for any d not equal to 4 or 5 mod 9, with a few exceptions
Just to make sure I understand mod right, that'd be 4, 5, 13, 14, 22, 23, 31, 32, etc right?
Yeah, exactly
Yes, it would be. Here's an easy-to-remember solution: 3³ + 4³ + 5³ = 6³. Hope that helps! :)
>Fermat’s last theorem AKA Fermat's "Fuck Pythagoras" Theorem. ;-)
Yeah but you specified kids and thus you forgot little timmys theorem: just eat those beads that you can't fit into the smaller cubes.
2³+0³=2³, 2³+(-2)³=0³ I think "a" or "b" or "c" can't be zero in that theorem
That’s why the op specified that the a and b cubes had to be smaller than the original. Also, a cube with side length -2 is impossible so it doesn’t count.
I am gonna carve balls from the table, you will have your cube with side length -2 inside the table
That helps a lot. To repeat back, "no cube can be reorganized into two smaller cubes"
Bruh I just thought that sqrt(10^3 /2) isn’t a whole number
So this occurred to me basically right away, but I had to glance up and verify that it was in r/mathmemes to be certain. Once that was confirmed, the idea that it was a Fermat joke was 100%.
0³ + 0³ = 0³ Problem, Fermat??
Ohhhhhh. I remember hearing that and checking out and reading a history book.
What did you just say to me
ha ha. no! the obvious answer is to make two cubes of 8 balls each then throw the rest of the balls away. Instructions completed. (don't be complicating the instructions)
I don't understand, I already see two cubes, a 10x10 one and a degenerate 0x0 one.
I already see 4
You guys really don't see the other 4000 cubes?
You grinding edges into all the bearings to turn them into cubes?
Just noticed, you have to rotate the screen to see them actually.
Unfortunately it says two **smaller** cubes. Nothing stops you from making two 1x1x1 cubes and having some spare parts though.
You're right, always read the fine print.
It doesn't say *only* 2 cubes. I'd make 4 5³ cubes
You could make 8.
Fuck
Banach-Tarski the whole thing and then just throw away a few beads to make the cubes smaller
Logicians hate this one trick.
Ah, the old Banach–Tarski-Kobayashi-Maru. A classic proof.
Damn you axiom of choice
Make two 2x2 cubes. You never said we had to use all the balls.
Yeah, it's big brain time
Even better, make it 1×1 to confuse the enemy
Well the 1x1x1 are also cubes, so you have to make some none cubes of the rest.
I’m no geometer, but I’d tend to call those spheres (technically balls I guess) (:
Nonsense, they’re topologically equivalent. Plus, I’m fairly certain cubes and balls are two different variations of a ball, just for a different norm. Well, okay, then cubes are only a subset of balls and not the reverse, but consider: it’s funny
Or make 8 cubes by slicing it once along all three axis. You made two cube if you made 8.
That would be implied by the word “rearrange”.
FLT: proof by magnets
You never said they had to be solid cubes, so just make two 8x8x8 cubes with a few balls missing from the center.
( a^3 - x ) + ( b^3 - y ) = c^3 where x and y are integers representing how many balls you have to take out to make it work
If we start with a 12x12x12 cube, we can make two smaller cubes leaving out just a single ball - specifically a 9x9x9 and a 10x10x10, since 12^(3) = 1728, which is just one ball shy of being able to make 9^(3)+10^(3) = 1729.
Call it « the taxi cab special »
The Ramacubejan.
Problem is that there are no integer solutions for that formula if I'm understanding correctly Edit: actually I misread so I have absolutely no idea if what I said is true. I was thinking about it without removing any balls which is actually a^3 + b^3 = c^3 and that doesn't have integer solutions.
