Can you solve the alien probe riddle? - Dan Finkel
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- čas přidán 23. 09. 2018
- Practice more problem-solving at brilliant.org/TedEd/
Solution to Bonus Riddle: brilliant.org/TedEd3ColorCube/
Your team has developed a probe to study an alien monolith. It needs protective coatings - in red, purple or green - to cope with the environments it passes through. Can you figure out how to apply the colors so the probe survives the trip? Dan Finkel shows how.
Lesson by Dan Finkel, directed by Anton Trofimov.
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For the solution to the bonus riddle visit brilliant.org/TedEd3ColorCube/! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Get riddling!
833! That's a very exact number! 🙄
Answer 833?? 😅
I think answer will be 833
Weird number.
Here is the best deal $600 FOR LIFETIME
Wish Netflix and Amazon has the same
I like how in each video, I'm someone different. A spy, a scientist, a deep space explorer, a cop etc. I can fulfill my long held dreams.
It's also great that this channel assumes I'm capable of making a probe like that. Cant even boil milk without burning it.
Hoor Anum burn milk??
@@xiaoweiwang01 Yeah. Dont ask me how, though.
Why would you boil milk?
@@TheSpencermacdougall its for illegal purposes
@@TheSpencermacdougall hot chocolate. pudding. boiled milk
A riddle about an alien probe. Does that make it an alien probe-lem?
Yes it does
Ba dum tss
I'm glad I'm not the only one who thought that😂
Badum Tssss
Sparkles! XD
Step 1: paint the cube green
Step 2: call the cube “eyes”
Step 3: inform the extreme cold and extreme heat that you have green eyes
Lolololololol
Island geniuses prison REFERENCE
i love how complex you made this just so you could make the green eyes joke 😭
step 4: we forgot to add lightning too
step 5: realized that green can handle lightning
step 6: sit on the corner
Me having a green eye meme on all the Ted ed riddle vids: What this person said
I dont have to solve it Bc apparently i have a team of engineers.
Lucky u
Lol true
U may have a alien problem
Lollipop
Lol
I like how I always pause it as if I can even figure out the riddle 🙈
Science with Katie i pause for a while just to see ur vids
Science with Katie!
Eshketit
Do you watch and comment on every single science video on CZcams?
That's so sweet of you accepting things.. .. 😊
Oh man, I love the little easter egg at the beginning of the video: the name of the planet, RH-1729 is named after the Hardy-Ramanujan Number, which just so happens to be 1729.
A Hardy-Ramanujan number is a number that can be expressed as the sum of two different cubes in two different ways. That means that 1729 is not the only Hardy-Ramanujan number it is just the smallest Hardy-Ramanujan number. 1729 is also the most famous Hardy-Ramanujan number as it is the reason that these kinds of numbers are called Hardy-Ramanujan numbers (there is a story surrounding this). Sorry if you knew this already.
Wat
I'm sorry, what are you talking about and how do you know that?
Hey Zsauce, Zichael here
@@meggamiauw4266 There is a story that Ramanujan saw this number on the plate of a taxi and mentioned to G.H. Hardy when he came to visit him, that it is the smallest no that can be expressed as the sum of two different cubes. Read in my earlier class school maths book.
"How do you paint the cubes?"
Me being simple: Start with the red, like it shows, then rearrange it to be all white, paint it green, do that again and paint it purple
I legit thought (and commented) the same thing-
How would you know that the green coat won’t screw up the leftover faces for the purple coat?
My toughts exactly
@@chedddargoblin lol yes thats true
@@chedddargoblin maybe, by having some models? or all of the changes goes directly to the probe?
I literally don't even know what he's talking about, I just like the animation. ;)
And his voice is very soothing
I wonder if aliens could solve this
... of the aliens
Only for aliens
OMG, HEARTED BY TED-ED!
Call the andalites!
If they don't want to help us, bribe them with cinnamon buns!
Only if they know English
Ted-Ed: the probe should be completely green in the electric storms
Also Ted-Ed: a module flies out to send signal
Now the cube is malfunctioning, and the aliens are dead.
actually, while the cube is rearranging, it let the other colors exposed so tecnically, the rearrangement of the cube itself needs to be fast before extreme condition breaks it apart
The 27 cubes can be made by cutting a big cube six times: twice in the xy plane, twice in xz, and twice in yz. If you paint the outsides red and the inner faces along two cuts in each direction one blue and one green, that works. Colors can be changed by changing whole layers of nine cubes around cyclically along all three axes.
okay, this one is the coolest solution
simple and straight to the point
This is so elegant and brilliant. I was skeptical but it passes at least one check - if the green paint is applied to all six cuts in one direction, and the purple paint to all six cuts in the opposite direction, then two opposite corner cubes will be painted red-green only and red-purple only - and the center cube will be painted purple-green only. That gives you exactly the three "double-corner" cubes you need. Brilliant!!
It took me a little while, but once I figured out the pattern I solved it!
Here's the fun thing though: While I was solving this I realized that you can also paint a 2x2x2 cube to display two colors, just as here you can paint a 3x3x3 cube to display three colors. This got me curious, and with a little more thought I figured out how to paint a 4x4x4 cube to display four colors.
So now I'm really curious if it is infinite. Like, could you paint a 10x10x10 cube to display ten colors? 100x100x100 to display one hundred colors?
actually, i think it is infinite. there's a way to prove that too, which I got from another commenter(uiuiuiseraph). it's hard to explain it visually as it's more a geometric solution than a mathematical one.
I'm not sure if I could explain how I solved this anyway, since I was mostly visualizing the cubes in my head.
Cool!
Nice! At least to see the number of cubes available is always equal to the paint required, you can see that you have available 6 faces times N^3 cubes which equals N colors times 6 sides times N^2 cubes per side
One128 1x1x1 is trivial :P
*Brain.exe has shutdown*
Brain.exe has stopped responding, do you want to close it?
Close Keep waiting
@@ThatGreenLED ERROR 404 brain.exe not found
Close all window to stop it escaping
[error sound] Brain.exe has stopped working
[Close]. [Report error].
@@ThatGreenLED close
Same
I solved this riddle, but the solution as worded is pretty difficult to replicate unless you have a 3d modeling program handy. "Start by painting the outside red" makes it a trial and error process with the task of mentally modeling all 27 cubes at the same ttime. Some can do that I guess but it is a needlessly complicated method. This riddle is more reliably solved by making it a math/logic problem, which the video does do but it's hidden by the use of modeling and not easy to follow.
