As a security guard, I might very well be persuaded to let someone in if they showed me a neat math trick.

The second guard always requires a rousing recitation of a Lord Alfred Tennyson poem for entry. It’s multilayered security.

??

Theres any knows.,, how number toto Sidney 04-08-2021???

They are intensely annoying and distracting.

@@custardtart1312 cry about it

They're world class for sure

I love it. It essentially reminds you that even complex mathematics can be accomplished with just a pen and paper.

I hope Numberphile answers.

@@miggle2784 Or Tom Rocks Maths.

Tag him

It's mentioned in the extra footage video in the discription!

I assisted with the zoom interviews of students at York this year for physics undergraduates, and many academics ask questions similar to this. They're not after the correct solution, but more to see how a student approaches the problem, and if they ask sensible questions or not to get to the solution.

​@@topherthe11th23 "it's" is correct - that is a possessive apostrophe

@@CrashSable Its is always possessive, it needs no apostrophe

I think the number of WAYS you can do it for any given number N is another interesting question worth another video = )

i got the same rectangles as you: in a 10 by 10 square, one 4x8 and one 2x4 can fill one side, leaving a 6 by 10 space. then 3 more 2x4s sandwiched together like books on a shelf cut off the bottom of that area by 4x6, leaving a 6 by 6 area to be filled by two 3x6s

This is how I did it as well. Instead of fractions, I used a 10 x 10 square to start. Which is more or less what you did. And I got those same rectangles.

Cool! I'd be curious how we each approached this. My solution was different: two big 3/4x3/8, three medium 1/4x1/2, and two small 1/4x1/8. I started by covering one side with two equal medium rectangles. I then added two large rectangles to cover the other corners. The gap between fit another medium rectangle, leaving me a square to split.

I found the same solution as you Dane. My approach was to think "is there a way that we would have a center rectangle surrounded by the 6 others ?"

I’d be surprised if none of them were.

dude, i keep finding your commetns in science videos and checking your channel out, thinking you also make science videos, and everytime its that pokemon intro and sike deja vu hits

I don't

@@Ian.Murray k

I wonder what kind of arm tattoo Pascal would have gotten.

they don't

@@toniokettner4821 he said they do, at least in oxford

@@toniokettner4821 Dr Crawford literally said he’s used this question when interviewing people who have applied to study maths at the University of Oxford, so how can you say that they don’t? While they don’t use this _specific_ question any more, they still ask questions where the thing the interviewer is interested in is how the potential student reasons their way through the problem.

@@Ray25689 well i should have gone to oxford then

@@toniokettner4821 I think the pressure there is enormous

That's the same solution I found first.

When he said "the solution" :/ ... This is unique and also unlocks the n+5 property.

Yeah, same here. Posted the result to the Numberphile Reddit thinking I'd stumbled on something new, nope, appearantly there's an army of us.

@@polarisraven5613 Dozens of us, dozens!

@anomie nous Depends on what you define as a thought, further if you count how one treats it (explores it, acts on it, rejects it, forgets it, etc.). Does Sensory Input count as a thought? In that case, even the slightest change in viewing angle will offer a different thought. Certainly there are patterns as to how we react, but everyone's somewhat deformed or otherwise atypically molded in some way or other, and that influences the way we think. Maybe the vast majority of our thoughts have already been thought up by others, so what? Originality is over rated, it's not the uniqueness of an idea, it's how effective it is that counts. Now, uniqueness has it's own benefits, occasionally toppling entire meta's and traditions, but this is often done either by chance, or through a deeper study of the principles that make it all work, often a combination of the two. Whatever your goal, the likelihood that you're not the first to consider something has very little to do with the viability of the option, and even then, there are fields of study ripe with research to be done, combinations of thought yet undiscovered, original thoughts are still out there to be discovered, though the paths we take for granted that bring us there have been forged by the efforts of previous like minded (in goal) individuals.

@@TomRocksMaths 😍

:)

Glad you enjoyed it!

@@TomRocksMaths some people in the comments have solution for 7 rectangles. Are they right?

@@haikumagician4363 There’s at least one other solution that works. If the one given in the video is on a 6x6 grid, there’s another on a 10x10 grid, and that one does not have the largest rectangle full half the square.

