Feynman's famous long division problem

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[@wyattstevens8574]

You may have dome this one, but what about y' = y²+1? I know of *a* solution, but I don't know if it's the only one.

[@ingiford175]

At 9:40 it should be mentioned that c(x-y) can not be zero because if c is zero, then A is zero also and c and A are different. and we know that x-y is not 0 since x > y

[@ingiford175]

Should have waited 20 seconds... blah...

[@jkid1134]

"Well, you might say..." 😂

[@kylecoleman6221]

Either way I don't think it was proven yet that c can't equal to A even though its true that it cant.

[@IAlreadyHaveAKey]

@@kylecoleman6221 c not being equal to A is a requirement of the problem

[@goodplacetostop2973]

31:06 Non-English people watching this long division format 😡

[@Redditard]

Thanks a lot

[@goodplacetostop2973]

@@RedditardYou’re welcome 👍

[@davidcroft95]

Literally lol

[@reubenmckay]

The editing on this one was pretty neat. Also: Another nice problem to feature on this channel.

[@janami-dharmam]

A very similar problem is described in the Martin Gardener's Second Book of Mathematical Puzzles and Diversions, under "the Lonesome 8". ( see PAge 160). But the method used was more elegant.

[@huguesbornet1211]

« Half » countries, in Asia and UK, NA, display long divisions as we see here. Most of Europe and Africa, plus more countries display quite differently the dividend ( top left), divisor ( top right), quotient ( second line, below divisor). Leading to space restrictions if you want lots of digits after the decimal point for instance. Same display is used for extracting square roots in France. Extracting cubic roots is rare but follows same display.

[@d4slaimless]

For space restrictions in the second case ("Most of Europe...") you would need result with lots of digits after decimal point AND to divide a big number. Otherwise I see no problem as the lines of partial results will go down and from left to right the same way. I mean imagine this example from the video written with dividend top left. You will have 1 digit less space to write the result, which is nothing.

[@CosmicDoom47]

11:02 - If anyone else (like I was) is confused why Michael says A cannot be equal to 0, the reason is that the 3rd multiplication would be bAc * A, which gives you a 4 digit number (so A cannot be 0).

[@blue_sand6854]

I think this is, also, the reason y cannot be 0 (for x-y=5).

[@jplay9710]

The problem states that no dot is equal to A and one of the first steps is writing that the dot based on A-A = 0, so A=/=0 (or 1) from the first minute or so of the video.

[@factorization4845]

when testing the 9 cases, the faster way is to find a value (when x = 7 or 9) or two values (when x = 8) for A is to consider the last digit only in the equation Like for 79A = 82 (mod 100), since 9 times 8 is 72, we just need to test whether A = 8 would work.

[@tolberthobson2610]

Loved the editing on this video. Also very interesting problem.

[@kkanden]

loving the fancy editing on this one!

[@donwald3436]

Massive improvement in video editing techniques well done!

[@bazboy24]

This is the most fascinating video I have seen you make I could not stop watching from start to finish

[@mrphlip]

I tried to solve this one before watching the video, my chain of reasoning did things in a bit of a different order in the middle, but the basic ideas are the same... Rewriting my notes to use the same variable names as the video: * The second subtraction ends in A-A so the result there must have a 0, so A cannot be 0 * There are two separate multiples of the divisor that end in A, so: * They must be different since they are a different number of digits * Subtracting them, we get that c(x-y) is a multiple of 10 * Both c and (x-y) are strictly between 0 and 10, so one must be even, the other a multiple of 5 * A appears as the last digit of a multiple * so if c is even, so is A; if c is a multiple of 5 then so is A * But A can't be 0, and A and c can't both be 5 * So c must be even, and so is A, and x - y = 5 * No digit of the quotient is 0 or 1 (as no multiple is "0" or "•A•") * so y >= 2 * The third multiplier has more digits than the second, so A > y * so A is not 2, our only remaining options for A are 4, 6 or 8 * Third multiplier is bAc * A = •A•• * If A=4, the A*A will carry 1, which would make the hundreds digit odd, no go * If A=6, the A*A will carry 3 or 4, must be 4 (ie c >= 7) and then b*A ends with a 2 * If A=8, the A*A will carry 6 or 7, must be 6 (ie c

[@ed.puckett]

Good problem and super fun video work!