The first crucial step is identifying that there are no wasted spaces. The second is realizing that only one cube at a time can lack a particular color (i.e 26 cubes have to have green SOMEWHERE and the same for the other two colors). The third is realizing that the solution will have to use symmetry.
Put those together and you can realize that three cubes will be two-toned corner cubes (3R-3G, 3R-3P, and 3G-3P). This provides 2 corner cubes for each color (plus the middle hidden one) Then since each color requires 6 more corner cubes, you can deduce that 18 cubes will be painted half red/green/purple (3R-?, 3G-?, and 3P-?). At this point we have accounted for all the corners and th hidden middle.
Remember now from the original points, all these 18 cubes must have a pattern using all three colors The only way for each if them to have all three colors, when half the cube is already used up, is to make two sides some other color and one last side from the final color. If we do this symmetrically, that means the six cubes of 3G will be divided into three cubes of 2R-1P and three cubes of 2P-1R. And the same for the 3R cubes and the 3P cubes, with the corresponding colors. This gives us (for each color) six corner cubes, six edge cubes, and six one-sided cubes. So we take this step, we find that we have accounted for the corner, hidden, and one-sided cubes for all three colors.
But we still have the edge cubes. Thankfully this part is easier because we don't have that many left to work with. We have used up 21 cubes (three two-tones and 18 multi-tones) so we only have six cubes. And since we need six more edge cubes of each color, we simply paint each cube with two adjacent same-color panels. This gives us six edge cubes for each color and finishes the problem.
That was a lot longer than I realized while typing it out, so maybe not "simpler" than the given solution lol. But it is easier to do it this way IMO if you don't have a 3D modeling program to keep track of the cubes visually.
Yep, I started with separating each cube and then realised what a tedious process it was going to be so I was like, eh forget it not doing it this way
njstuckey I will read your comment later maybe because it will take an hour to do so. It's so big
I did it this way too ;) good summary
I’m impressed
@@soum8975 haha yes I realized this after I typed it. I did it on mobile too, my thumbs haven't forgiven me yet
me: oh this is easy i can do this.
also me: you have to think
my brain: i found the best solution already, press the spacebar
I have an idea for riddle.
Space station called the Sephyris.ypur the commander of the ship. Suddenly the plants that make the oxygen in the shop are infected with a parasite fungus that one scientist accidentally released.As you scramble to get the backup seeds, a solar flare hits the space ship, and all the seeds in the containers fly out and get mixed up.One plant seed produces oxygen at the cost of carbon dioxide, the second absorbs oxygen and releases deadly toxins, and the third absorbs carbon dioxide but gives off deadly hydrogen.you obviously need the first plant.But there is a way to tell them apart.Each plant has only 3 leaves
The first plant can come in a variety of green, yellow and brown leaf colors, and their leaves can be mixed color.For example , one leaf can be green while the other two can be brown, or each leaf is an individual color and etc.The second plant can come in variety of green, brown and orange leaf color.The third can be brown, yellow and red leaf color.Not all leaves have to have all colors on their leaves.So the second plant can have 2 green and one brown.But, you have a machine.You can test up to 4 plants at a time to tell which is toxic.But you can only use it 5 times before the oxygen becomes critical.There are a total of 40 plants in all.And as long as you get the same amount of non toxic plants to toxic, or more non toxic plants than toxic, you will win.So what's the best strategy to ensure your survival?Please use my idea :) took me a long time to think and write
I really Apreciate Your Work but it seems much harder to me than this Ted Ed One's
@@vinitdhandharia3478 well thanks, but tbh alot of Ted ed riddles are quite harder than this, the prisoner hat riddle, the green eye riddle, the three gods riddle etc.Thanks for your input btw
That sounds awesome
Alright, so, here's what I got. Immediately, you can discard all plants that have Orange or Red leaves, as they will never be the correct plant. If a plant has Green and Yellow leaves, it can only be a good plant, so keep it. Since you explicitly need to have more Oxygen than Toxin, you should then move on to check plants that are Brown and Green, and discard any that are toxic. You do not check the Brown and Yellow with the machine unless you run out of Brown and Green, because it is more important to discard Type 2, which actively wastes oxygen, than Type 3. After your 5 checks, you keep any remaining plant that has Green/Yellow, Brown/Yellow, and any Brown/Green that checked out clean.
Now, I have some problems with this riddle:
First off, you should better define the 40 plants. Tell how many leaves each one has and what the colours of them are so we can better refine a strategy for searching them. For example, my strategy does absolutely nothing if most of the plants show only one colour, which is possible. Also, it would be nice to know how many of each plant their were, so that the strategy could be "surefire success" rather than "best guess".
Second off, better define the win condition. In my strategy, I assumed that you need as many plants as possible, but by the riddle's standard, I could win with just one plant, making the whole riddle trivial, since you get 5 chances to find a winner immediately, which is pretty game-breaking.
Last, you could work on the writing a little more. There's a lot of spelling and grammatical errors, so it's hard to read. Also, the important information is hard to gleam while coming up with the solution, so, add more white space! (Look at 1:38 to see a good example of this done right. And no-- "This is part of the challenge" is not a valid excuse.)
All in all, it's not a bad riddle, but it could use some work.
@@thearcanist8568 Truth be told, I was typing this off a mobile device xD so my clumsy fingers got the better of me.Secondly, this is just my draft, I didn't revise it alot and thanks alot for your input and suggestions :).
i cant even understand the riddle let alone solve it
First reply but 107th like
2nd reply but 115th like.
3rd reply but 130th like
i like your username bro
oh yeah uhhh 5th and 6th to reply but 142nd to like
No. But I can solve 3 by 3 Rubiks Cube.
Which has nothing to do with the video. XD
I can't even do that
it helps a bit. you realise faster that you need to think in corner edge and center pieces
@@owendeheer5893 when i started learning how to solve cube i understood that centre never moves,middle cubes have only two colors,and corners 3 that need to be put in their exact placd
Same
3x3x3. A 3x3 is just a square
Step 1: Confirm you have green eyes.
Step 2: Ask the dangers to leave.
Step 3: Cover it all in red.
No
What is up with the green eyes?!
@@iciclesunshine4935 reference to Ted Ed's green eyes riddle
@@sayoriiiiiiiiiiiiii obvious
Step 4: Make the pirates agree with your plan.
I figured it out with my own method that with some trial and error got me to the correct answer.
1/ I figured the pieces needed for example: Color A, were:
* 8 Corners: 3 A Faces
* 12 Edges: 2 A Faces
* 6 Faces: 1 A Faces
* 1 Center: Any
2/ Then I figured the combinations needed to be symmetric in some way so that it works the same way for all colors, so I thought:
x.- 3 A + 2 B + 1 C
y.- 3 A + 3 B (I figured there couldn't be more than 1 of each of these (so 3 in total) since that would mean I wouldn't be able to make a cube where all faces are the same color)
z.- 2 A + 2 B + 2 C
* Made all combinations of colors for each.