It's a numberphile video. You shouldn't be surprised it's interesting.

@@General12th I didn't find this one interesting at all and finally gave up on it. I don't like "proofs". There's way too much going over my head and so I lose interest after about 1 minute.

Then what was your n+4 method?

​@@harveyrice8504 One alternative way is to do: Have two 1/3-by-1/6 forming a 1/3 side square on the top left corner, then add a 2/3-by-1/3 to complete the top 1/3 of the square. then put a 1/3-by-2/3 on the bottom right, and you are left with an empty square of 2/3-by-2/3 at the bottom left. So you can shrink your original solution by 2/3 into the bottom left and put those 4 rectangles around.

@@harveyrice8504 that was my point, I didn't have an n+4 method. Hence I wouldn't have come up with a complete solution without the video

@@dylanwinestone4625 I bet you would. You just with a bit more time. The tricky part was the n=7 I think.

@@dylanwinestone4625 You said you came up with the n+4 method from the start, then realized you hadn't come up with the n+4 method. I don't understand.

What path did you take?

Because a lot of stuff numberphile shows is really surface level that is meant to inspire the pursuit of learning. But deep level learning is not so fun all the time. There is a reason that historical documentaries are easier to digest than sifting than through loads of the actual historical texts/analysis.

@joseph crosby mecham -147?

You can pick up a graduate level physics or math book, in sure you'll start to dislike then again 😃 (Pl. Take it for Fun)

??

Yes x 1000. The stopmotion on this video isn't getting nearly as much praise as it should in the comments.

:)

I LOL’d so hard

XD Same!

nah, I figured 3+4=7 and therefore it should work...out of the depth of my guts...

For all ratio X:1, You can increase the number of rectangles by 4 each time, and go on forever. By "Going up" and "Going down", you could do modular arithmetic each time until you get to your sequence of a length you could go on forever. A solution to this problem will require looking at unique solutions for squares. edit: Wrote this comment like 3 times tripping over myself, Lol.

@@KiLLJoYKZfaq So proving that you can make all numbers over a certain size shouldn’t be too hard. The challenge, then, is to find out which small numbers are impossible for each ratio, and if there is a pattern to it.

@@KiLLJoYKZfaq How can you increase by 4? If you use the going down method you increase by X²-1 (for a ratio X:1), because you can choose any one rectangle, split it into X squares, and split each square into X rectangles. The going up method increases by 4(X-1), because you need four quarter-edges of dimensions X:(X-1) each to make a larger square. Now, there is one other thing we can do. If we split our initial square into four smaller squares, we can solve each sub-square to get a bigger solution. This is, if n, m, j, and k are solvable then n+m+j+k is solvable. In particular, if n is solvable then 4n, 7n, 10n, 13n, 16n, and so on are also solvable. Then, if n is solvable for some X > 2, we have that n+4k(X-1), n+k(X²-1), and n+3kn are solutions for all positive k. Interestingly, we are no longer guaranteed that every sufficiently large n is solvable: for X = 4 our derived solutions are n+12k, n+15k, and n+3kn, which can only cover differences which are multiples of 3. Of course, having a guaranteed n+4k would solve this problem for X = 4, and I think for all values of X.

Ah! You can also make a different going-up by building a 2:1 rectangle out of your X:1 rectangles, and then putting four of these 2x1 on the edges. You need at most 2X rectangles to build each of these, for 8X total, which means that: For an X:1 ratio, if there is a solution for a value of n then there is a solution for n+4k(X-1), n+k(X²-1), n+3kn, and n+8kX for any positive whole number k. For even values of X, the greatest common divisor of X²-1 and 8X is 1, so all sufficiently large n have solutions.

@@TheBasikShow you can split a rectangle into 4 smaller rectangles of the same ratio. Giving you +3 new ones. The +4 is from 4 new ones. Not entirely sure why you would quadruple everything

Dr. Stone?

In Moriarty the Patriot the MC is a maths teacher, but we never see him teach

I'd watch

@@dandre8019 Follow your own advice.

@@aminulhussain2277 i deleted my childish comment. thanks for reminding me!