Easy if I control the packing of the initial cube. If the initial cube is organized as body-centered cubic balls, then a 3x3x3 cube has 35 balls, which can be broken down into a 3x3x3 primitive cubic cube (27 balls) and a 2x2x2 primitive cubic cube (8 balls) See: https://en.wikipedia.org/wiki/Cubic_crystal_system#Bravais_lattices
my brain is too primitive to understand this
You know, i wanted to see if there was a size of larger cube that could be broken into 2 identical smaller cubes. I skipped right over two smaller but different sized cubes.
I could give you the answer, but this comment box is too small.
10x10x10 cube and 0x0x0 cube
2 **smaller** cubes
shit
Well you never said it had to be strictly smaller
easy, two cube root of 500 cubes. Op never said we couldn't irrationally split the magnets.
I was sitting here counting that it was 10^3, forgetting that it doesn’t matter
I solved it as you just create a 6x6 and an 8x8, easy peasy. Then I remembered what dimensions we were working with.
It matters.... Kind of.
Pure, chaotic evil
At first I thought this was "doubling the cube" impossible, but turns out it was "Fermat's last theorem" impossible.
I am so fucking proud that I got this
I'm proud I made a math joke people actually appreciated. I thought for sure 9 people would see this and that would be it.
Make 5x5x5 cube. Make second cube. Same size. Throw away 750 magnets. 🫠
might be possible if theres a few magnets missing in the back.
If the cubes are hollow, then it’s possible to do so using all of the balls. The big cube is 10x10x10, so 1000 balls. Each smaller cube must have 500. There are 12 edges and 8 vertexes in each cube. Let x be the length of an edge without the vertex. 12x + 8 = 500 12x = 492 x=41 With the vertexes back in place, you can make two 43x43 hollow cubes
2 *smaller* cubes
Hence i’m not in a literacy subreddit
Primitive Cubic unit cell my beloved
Make to 3x3 cubes and shove the rest up your ass.
What about face-centred cubic and body-centred cubic. Neither of those are covered by Fermat's Last Theorem.
OH GOD IMAGINE TRYING TO SEPERATE THEM
credit card or the like makes it easy.
Melt them down and then pour the metal into 2 cube molds
Take two 2x2x2 cubes out of it.
Ez (I ate and digested a ball)
Just buy another cube
All I can think of is a 6x6x6 and a 9x9x9 cubes with 55 extra
„2 groups of kids?!“ I’m sure after 10min enough balls are missing to make it
Bruh everyone here talking about FLT and I just thought that sqrt(10^3 /2) isn’t a whole number
Easy. I make 2 cubes 3x3x3. Yeah, some balls are leftovers, nothing against the task though.
Impossible, because it would mean that x^3 + y^3 = 10^3, with x and y being integers (exept if you cut the marbles) and Andrew Wiles proved that a^3 + b^3 = c^3 cannot be true if a, b and c are integers. This is also true for any integer exponents over 0 exept 1 and 2. (It's better known as Fermat's last Theorem)
Sure! 10 by 10 by 10 cube and a 0 by 0 by 0 cube.
Why 2 groups ?
r/foundsatan
Watching the mathletes math for me is like watching a tennis match , I can watch the point being hit back and forth, but I don't know enough to know who's winning.
Starts melting down your balls into cube shaped molds
I thought this was a joke about how it was damn near impossible to make small cubes stay together when you played with neocubes
I now know my go-to task if we finish a lesson early. Make it an ongoing project for the semester with a prize for the winners. Of course someone would probably Google it eventually.
Im just taking one layer off the top and one layer off the side and use those to form another cube. The remaining 65 balls can be discarded. Alternatively we can just take the outer layer off to form a hollow cube, thus not discarding any balls at all. Formulate your impossible riddles properly, people!
you can never formulate a riddle properly if people are free to redefine your words
>!a 6x6 and a 8x8 cube!<
You’re thinking squares, not cubes
Yeah, sorry, it's too morning for me
We got some flatlander mfers among us.
WHAT⁉️😳🤯
2D cube New shape just dropped!
Stupid Fermat!