3/ Then went through trial and error testing with different proportions of each group until I could make JUST the red cube with that, since I always kept the proportions symmetrical I only needed to test one color, and it took me less than an hour to solve even with all the preparation and the trial and error or even the fact that it took me some tries to even consider the idea of "y" and "z", since I started only making combinations with "x".
I loved this one.
This is really good! Right on!
“Bonus riddle: How many _more_ ways can you solve the riddle?”
Objection! Question assumes facts not in evidence.
I really struggled with this and was initially certain that there was only ONE way to solve the puzzle. Every part of the solution seems logically necessary!
Then I realized what they meant. Technically, there are 7 different ways to solve the riddle - really, 7 minor variations on the solution.
Most of the solution stays the same, because it is logically necessary:
-There must be 3 double-corner cubes (red-green, red-purple, and green-purple)
-There must be 6 triple-edge cubes (all identical to each other, 2 faces in each color)
-There must be 18 single-face cubes (1 face in one color, 2 faces in another, and 3 faces in the last color)
-Among the 18 single-face cubes, 6 of them must have a single red face, a different 6 must have a single purple face, and the last 6 must have a single green face.
Here's the tricky part: within each single-face color, the balance of corners (3 faces) and edges (2 faces) in the other two colors can actually vary! E.g. among the 6 single-red-face cubes, there could be 3 purple corners and 3 green corners. I initially assumed this *had* to be the case, that it was always 3 vs. 3, but it doesn't! There can be 1 purple corner and 5 green corners - or 4 purple corners and 2 green corners - or even 0 purple corners and 6 green corners! The key is that this has to be balanced out by the other single-face colors. For example, if I want *all* of my single-red-face cubes to have a purple corner (3 purple faces) and a green edge (2 green faces), I can do that. I just then have to make sure that the single-green-face cubes have NO purple corners (I've already used all 6), and that the single-purple-face cubes have ALL green corners (I'll need all 6!) It all balances out.
If we code the different types of single-face cubes by face-edge-corner, e.g. RGP means red face (1 face), green edge (2 faces), purple corner (3 faces), here are the 7 different ways of painting them:
0 RPG 1 RPG 2 RPG 3 RPG 4 RPG 5 RPG 6 RPG
6 RGP 5 RGP 4 RGP 3 RGP 2 RGP 1 RGP 0 RGP
6 GPR 5 GPR 4 GPR 3 GPR 2 GPR 1 GPR 0 GPR
0 GRP 1 GRP 2 GRP 3 GRP 4 GRP 5 GRP 6 GRP
6 PRG 5 PRG 4 PRG 3 PRG 2 PRG 1 PRG 0 PRG
0 PGR 1 PGR 2 PGR 3 PGR 4 PGR 5 PGR 6 PGR
In each of the 7 columns, if you count up the number of single-faces (first letter), edges (2nd letter), and corners (3rd letter), you'll get exactly 6 single-faces, 6 edges, and 6 corners for each color. That means all of these will work! However, they are technically different, "unique" solutions - e.g. the far left column and the far right column actually don't share any identical cubes!
Lol I'm sorry, I just realized I totally misread your comment and missed the joke. You're saying the question assumes you've already solved it at least one way, when maybe you haven't. My bad for misreading and going overboard 😂
@@CrescendoAmpio No, this isn't possible. Having 4 or more sides of the same color means some the sides will be hidden away and unused. Every side of the color must be visible outside. These unused colors will take up sides that could be used by other colors. Therefore, there's only one solution.
I.e. if you can a cube with 5 green and one purple, it can be either a corner piece or a side piece when green sudes are necessafy, but not both.
The one solution: 27 cubes. 3 3+3 cubes, 6 2+2+2 cubes, and 18 3+2+1 cubes, with colors distributed symmetrically.
@@MissMiserize I think you misunderstand me! I agree with everything you wrote, no cube can ever have more than 3 sides of the same color. The “7 different solutions” thing is about 7 different ways of distributing the 18 3+2+1 cubes. These cubes *do* need to have a balanced total number of colored sides, and always 3+2+1, but the ways that you pair the “3” colors with the “2” and “1” colors can vary. If you read my comment carefully, hopefully it will become clear. (Sorry, the verbal description of all this is a little tricky, so easy to get confused)
@@CrescendoAmpio I see what you mean now.
Anyone who can solve the alien probe riddle should work for NASA.
Thats not how it works or does it ?
Well I could have solved the 1st question when I was 14 yrs old.
And I know many others like me.
hmm that's the reason why NASA is hiring so many people from my country.
take my application
@@kudoamv which country are you from bro?
Eh honestly most computer scientists could solve this pretty easily. This is simply a logic problem and that's what we do best.
Alright look. You color 18 boxes with one colour on 3 sides, a different one in two sides and the third one on one side. You colour 3 boxes with one colour on 3 sides and another on the other three. You color 6 boxes with one colour on 2 sides, another on 2 and the third on the last two. You need 6-oneside 8-3side and 12-2sides for each color. It does not matter where you put them so you color 3 sides of red until you make sure you have 8 of them and the pieces fall together. It's simple when you think of what you need and especially because they can be rearranged to suit what you need from them. Basically 18(3-2-1), 3(3-3), 6(2-2-2). The reason there's so many solutions is because I can take a 3(3-2-1) and make 1(3-1-1-1), 1(3-3) and 1(2-2-2). Essentially resulting in the same number of kinds of faces. *nerd out*
Alice Fitzgerald I worked out the same flow of thought as you mentioned, but was still surprised TedEd used such 3D image demanding solution in this clip. Glad to see accompany here!
After solving I didn't want to think about all other solutions. But now I see the idea of.just breaking down the solution we found into its parts and permutations
The art of this video was so crisp and perfect. Mad props.
There's a *really easy* way to solve the riddle in this video.
Take the cube, remove the top layer, and without changing any orientation, put it on the bottom. The red faces from the top should end up touching the red faces from the original bottom.
Then do the same left to right
And again front to back.
You should end up with all the red-painted colors facing each other along three internal planes in the cube.
Paint the outside purple now... then do it all again. Now all the red and purple faces should be in the two internal sections, leaving all the final white faces on the outside.
Paint the final color.
Fun fact, you should be able to extend this, painting four colors onto an assembly of 64 cubes.