[@user-rk3su9tm1y]

I hope there will be more videos with similar long division problems

[@stephenhamer8192]

Incredible that you can get to a solution with such meager initial information

[@gregrice1354]

As a non- mathematician onlooker, I bet that's the lesson that made Dr Feynman see value in sharing it one gets to practice the dance steps in order to learn about the relationships of everyone at the party. This results in appreciation for various encounters with other groups of information symbols. With his great gifts and skill in communication of ideas, like notation system representation of quantum phenomenon more readily, we get nice look at the clockworks of these quantitative signs, in qualitative terms. Information design recognition?

[@cambridgehathaway3367]

I cant believe i watched this whole video. Very well communicated and a fun puzzle!

[@laogui2425]

Writing the divisor as bAc, and the result as defg, we see that c and therefore A are both even and different and non zero, so Ac is one of {24,26,28,42,46,48,62,64,68,82,84,86} and d-e = 5, so de is 94, 83,or 72. Then look for combinations of Ac and de which give the xxAA and xxA patterns and we get only bAc = {186,184,284,384,484}, so A=8, and only 484*8=3872 also satisfies the penultimate xAxx pattern, so defg is between 7283 and 7289 and we are almost there. However pure logic didn't solve it, I needed to use a spreadsheet plus logic. Oh, I didn't notice that the puzzle statement includes defg = xxAx, and therefore that is additional and unnecessary information. Of course 7289 is the only result amongst 7283-7289 which gives the pattern xxxxAxx

[@9adam4]

When I worked the problem myself, I made a 9 by 9 grid of digits for multiplication. Helped me narrow down the possible digits. 😊

[@CM63_France]

Hi, 0:41 : but are not equal to each other, unlike the A's . You have a very good assistant and a very good editor!

[@Daniel-ef6gg]

The simplest way to solve (10b+c)A = (10d+e) mod 100 would be to first look mod 10, solve cA=e mod 10 with A = e c^-1mod 10), which gives a possible single digit value for A. Then plug that value back into the original mod 100 equation and see if it solves that as well.

[@jongyon7192p]

Allow any blank to equal A and it'll still result in the same 1 solution. Will try again with leading zeros allowed

[@zygoloid]

@15:56 A slightly quicker way to rule out c=6: mod 5, y•bAc = ??A reduces to y=A. But that leaves only two possibilities for A, which are A=y and A=y+5=x.

[@InigoQuilez]

I never saw a long division written this way, the location of dividend, divisor and quotient are flipped vertical and horizontally with respect to the "natural" arrangement used in Europe. By "natural" I mean that it matches the order of the sentence "a divided by b is c". With the arrangement in tbis video you have to think of the division in a twisted ways, something like "c is how many b's you find in a" or what is it? It's also not optimized for locality of computation. Vert interesting, I'll research the history behind this.

[@normanstevens4924]

Not sure what you mean but this is the way that I learned it in the UK 60 years ago.

[@davidcroft95]

This way is useful, because it leave some room from calculations (from experience, the "natural" method don't); but as you said it's a little confusing in the "order of algorithm" (probably just because we were not trained this way)

[@iGeen7]

​@@davidcroft95it does not leave the room for quotient... also, you completly waste the space under quotient

[@edwardlane1255]

oh, took me a while to realise why at 12.09 you can't declare that y can't equal 2 because c is 2, because y could be any number including c

[@AlfonsoNeilJimenezCasallas]

More than a problem, it's an algorithm, impressive!