Cool. I have been exploring triangle based versions of the original rectangle puzzle. The interesting variant is indeed the question of fitting N equilateral triangles into one larger equilateral triangle. My proofs, so far, show this is impossible for N=2 and for N=3, not yet determined for N=5 or for N=8, and proved to be possible for N=all other positive integers.

@@millylitre 8 is possible. Take one large triangle with a side length of 3/4, put it in the corner, and fill in the remaining strip with 7 triangles of side length 1/4. This method should work for any even number, where the small triangles with side length 1/n will give 2n triangles. And then of course you can subdivide one triangle to add 3 more and get any odd number (greater than 5). I suspect 5 is impossible but I'm not certain.

@@evanhoffman7995 The impossibility (for Square and triangle) is probably due to there being a minimum size that is needed for the sub-size or supersize internal or external number of shapes. One might consider for why filling a circle with circles is both impossible (adjacent whole circles leave little gaps where they join) and (only) infinty ? - filling a circle with another immediately inside- I think the ratio remains the same, then another inside that and so on ?

You and me both :)

:)

Then it'll have a hidden conjugate triangle people think is safe but was part of risky rectangle earlier

Oh no, we lost a vertex!

I found the hidden 180° angle the government doesn't won't you to see!

Ever consider that there might be a complex reason, for which i might be involved?

false.

:)

Indeed how can you python code or test this ?

@@highpath4776 Problem solving is also part of an interview.

@@NoNameNoLastName Pretty much everyone in CS watches numberphile 😂

:)

I got the same solution.

I did this one too! I paused the video before they gave the solution and was surprised when they did it differently!

@@doctajohn07 Me too.

Same solution here. Two 18 unit rectangles, three 8 unit rectangles, two 2 unit rectangles, add them all up and you get 64 units.

Yep, I had the exact same solution.

Glad you enjoyed it :)

This doesn't add up. Your rectangels would form a 34*28 rectangle.

@@Zwiezwerg92Whelp you're right, those numbers didn't make any sense. I plead the sleep-deprived college student at the time. I redid my algebra and got the same sort of pattern I remember seeing originally, but with different numbers. 1x2 (2) 2x4 (3) 3x6 (2) 3x6 goes vertically in lower left and upper right corners, 2x4 goes horizontally in the upper left and lower right corners, other 2x4 goes vertically in the middle, and 1x2 fills the gaps. I just noticed your comment and felt I had to dive in and figure it out once and for all. I remember being so determined to find a solution with a rectangle centered on the board. Thanks for double checking me!

Interview questions at Oxford aren't really timed. The interview itself will last 20-30 minutes and you'll explore two or three problems in that time. Also, it's not just that they say "do this problem" and you do it and that's it. You will talk through your thinking and what you try with your interviewers. That's what they're interested in seeing: how you tackle new problems and things you've never seen before. Not whether you get the right answers to some specific list of questions.

Robert is right - this is an interview question. But for the formal exam, a science exam will typically be 3 hours, have 10 questions on the sheet, and you'll be asked to answer 6 or 7 of them. And you have to show how you get your answer. And note: you'll be taking more than one exam for each subject (so one classic division for maths is a pure maths paper, and an applied maths one). Lastly, Oxford and Cambridge entrance exams tend to look for "something more" than just regurgitation of facts or repetition of a standard proof: they're looking for evidence that you can, in fact, think - not just remember.

Consider a ratio `a:b`. Breaking down and building up still work and give you n + 3 and n + 4 if you have n. So you know that if you can do n, you can do everything strictly greater than n + 5 (Frobenius applied to 3 and 4 gives 3 * 4 - 3 - 4 = 5). So the situation is like this: There's a bunch of stuff you can't do. Then you can do one of them. Call it n_0. Then maybe you can do some of `n_0 + 1`, `n_0 + 2`, `n_0 + 5`, maybe not. You can definitely do `n_0 + 3` and n_0 + 4. Then you can do anything >= n_0 + 6. The only uncertainty is about what is `n_0` and whether you can do `n_0 + 1`, `n_0 + 2`, `n_0 + 5` To determine `n_0`, let's first consider the simpler case of ratio `1:b`. You can definitely do it with `b` rectangles of size 1 x 1/b. And you can't do less because a rectangle has sides c x bc (or bc x c) for some real c and so bc

@@yaeldillies I'm not sure you can say you can do n+3 or n+4 when the ratio is different. Those builds are using the ratio of 1:2...