This method can also be used for any number of dimensions with any subdivison size.
Here's my solution, which I think is shorter and slightly easier to understand (hopefully):
1. Notice that no space can be wasted.
2. Assume symmetry
3. There are 3 types of cube paint patterns: 6 faces, 8 corners and 12 edges.
4. Realise that each cube cannot have 2 different patterns of the same colour (e.g. you cannot have both a green edge and green face on a cube)
5. Thus, realise that there are only 3 unique pattern combinations: face-edge-corner, edge-edge-edge and corner-corner
6. since only the face-edge-corner pattern has faces, we must have 6*3=18 of those.
7. We need 2 more of each colour's corner to get 8 corners, so we must have 3 corner-corner cubes that are coloured RG, RP & GP, where the center unseen cube would be one of these corner-corner cubes that do not have the appropriate colour.
8. Now we only need edges, so the rest of the 27-18-3=6 cubes will be edge-edge-edge cubes.
9. Now that you have the types of cubes, colour them according to the previous rules (symmetry and no repeat of colour)
I loved this one. Even though I didn't have the tools to make it, I discovered a similar method. Beautiful riddle!
I love these riddle videos! Please do them more frequently!
Next: Can you solve the Brexit riddle?!?! 🤔
Leaked: Can you solve the Brilliant Riddle? (sponsored by Brilliant) * going meta here *
No, no one can!
Easy. Break the modules apart. Paint Northern Ireland green, paint Scotland tartan and paint Wales dragon-coloured. Doesn't matter what colour the centre part is because nobody's ever going to hear from it again.
Lugh Summerson you mean england
Lugh Summerson I'm in England so I'm dead
This has nothing to do with the video, but the animation and interface is so pleasing to watch.
When we're painting the purple and green faces, all I can think of is "HULK SMASH."
Maybe I dont get it, but it seems pretty easy.
Imagine it beeing picked apart like at 00:00:44 . Now you can paint every layer. From bottom to top: Paint the most upper layer of cubes red on top, the second layer blue on top, the third layer green on top. Do the same from bottom to top: The bottom layer red on its bottom, the middle layer green on its bottom, the third layer blue. (I switched the order of green and blue, so you dont get boxes with blue on top and bottom.)
Do the same from front to back and back to front. The do the same from left to right and right to left.
If the probe now needs red, it can rearrange the top to bottom layers, so its red on bottom and top. (Every front/back and left/right layers stays the same by doing so.)
Then do the same rearranging for left/right and front/back.
It should work. :O
Wow. The only one who's thinking outside the box! I really want this to work! First test: are there three and only three bichrome miniboxes? Yes! One non-red in the middle; one non-purple in a corner; and the last non-green in the opposite corner. Perfect. The rest, sorry dude, I just can't visualise it. Someone get your blocks out quick!
Test two: do any blocks have the same colour on opposite sides? (This disqualifies.) No! So far so good.
Awesome! The old 3D visualisation capability is kicking in here. Test number three: are there six and six only boxes with only one red side? Ditto purple and green. This is another must.
Answer: yes yes yes!
As far as this ol guy can see...
Of course six red... Six centre face blocks in the original cube.
And six purple, two each stretching up, to the right and forward from the no-purple corner block. (I'm developing here on the x, y and z axes.) And six green, again two each stretching down, to the left and back from the no-green corner block.
These blocks with only one red side or only one purple side or only one green side are obviously 3-2-1 blocks. They can't be 4-1-1 or 5-0-1. Just imagine you had 4 red 1 purple and 1 green on a block. That's a no-no. You can't afford to waste red on the inside. You can allow yourself only 54 red sides, total. And, by the same token, you need all your hidden sides to sum 54 purple and 54 green.
So, well done Seraph. So far it looks real good. We've got our 3-3 blocks, three of them. We've got eighteen 3-2-1 blocks. And of these definitely six with 3 red, six with 3 purple and six with 3 green. I'm guessing the other colours are regularly distributed, eg three 3r-2p-1g and three 3r-2g-1p. This all fits in with my theoretical model, leaving just six more blocks of the 2-2-2 configuration.
Another great riddle! Thanks for supplying us with these entertaining and educating videos!
There are 6 distinct ways. I used a linear algebra approach
First construct the 10 vectors corresponding to each way to color the die, where [x,y,z] is [red faces, green faces, purple faces]. (For example a=[3,1,2] ... g=[2,2,2] ... j= [3,3,0])
Now construct 10 equations that explain how these vectors add.
#of red 3’ =8, so all vectors with a 3 in the first slot added must equal 8. This gave me the equation a+b+h+i=8. Do the same for green and purple.
# of red 2’s =12, so get that down followed by green and purple
#of red 1’s =6, so get that down followed by green and purple
Final equation is that all of them added = 27, as there are 27 dice.
Put these equations into a 10x11 matrix, where the first 10 columns correspond to # of a given vector in an equation, the last column corresponds to the total in that equation, and the rows correspond to equations.
Finally, reduce this matrix using gauss Jordan elimination. You get 1 free variable(call it x), 6x[2,2,2], 1xeach of the [3,3,0]’s, and the rest being either x or 6-x.
This gives 6 solutions, as that is the number of values for x that make the count for each die legal values
If you don’t want to do the elimination yourself, matrix.reshish.com works beautifully. Just remember to put in the zeroes as well as the 1’s.
When I was little, 4 to 6, I had such cube made out of wood, my father made it and painted it. I used to play with it a lot. There also was a book with it with different tasks and 3d patterns that should be assembled. It boosted my spacial thinking, and for that i am grateful. Although this video was not a challenge to me, it was a nice reminder
I love these riddles but they seem to be getting more math based. Could you guys do some more involving logic or statistics? Those are my favorite!
I didnt watch the solution yet, but this one didnt seem too math based. Just reasoning about how many corners, edges, and faces you need and how to get there 🤷♂️ i did it in my head while lying in bed unable to sleep
I love your riddles, they make my day!
Well that was pretty simple and I usually can never figure out these problems. After calculating the total number of faces needed (3 corner faces will need to be painted 24 times (72 faces), 2 edge faces 36 times (72 faces), and 1 side face 18 times, total of 162 faces that must be painted and 162 faces total) I found there was no possibility of waste, so the center cube would always have to contain only two colors, since if it had a third color it'd be wasted. So you can do that three times--three cubes will occupy corners 2 times and the center 1 time each (3 cubes with 3-3-0). Every single other cube will be facing out all the time, so they will have 3 colors exactly, so you don't need to track the colors.