[@sfurules]

I feel like you just secretly taught everyone about encryption.

[@derciferreira2523]

Thank you! you improved your presentation. It´s lighter and cleaner. 🤗🤭🤫

[@jakehobrath7721]

I think I’ve seen BAC and AA written together before, but this feels less tragic

[@mspeir]

Well, my head now hurts! 😵‍💫🤣

[@tsriketwm7274]

Interesting method. Would that work for the puzzles in the daily Feynman long division puzzles blogspot?

[@andersama2215]

They give that 0 is the last result, meaning bAc evenly divides into the four digit number xyAz, your last step you can "subtract one" from xyAz (pretend z is 0), divide by bAc, you'll get a divisor you can multiply by bAc to get just short of xyAz, then add bAc to that result and you'll have "solved" for z.

[@HarshColby]

In the bAc * A = •A•• case, A must be at least 2 to get a four digit answer. In the bAc * y = ••A case, A must be at least 4 because c*y can't be less than 4 without using the A digit in c or y. (1*2=2, 1*3=3 specifically). Brute forcing bAc * x = ••AA, gives 22 solutions. I feel it's okay to brute force it, since the Mod solution requires quite a lot of brute force test and check itself. If you have a square times table handy, you can scan across the rows pretty quickly to find the few that might work, so it doesn't take as long as you might think. Brute forcing those 22 possibilities against bAc * y = ••A leaves only 4 possibilities and all have A=8, c=4, y=2. Check those 4 against bAc * A = •A••, leaves b=4. Checking bAc=484 against the 22 results in the first brute force list, shows x=7. Next, check (as was done in the video) the last digit z by multiplying just the first 3 digits to see if the divisor has an 8 in this hundred's digit. Complete the long division, since all numbers are now known. It can be done just about as fast (er, slow) as the Mod method, just in case Feynman's mom wasn't up on her modular arithmetic.

[@alexpotts6520]

I just like to imagine that Feynman goofed off and cooked up this problem over a long lunch break, while his colleagues were frantically trying to beat the Germans in the nuclear arms race.

[@Loki-]

In between him picking their cabinet locks because he was curious.

[@IntegralKing]

great video also, that's the cleanest chalkboard I've ever seen and the brightest chalk. Is that that chalk that's no longer made by that Japanese company?

[@bcjames2011]

The language some people use for division truly baffles me. I appreciate "it's just a convention" (and therefore the important thing is that the grammar is agreed upon, rather than strictly logical), but I still hit a mental barrier when people use the form "dividing X into Y", where X is the divisor and Y the dividend (Y/X). Consider this nonsensical grammatical example: • "Dividing three cups into seven litres". This makes no sense because it is not the cups (the divisor) being divided, but the water (the dividend). In English, the correct grammatical form for this example is obviously the inverse (dividend first): • "Dividing seven litres into three cups". This latter form is even logical when the divisor is omitted completely: • "Dividing seven litres" - perfectly meaningful. In the UK, when talking about mathematics, we are (or at least were, when I was in school) taught to use another form: • "Dividing Y by X". This doesn't have a strict grammatical counterpart (we wouldn't typically say "dividing seven litres by three cups" these days), but it does still preserve the fundamental fact that you are dividing the dividend and not the divisor. I don't expect anyone to be converted by this argument; "just saying" that even the simplest part of the problem (communicating it) can still be confusing. Now who wants a flame war about the superiority (or otherwise) of the Oxford Comma?

[@gregrice1354]

I like your degree of detail, explanation and example. I feel your type of sensitivity to the logic and reasoning used in teaching math may be why I didn't advance to calculus . Your brief example set against the text transcript of this teaching (with board graphics) would serve educational book and curriculum Design consultants well.

[@OrenLikes]

~ 18:00 SIX cases! each possible value of c eliminates one ordered pair of (x,y) !!!

[@KF888888]

I disagree. Nothing in the problem says that c can't equal x or y, only that none of those three digits can equal A.