@@shapiroyaacov Using a ratio of 1:2 , min is 2, 1:3 surely min is 3 1:4 is min 4 and so on.

@@highpath4776 I completely agree with that. But what about the rest of the numbers?

@@shapiroyaacov No, the n+3 and n+4 techniques are still valid. Any a×b rectangle can be broken into four rectangles, each with side lengths a/2 and b/2. That takes care of the n+3 case. For the n+4 case, the "build up" technique still works. It's just a matter of finding the ratio needed to wrap four rectangles around a square. Looking at 3:1 as an example, every n can be ruled out except 1, 2, 4, 5, and 8. 1 and 2 are clearly impossible, leaving only 4, 5, and 8 to investigate.

:)

there will be a smol fee and a smol donation :)

"Rec tab"?

@@jursamaj recent

@@jursamaj Recommendations?

@@dayawalker That makes more sense than 'recent', altho I don't see anything for either that I'd call a 'tab'.

@@jursamaj record

This was my approach, too, though it took a moment to realize that I could divide by numbers other than 2.

@@1980rburgess makes me wonder what the limitations are given N different rectangle sizes. for N=1, all the squares must have even area, and therefore even side length, and therefore the number of rectangles must be an even number; other numbers get impossibly complicated very fast

:)

if you're ever in Oxford, come say hi :)

@@TomRocksMaths Aaah omg🤩 I take your word for it!🛫

:)

:)

How what works? You have to specify what you want to do and not just say "more dimensions" 😅

@@Ray25689 You’re a very uncreative person

@King Pistachion i think that dude is in gen z or something

@Michael Darrowyup. And the answer very much depends on what you choose

@Michael Darrow Yes, that's what I meant. If you make it 1×2×2, wouldn't the pattern be the same? But if it's 1×1×2, the first number you can do is 4.

The only relation I can come up with is that 3 and 4 are the 'magic' numbers which you need to solve it for all other numbers except 7. But if that is the true relation I don't know.

I was gonna say the same. The n+x works only with x=3 and x=4. But if n+x=3 works it would mean n=0 or n=-1 which is impossible. Same goes for n+x=4 I means that n=0 or n=1 who. Again is impossible.

@@hadrienlart but what if n+x=2

@@hadrienlart , proving n+3 doesn't imply n-3, because n has to work for n+3 to work. Meaning this doesn't apply to 3 or 4, because neither of them work. By your logic, 2 and 7 also shouldn't work, because neither of them are 3 or 4 more than a possible n. You have to have a base case, but, as we see here, there can be more than one base case.

You could take the 7-rectangle solution at 15:06 and replace the rectangles labeled 4 and 5 with any square, creating an n+5 rule, so it's probably coincidence.

That only gives 1/3 of numbers.

Well, once you've discovered the means by which all answers can be reached, there is no point in enumerating them, since, you know, they are infinite.

@@TomRocksMaths

The corner remark (along with some arithmetical argument maybe) shows that there are only finitely many ways!

@@yaeldillies I had not spotted the corner element, one could indeed split this into a vertices matching (topography?) method

Seconded

Nah

Focus on the maths, Edwards!

Eewwwww

For m:n with positive natural m and n, you can always add 4 by building up and add m*(n^2) - 1 by breaking down (first fill 1:n by n^2 of its copies, then copy the result m times). If the breaking down number is not coprime with 4, we won't be able to reach all numbers this way. So almost all natural numbers for even:odd ratios and at least almost all numbers in some mod 4 modulo classes for other cases.

Came here looking for this comment lol

:)

In which way?

@@Ray25689 in divinding objects in half and the number of objects in the cantor set still being infinite

The common bit is the fractal nature of both sets.

@@TheGarbageMann yeah, but how does this occur here? All you do is dividing things up, but nothing with infinity.

@@skebess that's what I was thinking as well, but it feels like there's more than just the fractal nature, as the cantor set is mainly about dividing a whole into infinite fractions