That leaves 18 cubes that need three corner faces painted separately, which is the same as the number of 1 side faces that need painted, and will leave 2 edge faces that must be painted. Since there's only 3 remaining faces on these cubes, and you cannot paint them with another 3 corner faces, you MUST paint them 2-1 (giving you 18 cubes with 3-2-1). That leaves 6 cubes that are still blank, so you paint those with the remaining 18 applications along 2 edge faces for each (6 cubes with 2-2-2).
So far as I can tell this is the only configuration possible not counting any mirrors or rotations. Given that 3 cubes MUST contain only two colors each, and every other cube MUST contain 3 colors, then the 18 cubes with 3-2-1 can only be painted that way (since you'll always have 3 remaining blank faces after coloring in the corners and can't fill them in with another corner) and the 6 2-2-2 cubes can only be colored that way (since by necessity that's all you have left and they must have three colors each)
this is one of my favorite riddles you made because i like the animation and because the concept isn't a disasterous situation, but it's a discovery.
but could you find all the distinct ways.
Imagine humans finally found a way to contact aliens, and the first message we recieve is just "♡."
Which means 💩 for them
Now I want to have a cube like that as my personal "rubix cube" that changes color _completely_
Thank you for doing this, I really love watching these, dont stop doing them!!
I solved it very differently, more intuitively. I realized there's only 3 combinations:
- Cube with a corner, an edge and a center (I called it 321, 3 faces of one color, 2 faces of another, 1 face of another)
- Cube with two corners (I called it 33)
- Cube with three edges (I called it 222)
Then:
- We start with 8 corners, 12 edges and 6 centers for each color.
- There can only be 6 cubes of 321 for each color, because there's only 6 centers. They have all 3 colors. We're down to 2 corners, 6 edges and 0 centers for each color.
- With 3 cubes of type 33 we have all the remaining corners. We're down to just 6 edges.
- All the rest of the cubes are of type 222, they all have all 3 colors, so it's 6 of them.
My head twists and turns when I try to understand riddles, now add aliens.. head explosion
So hard every time I try I fail, yet I ❤️ This channel
I love these riddle videos, keep em coming!
So what I did was I first considered optimal arrangements. I eventually thought of a module that acts as a face for one coating, an edge for the second coating, and a corner for the third. It took me a few moments to realize that how many of this module configuration I needed must be the number of required faces for each coating. 6 faces x 3 coatings means 6 with 1 R face, 6 with 1 P face, and 6 with 1 G face. This left me still needing more edge and corner coatings, as I only had six edges and corners for each coating, and I knew I need 8 corners per coating and 12 edges per coating.
From there, I found the configurations for how many more edges and corners I needed for each coating. Ultimately, I ended up with what I call “edge modules”, “universal modules”, and “corner modules”. Edge modules are modules that act only as edges. Corner modules act only as corners. Universal modules act as an edge, a corner, and a face.
Edge modules are:
• (2R, 2P, 2G)x6 = 6 edge modules
Universal modules are:
• (3R, 2P, 1G)x6 +
• (3P, 2G, 1R)x6 +
• (3G, 2R, 1P)x6 = 18 universal modules
Corner modules are:
• (3R, 3P, 0G) +
• (3R, 0P, 3G) +
• (0R, 3P, 3G) = 3 corner modules
---
Total = 27 modules
Note:
The universal modules can be modified so that instead of the sets I listed above, you get sets with (3R, 2G, 1P), (3P, 2R, 1G), and (3G, 2P, 1R).
TED-Ed: The green shield protects the probe from electric storms.
Me: On earth we call it "lightning" 😆.
Wonderful riddle
Nice Animation by team TED ed
One special like for TED ed 👍👍👍
There is another solution. I'll be using this kind of notation - 3R, 2P, or 1G to mean 3 adjacent red faces, 2 adjacent purple faces and 1 green face, respectively. So here are the 27 cubes: 6 X (3R, 2P, 1G), 6 x (3P, 2G, 1R), 6 x (3G, 2R, 1P) -- these are 6 of 8 corner cubes, 6 of 12 edge cubes and 6 of 6 center cubes for each color -- 6 x (2R, 2P, 2G) -- the other 6 of 12 edge cubes, rotate as needed for each color -- 1 x (3R, 3P), 1 x (3P, 3G), 1 x (3G, 3R) -- the final 2 of 8 corner cubes for each color. The one that's not needed for a particular color goes into the center. Hope I haven't messed up something along the way.
Started with the same observations but took a different route that doesn't require the cube to be assembled while painting. This also hints at the number of possible ways to paint them.
Each color will need 8 corner pieces (3 sides painted), 12 edge pieces (2 sides painted), and 6 center pieces (1 side painted). The number of available faces, including the center cube faces, exactly matches the number of required faces (162 faces). That means that no faces can be wasted. That eliminates some combinations since they would either waste space (like just painting two sides one color and 3 sides another color) or double dip on colors (using the same cube for 2 center pieces plus 2 edge pieces would cover all faces but use one color twice).
The only ways to paint the cubes so that no faces are wasted are by painting them in one of three configurations:
1.) One color painted on 1 side, another color painted on 2 adjacent sides, and the last color painted on 3 adjacent sides.
2.) Two different colors painted on 3 adjacent sides each
3.) Three different colors painted on 2 adjacent sides each
Since the only way to color the center pieces is method #1, the use of 18 cubes is already decided (3 colors and 6 centers for each color).
Six cubes will have Red on 1 side with Purple on 2 sides and Green on 3 sides (alternatively Green on 2 sides and Purple on 3 sides)
Six cubes will have Purple on 1 side with Green on 2 sides and Red on 3 sides (alternatively Red on 2 sides and Green on 3 sides)
Six cubes will have Green on 1 side with Red on 2 sides and Purple on 3 sides (alternatively Purple on 2 sides and Red on 3 sides)
Note that using an alternative color scheme for one cube requires using the alternative color scheme for one of each of the other cubes. For instance, a cube with 1 Red, 2 Green, and 3 Purple would mean you have to have a cube with 1 Purple, 2 Red, and Green and another cube with 1 Green, 2 Purple, and 3 Red.
At this point, each color has their all the center pieces they need, but they each still require 2 corner pieces (3 sides painted) and 6 edge pieces (2 sides painted). These can be taken care of with the two remaining cube color configurations. Three of the nine remaining cubes will be used for corners and the other six will be used for edges.
Three cubes will have 2 colors painted on 3 adjacent sides each. The combination of colors will rotate for each one to give:
One cube with Red on 3 sides and Purple on 3 sides
One cube with Red on 3 sides and Green on 3 sides
One cube with Purple on 3 sides and Green on 3 sides
The remaining six cubes will have all 3 colors painted on 2 adjacent sides each.