[@talastra]

I might actually have a reason to try to oput this to use.

[@warrengibson7898]

I wonder how Feynman concocted this problem, making sure it was solvable.

[@sobertillnoon]

It must have been so hard to be close to Feynman.

[@NaHBrO733]

Case by case checking can be reduced and simplified Using notations in the video, and assuming no leading zeros as pointed out in video: Obvious: A!=0,1 (bAc * A is 4 digits) ; x>y>=2 (y=0 A=y; y=1 A=c); c!=0 (A=c=0) c(x-y)=0(mod10) => (c=5 and x-y even) or (c even and x-y=5) c!=5 (else a odd A=c, a even A=0, both contradiction) c even and x>=7. From bAc*x = _ _ AA, A is even c={2,4,6,8}, x={7,8,9}, A even. Ac * x = _AA, we can say that c * x = 20k+A (c, x) = (4,7), (6,7), (6,8), (8,8) , check Ac * x =_AA to know c=4,x=7,A=8,y=2 b,z solved with the same method in the video

[@rpa41985]

in the end when z=0 484*z+520 = 5 (mod 100) not 0

[@waynebennett745]

Anybody else unable to differentiate the chalk colors unless he says what they are?

[@fredfred9847]

Nice edits

[@fuxpremier]

It could be useful to introduce your formalism before going into details. I've never seen a division set this way and the beginning of the video is extremely confusing. It took me a while to understand why there are substractions in a division in the first place...

[@valseedian]

I got to 9:03 before I did the math and decided that it's only 100b perturbations in 32bit. wrote a quick js script to find all combos of 3/7=4 digits where A is in the right places knowing A can't be 0.

[@charlesgantz5865]

But did Feynman's mother manage to solve the problem.

[@georgehs3386]

I like to imagine her reaction being something like "Pff you are a thousand years too young, my son, these is a rookie problem"

[@geoffhazel2]

Multiply an into b. I have used into with division. Multiply a by b. The into was somewhat confusing

[@marsgal42]

Fun problem, interesting editing. You know you need to get out more when you start to wonder how many As must be present for the solution to be unique? What conditions are necessary for a solution to exist?

[@AliceYobby]

this is the question that really interests me! what was the underlying logic that allowed feynman to discover this puzzle in the first place? I was hoping this video would go into that a bit! but it seems obvious it has *something* to do with modular arithmetic

[@jean-baptiste6479]

Very difficult to understand when you have leared division in french school

[@reedjasonf]

At roughly 5:50 when you do the expansion of the right side, why is there just 100*box_pink? Where did the other unknown digit go? Maybe I haven't gotten to where that makes sense yet but I am not immediately understanding why you can make that step while you are doing it.

[@gloowacz]

I also don't follow. There are 4 digits, so the right side should be 1000[] + 100[] + 10A + 1A.

[@KF888888]

The colored boxes don't always represent numbers in the range [0,9]. In this instance, he combined 1000[]+100[] into 100[pink] where [pink] is in the range [0,99].

[@gloowacz]

@@KF888888 oooOOOOOOoooh, that makes so much more sense now. I missed that the boxes are not digits but numbers now, on the right side. Ok, now it makes perfect sense.

[@RigoVids]

9:35 aren’t you forgetting the case where c=0 and the multiple c(x-y)cong0(mod10) has a trivial solution? Why is c forced to be nonzero? Why are x and y not allowed to be equivalent?

[@RigoVids]

Jesus literally like 20 seconds later 💀💀💀

[@weleiful]

I watch this over and over and over. Finally I still don't understand.

[@Tiqerboy]

Is the goal to simply find the value of A, or do you need to find the 3 digit number (divisor) and the number you're dividing (7 digit number)? It would be nice to state what you're supposed to solve for.

[@demetriuspsf]

Tried to solve it myself before watching. I was able to do it, but it took me about an hour to figure it out. And my method was not nearly as neat as the video. Awesome edit on this one btw.