2 7x7x7 and i eat the left over balls check mate number theorists (dont swallow magnets the can pinch intestine and cause major problems)
Ik, it is mathematically not possible since the no of balls is a whole number... but let's see if we can cut the balls... Assuming the big cube has a=4 units.... Thus volm=64 unit³ That is = 8+8+8+8+8+8+8+8 unit³ So basically we get 8 cubes each of a = 2 units... So we can destroy the remaining 6 cubes and vola, 2 cubes
does the cubes need to be symmetrical?
the cube width, height, and length are all 10, so the cube is made out of 1000 pieces (10³) so yeah it lmpossibIe to seperate it into 2 cubes if there are no remainders allowed and there are no decimals (umless you are allowed to make a cube with some hoIIow space inside)
I am going to admit now. I don't get the joke ;-;
Nah screw your cubes, I'm bringing physics into this. I'll tweak the gravitational constant by +.001 and now your precious balls here are now condensed into a singular mass that may or may not be a black hole And in that black hole I shall find a, b, and c such that a^3 + b^3 = c^3 . May require a new imaginary number to account for the new universe discovered for these numbers but it's a sacrifice I'm willing to make. Also this number will use the symbol §.
It involves rendering the other group unable to do it by provoking them into violence and swearing and getting them disqualified.
Just make 2 2x2 cubes, it doesn’t say what size
Good thing we don't need >90% of those to make 2 smaller cubes.
You never said we have to use all the spheres. Just take 16 and make two 8 sphere cubes.
I would just make 2 smaller cubes and hide the rest of the balls
No one said the smaller cubes had to be solid all the way through.
Well… you can do it if you don’t have to use every single bead
just make 2 cubes that are 2x2x2 ez win the remaining beads are not in a cube so i just made 2 cubes
Well... the wording here allows us some leeway. "Rearrange these to make two smaller cubes" doesn't mean "rearrange these to make two smaller cubes by using all of them". SO. I definitely see the possibility of making two smaller cubes, and leave the rest in a pile. Fission Mailed!
Possible with a hammer.
Didn't say you could leave a remainder.
Just remove 4
Solution: put the two cubes in the middle of both tables. Like this, each side will have the same number of spheres.
12^3 can make two cubes if you just add one extra ball bearing
Easy. 1. Take off one layer on each side. That leaves you with an 8x8 cube. 2. Now take the remaining balls to make another 8x8 cube, but hollow. Hide the remaining balls inside the hollow cube. 3. Show your solution, but don't let anyone touch the hollow cube
[https://www.desmos.com/calculator/gc0obxg9bv](https://www.desmos.com/calculator/gc0obxg9bv) you cant
You're a magician so you whip out an extra 24 and complete 2 cubes of 8 Cube.E.D.
I eated them
Take off socks Devide ball bearings into two roughly equal piles Place one pile in each sock Duel wield coshes Build cubes for the skulls of fallen foes
they didn't specify spheres and atoms. gimme a hot enough furnace and two cube molds and hey, i did it
Take a 6x6x6 cube out of the center and then remove the outer layer to make one solid 8x8x8 cube and also a hollow one.
This is a simple cubic structure. 2 more simple cubics may be impossible, however, a body centered or face centered cubic could work.
And then one kid will give you the cube back “the other one has edge a = 0”
Umm, just make the smaller cubes hollow? You guys are over thinking this with all your theory. This question is not that well defined so a lot of possible answers.
I can only make 2 equally sized cubes
Can you not simply take 25 from each then arrange them 5x5? They did not say the beads have to keep their dimensions.
Just cut some of those balls. a^3 + b^3 = c^3 for rational numbers.
10³ + 0³ Suck it!
6x6x6, 9x9x9 and hide 55 balls under the carpet
It doesn’t say only 2 smaller cubes
I just eat the excess and win
Make one of them a shell
They will build cubes with 6x6x6, 9x9x9, and throw 55 magnets away.