All this should give you enough to find out the number of different ways you could paint the cubes. Wasn't sure if the last riddle was asking how many ways you could rearrange the colors (as in the alternatives I mentioned) or how many ways cubes could be painted to give required colors (as in every cube can be painted in one of 27 ways).
Scientist 1: let’s name a planet
Scientist 2: Smashes head on keyboard
Scientist 1: Perfect!
Actually no, the planet is named as Rh-1729, referring to the Ramanujan Hardy Number.
1729 refers to the first number that is the sum of 2 different types of cubes, therefore, it can be expressed as (1)^3 + (12)^3 = 1729 and also (9)^3 + (10)^3 = 1729
Hope I have confused you more...
3blue1brown: "The hardest problem on the hardest test"
TedEd: "Alien probe riddle"
*Honesty vs Clickbait*
Ted-Ed has a different goal with their videos. An obvious way to see that this is edutainment is tje animation and visual interest provided in these videos. 3blue1brown is utalitarian.
Do you mean, they both use "honest clickbait"?
3b1b's problem was definitely harder than this one ( even though I didn't solve any of them) . The title was a clickbait and Grant himself said that in his tedtalk.
I’m late but the title wasn’t clickbait. It was the hardest problem on the hardest test…
I tried this problem 4 years ago and couldn't solve it. Today I tried it again and it felt kinda like a sudoku, you know enough information to fill the gaps, piece by piece. I actually got to the exact same solution on the video, and also solved it in the same order. I don't know if I can do the bonus tho.
I solved this using a different method. I first defined the number of roles that the complete cube needed to have: 8 corner cubes, 12 edge cubes, 6 face cubes, and 1 center cube. The corner cubes needed to have 3 faces the same color. I then realised that you can't hide more than 1 cube at a time, so it didn't make sense to have more than 3 cubes that were 3 sides one color, 3 sides the other color. Still this meant that for each outside color configuration, I only needed 6 more corner pieces, because these pieces needed to have 3 sides one color and also needed all three colors, I called these 3:2:1 cubes (three faces one color, two faces the second color, and one face the third). For each color I needed 6 3:2:1 cubes. This left me with space for 6 more cubes, and all of these needed to function as edge cubes. Thankfully the is an easy way to make edge cubes with 2 faces of each color or 2:2:2 cubes (making sure to keep the same color faces adjacent).
Summary,
3 x 3:3 cubes that have 3 faces of one color and 3 faces of another color:
RED-BLUE, RED-GREEN, GREEN-BLUE
21 3:2:1 cubes
6 x BLUE:GREEN:RED
6 x RED:BLUE:GREEN
6 x GREEN:RED:BLUE
6 x 2:2:2 cubes
As for the bonus question, I imagine that the answer has something to do with the number of 3:2:1 cube color permutations you could have - and then the number seems like 2.
No. There are 7 solutions
This was literally the first riddle on this channel that I got right😂😭
Omg first riddle solved!
The answer can be confusing when you don't have a way to keep track of your cubes so I did mine in a spreadsheet.
It's best to start with the minimum requirements: For each color, the are 8 corner cubes that needs 3 sides of the same color. Initially, it would look like you need (3x8) 24 corner cubes in total, but that would turn out to be a tad too many. I then recommend you allocate number of color faces across 27 individual cubes on a spreadsheet. I feel it's more manageable that way.
Half of 27 is 13.5, and half of 13.5 if 6.75.
6.75 is almost 7, so I say the first 7 is green, next 7 is red, then purple is last. However 4 remain.
First one is red, then second green, then last purple. The remaining cube will be green at the top, purple at the bottom, and red on the sides.
Red will be on the outside of the cube, purple on the outside of the middle, and green sitting in the middle.
The outer red will fall of as heat approaches, allowing purple to shine, next as the purple shell feels the green tough surface of this odd planet, it cracks off. Then green finishes, allowing green to finish everything off, with a nice camera located on its extra little cube.
3:18
Me: I'm already lost.
Ted: It should be simple.
Me: Then I must be SiNgLe aS a PrInGlE
Almost got it!!!
Just gonna copy-paste the spreadsheet I used to solve this here. Also I definitely think that my work with rubiks cubes helped me solve this one, made conceptualizing the whole thing much easier:
"Rules:
Can paint each of the faces one of three colors (red, purple, green)
Each face can only be one color
Cubes must be able to be rearranged so that all outer faces are solid red, or solid purple, or solid green, without repainting (must be able to be any of the three)
First Things I Notice:
First things first - only one cube can be fully obscured at a time
There are 12 edge pieces, 8 corners, 6 faces, and one center
Multiplied across 3 separate cubes to total: 36 edge, 24 corner, 18 face, 3 center
Since there's no repainting, there can be no overlap within a piece between what sides are used for a corner, edge, or side
My Solution (wokred out as I typed it):
Easy combinations include triple-edge and corner-edge-face
Start w/ 18 corner-edge-face - remaining pieces: 18 edge, 6 corner, 0 face, 3 center
(None of those 18 CEFs have to - or can - be center at any time for our purposes)
9 actual cubes remain - this is also important, I realized
54 remaining sides and 54 sides that need to be covered mean that all remaining cubes must be maximally efficent on side usage
Use 6 edge-edge-edge pieces to remove all remaining edges - leaving just 6 corners to be covered
3 corner-corner pieces lets me place one inside at any time, while two others fill up the corners not covered by the CEFs outside - perfect allotment!
LOL I used the exact same terminology as they did - blame speed-cubing (we say center there instead of face but it was clear I needed a different word here)"
Once you painted the large cube with red, we have 4 kinds of small cubes:
1. 3 painted faces (8)
2. 2 painted faces (12)
3. 1 painted face (6)
4. 0 painted face (1)
Each small cube has 6 faces and we can have a cube painted in 3+2+1 configuration. (3 of first color, 2 of second and 1 of third) For example, we can paint top, front and left with red, right and back with green and bottom with purple. ------- (lemma 1)
There are 6 centre pieces (except the inner one), 12 edge pieces and 8 corner pieces, minimum of which is 6. So, we can have at most 6 groups of 3 cubes, where each group of 3 cubes will have first cube painted on 3 faces, second cube painted on 2 faces and third cube painted on 1 face. By lemma 1, we can paint the 3 cubes in each group in 3+2+1 configuration.
In a nutshell, there are at most 6 groups of corner+edge+centre cubes possible and we can interchange them for each colour.
After this, we still have:
1. 2 corner cubes painted on 3 faces.
2. 6 edge cubes painted on 2 faces.
3. 1 inner cube not painted at all.
The 6 edge cubes can take care of themselves (2 faces of each color).
We are left with 2 corner cubes and a single inner cube. Painting them in red-green, green-purple and purple-red configuration helps in utilising all the remaining faces.
anyone else here have no idea what he is talking about?😂 😂 😂
Me. Something about cubes and colours. After that I don’t know what he’s talking about. Now I think I’ll have a few beers
Same XD
same
I actually figured it out to a point where I knew you had to separate the 8 corners into the inner 3 faces being painted green and purple, but the rest of it I couldn't figure out
David Wallerstein me LMAO
I pause the video to solve the riddle but hey who am I kidding.
I can't solve single one of them fml
This riddle is way easier to solve than the trial and error solution.
1. Align the cube in an x-y-z grid.
2. Paint the 3x3x3 cube red.
3. Cyclic-shift the 3x3x3 cube by one layer, in each of the x, y and z directions.
4. Paint the cube purple.
5. Cyclic-shift the 3x3x3 cube by one layer, in each of the x, y and z directions.
6. Paint the cube green.
The same algorithm can assign N colours to an NxNxN cube.
Wonderful riddle! We had fun spending an hour trying to solve it
Or you can mix the paint before applying
I managed to solve this one in my head before the riddle
...so yay i guess
the only ted-ed puzzle i’ve solved so far :)
Here is the reasoning process that led me to a solution.
1) Notice that there can be no waste.
s=side, f=face(1s), e=edge(2s), c=corner(3s)
8c+12e+6f = 54s * 3 configurations = 162
27 cubes * 6s = 162
2) Realize that any cube that takes a turn in the center must spend both other turns as a corner (remember no waste can be allowed). There are 3 turns so 3 cubes must spend 2 turns as corners and 1 turn as center. This means we can forget about 2 of the corners.
Now we have 6 corners, 6 faces, and 12 edges left.
3) Any cube can be an edge 3 times in a row (2+2+2=6) so we can leave as many edge cubes in place as we want.
4) None of the 6 corners remaining can take another turn as a corner because it must keep at least 1 face free to participate in the third turn. Also none of the 6 corners can take more than 1 turn as an edge (3+2+2>6). So how about having the 6 corners be 6 edges next and then 6 faces?
Solution:
3 cubes rotate through center->corner1->corner2 (supplies 2 corners and 1 center each turn).
3 sets of 6 cubes rotate through corners->edges>faces (supplies 6 corners, 6 edges, and 6 faces each turn).
all other cubes remain in place as edges throughout (supplies 6 edges each turn).
2c+6c+6e+6f+6e = 8c+12e+6f so all corners, edges, and faces are supplied each turn.
*Oh yes! 2 Riddles in 1 Video to Boggle Your Mind!*
and the solution for 2nd one is a phrase. The phrase it brilliant.org/gocommitdie
I got my teacher to show these in math!!!
I could legit listen and watch these riddles for hours on end
Love these Ted-ed riddles. Gets the brain juices flowing
1:37 is when I realized I couldnt solze this lol
"It can also break itself apart and reassemble into any other orientation" as long as orientation of cubes are constantly changing with spherical motions and unlimited speed (no defined speed of changing orientation) the heat will be dispersed due to only one side of each cube being red. As long as each cube has at least 1 of each color their orientation does not matter if they are constantly changing with spherical rotations and unlimited speed.
Trying to solve ten ted-ed riddles in a row - so hard, but that was my fourth so i"m making a progress.. the beauty is that i started solving those riddles just after i solved one and forced myself to maintain the sequence.
Great explanation could not get get to the solution but at least learned a way to look at at complicate problems
So we're smart enough to develop a smart probe that can rearrange themselves but can't write a program to do a simple calculation. Adds up.
Even though I’ve watched all of these videos, I’ve solved none of them.
I have a super simple awnser!
Paint the outside red, get is to reshuffle into only white, paint purple, redo step #2, paint green! Done!
Wrong. After you paint red and purple, you have to paint the rest of the unpainted sides green since there is no space that you can waste. And since your method involves guessing, you likely just painted the sides so that there is no way to get the whole outside to be green.
For the bonus riddle, I got 7. The only cube type that can be painted multiple ways is the one that has 2 of each color. There are 2 different ways to paint the cube so that it 2 of each color. You need 6 cubes that have 2 of each color. So, you can choose any number from 0 to 6 to be painted a certain way, so it is 7.
I also got 7 for the answer to the bonus riddle, but I disagree with your description of why. Actually, there is only one way to paint a cube with 2 faces of each color ("triple-edge" cubes). 2 adjacent faces must be red, 2 adjacent faces must be green, and 2 adjacent faces must be purple. Although it might seem like you can paint these faces in different ways, any triple-edge cube is identical to any other triple-edge cube if you just rotate them both the right way.
It's actually the face-edge-corner cubes (1 faces of one color, 2 faces of the next color, 3 faces of the last color) that allow for 7 different variations! There must be 18 of these cubes, with 6 having a single face in each color. For each of those groups of 6, there can be anywhere from 0-6 corners (groups 3 faces) in a second color. E.g. the 6 single-face-red cubes can have 0, 1, 2, 3, 4, 5, or 6 purple corners. Once you choose a number out of those 7 possibilities, though, everything else about the 18 face-edge-corner cubes locks into place. So there are 7 unique solutions.
Actually, I figured out that you two both are right and the answer should be 49. There are two ways to color triple edge cubes (These two cannot be made the same by rotation) , that means 7 ways to color triple edge cubes. And for the red face cubes, it is true that you can choose to have 0-6 purple corner cubes, which gives you 7 again. That gives 49
Any math/logic content: * exists *
Brilliant: * takes it *
Me: *ANGERY*
Just paint them all brown and it will work.
My method is like this:
Each cube have 6 faces( hence all 6 faces should be used in the three phases, in one of these 3 variations 1-2-3, 2-2-2 or 3-3-0, zero means it's a center cube in one of the phases)
1- we have 3 center cubes at the three phases so 3 cubes should be painted with the 3-3-0 formula ( 3R-3G-0P, 3G-3P-0R and 3P-3R-0G)
2- we have 6 one face cube at each phase so a total of 18 cubes with the formula of 1-2-3, color coded like this 6*(1R-2G-3P, 1G-2P-3R, 1P-2R-3G).
That's a total of 21 cubes so far, the last 6 cubes will be painted with the 2-2-2 formula, 6*(2R-2G-2P)
To check if you have the correct pattern of painting you can calculate it like this:
At each phase the cube have 8 three faced cube(8*3), 6 one faced cube(6*1) and 12 two faced cube(12*2), multiply by the 3 phases we get( 24*3, 18*1 and 36*2).
So our paint job should have the same numbers, simply calculate the number of each painted part from the formula multiplied by the number of cubes used in each formula:
A- to calculate the 3 faced parts multiply 3 by 2 from step 1 and 18 by 1 from step 2 and 6 by zero from step 3 then add them together which gives 24 three sided parts ( 24*3)
Do the same for 2 sided parts and 1 sided parts
B- 3*0=0
18*1= 18
6*0=0
Which adds to 18 one sided part (18*1)
C- 3*0=0
18*1=18
6*3=18
Which gives 36 two sided part (36*2)
If I had a pen and paper my explanation would have been clearer and simpler, and finally sorry for my bad English
The solution I thought of was to make 12 two-tone corner pieces. 3 sides would be one color (all sides touching one corner) and the other 3 sides would be another (also touching at a corner).
If that's confusing, picture a die. Paint the 1, 4 and 5 sides one color. Then paint the 2, 3 and 6 sides another.
These 12 cubes are made in this way. Make 4 (purple and red), 4 (purple and green) and 4 (green and red).
The remaining 15 cubes are made as 3-tone cubes. 2 sides (that share an edge) painted 1 color. Another 2 sides (that share an edge) painted another color. The last 2 sides painted the last color.
Using dice again, (for example) that would be 1 and 4 purple, 5 and 6 Red, and 2 and 3 Green.
These last 15 cubes are identical to each other.
So to sum up, that's 12 two-toned corner pieces and 15 three-toned edge pieces.
That should serve as a solution.
This solution unfortunately doesn't quite work.
Let's try to make the big cube show all red with the pieces you've described.
First we use the 8 two-tone corner pieces that contain red to make the 8 red corners. So far, so good.
Now, what can we do with these 4 purple-and-green two-tone corner pieces? They have no red paint at all. One of them can be used as the center piece (no red paint needed), but the other 3 cannot be used because they have no red faces to point outwards. That's 3 holes in our cube already.
Moving on anyway, we can use 12 of the 15 three-toned edge pieces to make the 12 red edge pieces of our cube. Great! We have 3 left over, so we can use those 3 to make 3 of the 6 red "faces" of our cube - but one of the red faces on each of these cubes will be wasted, facing inside. Unfortunately, that still leaves 3 faces unaccounted for.
So to sum up, the best you could do is a cube with 3 exterior faces that can't possibly be red (these pieces only have purple and green), and 3 red faces wasted as they face inside. Close but not quite!
@@CrescendoAmpio Good point. Thanks for the feedback. Now that you said that it seems obvious that there would always be 3 cubes unaccounted for in this model.
But in my own discovery...
*I DON'T EVEN UNDERSTAND THIS PROBE-LEM* 😔
I'm Morty nice pun
BOIIIII
My reaction after watching this video: BOOM 💥
Each cube can be de described as (#red faces, #green faces, #purple faces), so to get the three cubes you want you need:
6 x (3,2,1)
6 x (1,3,2)
6 x (2,1,3)
1 x (3,3,0)
1 x (3,0,3)
1 x (0,3,3)
6 x (2,2,2)
If we want to construct a red cube we use:
- (3,2,1) and (3,3,0) as corner pieces
- (2,1,3) and (2,2,2) as side pieces
- (1,3,2) as center pieces
- (0,3,3) at the center of the cube
Plot twist: The aliens only likes circles and sees you cube as a sign of aggression.
The FBI had a hard time solving this
FBI hi
Im not a criminal or anything
THAT MEME SUCKS
Daniel Cuevas I’m arresting u on suspicion of being a criminal
No one’s talking about how the monolith is clearly the one from 2001?
I can't stop watching this series..
“You can start by painting the outside of the cube red, so that it can withstand the cold, alien planet”
Wow, that’s some strong paint.
No, I can't.
2:00 this was much easier to solve seeing the visual. Reminded me that I have to use all but one cube every time. So the obvious of splitting up 3 that were 50/50 hit. Then I just looked at what component was needed the least( single face) and calculated the 3,2,1 series based on that with enough of each of the 1 of any colour totaling to 6. That left me with only 6 left that I could use to complete the most common case, the 2 sided, which was just the amount I needed to have 12 interchanging 2's once I added it with the 6 3,2,1's that existed where the 2 landed on the needed colour.
Edit: Your example doesn't really demonstrate Why your method works, you just go through doing it and expect that it makes logical sense...
This is the best animated one yet
I used algebra to solve it.
Assuming you use no waste (as was proven in the video), the most of one color on any cube is 3. This leaves 10 different types of cubes
A- 3R 3G
B- 3R 3P
C- 3G 3P
D- 3R 2G 1P
E- 3R 2P 1G
F- 3G 2R 1P
G- 3G 2P 1R
H- 3P 2R 1G
I- 3P 2G 1R
J- 2R 2G 2P
The question is how many of each cube to have. As explained in the video, if you don't want to have any waste, you must have exactly 1 of type A,B and C.
Also note the following:
A+B+D+E=8 (there should be exactly 8 corner cubes for red)
A+C+F+G=8 (there should be exactly 8 corner cubes for green)
B+C+H+I=8 (there should be exactly 8 corner cubes for purple)
Adding the left sides to the left and the right sides to the right we get
A+A+B+B+C+C+D+E+F+G+H+I=24 or
A+B+C+D+E+F+G+H+I=21 (since A,B and C all equal 1)
Since A+B+C+D+E+F+G+H+I+J=27 (All the different types of cubes added together must be the 27 cubes)
27-J=21 or
J=6
Note the following:
F+H+J=12 (there should be exactly 12 cubes that have 2 red edges)
D+I+J=12 (there should be exactly 12 cubes that have 2 green edges)
E+G+J=12 (there should be exactly 12 cubes that have 2 purple edges)
or because J=6
F+H=6
D+I=6
E+G=6
Also
G+I=6 (there should be exactly 6 cubes that have 1 red edge)
E+H=6 (there should be exactly 6 cubes that have 1 green edge)
D+F=6 (there should be exactly 6 cubes that have 1 purple edge)
also using the corner equation and subtracting out A,B and C (since we know their value is 1) we get
D+E=6
F+G=6
H+I=6
So
D+I=D+F=D+E or I=F=E and
E+G=E+H=E+D or G=H=D
There are 7 possibilities. E, F and I can be 0,1,2,3,4,5, or 6. G,H, and D would then be 6-E.
Hope that clears things up.