[@charleyhoward4594]

Michael is the kind of intellect I'd wish I was.

[@pepefrogic3034]

You are a very modest person indeed, you should aim higher

[@mihaleben6051]

Dear god. Long division curses me again.

[@simonmasters3295]

"Multiply into" ?? What is all that about?

[@gregrice1354]

That is the same , standard, intellectually lazy (disrespectful to listeners) American verbal usage patterns. An annoyance of mine as a citizen, as it reflects so poorly on the speakers, and every American wants to be a credible expert, ya know? Example outside math: "Anymore, the optics on teachers, as regards, math, gives the ask of students, irregardless of if they be in the know." That is university level jargonism, to coin a term. It is embarrassing for masses of Americans who have not been able to pursue higher education. It's a repudiation of care for successful communication, and of respect for listeners.

[@slk2650]

Cool video😊😊

[@la.zanmal.]

I'd be more interested in knowing how to *create* the problem - i.e., figure out how many constraints are necessary, and where, in order to end up with a unique solution.

[@Happy_Abe]

@1:25 why can’t it be a 0? If a 4 digit number starts with a 0 then when subtracting a 3 got number we can end up with a 3 digit number as long as the second number was a smaller 3 digit number.

[@martinplesinger]

There are no leading zeros

[@backwashjoe7864]

Can A really be 0, if the remainder is given as 0? In puzzles like this that I have seen, any letters cannot be equal to any of the given digits, if some are given. Also, if A = 0, it seems like he would’ve written the remainder as A, not 0.

[@huguesbornet1211]

Betraying my age, I attended Dick Feynman’s classes. What was most special in his form of intelligence, and the resulting ability to explain things, may have stemmed from his imagination. Per the definition Einstein gave of intelligence.

[@carlosmirandarocha8905]

I loved the part where Professor Penn multiplies 8 times 84. Loved this puzzle too. Thank you professor Penn.

[@pappaflammyboi5799]

When whatever variant of ChatGPT solves this puzzle, we will truly have AGI. But, I'm not holding my breath...

[@19divide53]

Since Feynmann's puzzle has already been solved, ChatGPT could've just looked up the solution. But even if one varies up the problem, oculdn't a computer (or AI) simply go through all the long division of the same size (number of digits of divisor, quotient, remainder, etc.) and then select the ones that give the required pattern?

[@pappaflammyboi5799]

@@19divide53 Solving is not looking up the solution, neither is it brute force guessing, hence my claim of AGI.

[@alexpotts6520]

​​​@@19divide53 A computer program trained to do maths (eg Wolfram Alpha) could probably do that. But not a natural language processor like ChatGPT. Indeed, there is a sense in which it would be wrong for ChatGPT to answer such a complex question correctly, since it's built to emulate human writing and most humans couldn't answer this question correctly either. The point OP is making is that if a natural language processor can do complex maths then its intelligence is highly general, much like humans, it would certainly be a step closer to AGI than we have right now.

[@ferreiraperaltamiguelangel6873]

Está bien chida la edición

[@stevejohnson1685]

But wait - didn't Feynman's mother know about modular inverses?

[@jimallysonnevado3973]

there's a lot of case work is there a way to solve this without needing so mucb cases

[@OrenLikes]

~ 11:15 why can't A=0? You didn't explain...

[@unknownstoneageman81]

🔥

[@pepefrogic3034]

Yeah many cases, you could just write a line in code to check all milliom possibilities and get it result in a second. Sure instead of going through few dozen cases you go over a few dens of thousands cases more but with computer its faster. You can ask chatgpt to write you that line of code, and voila

[@gp-ht7ug]

I have noticed that your videos are getting rarer and rarer….

[@OrenLikes]

~ 22:00 why did you skip c=4 and c=8?

[@KF888888]

Because the audience would've fallen asleep, lol. He checked all the cases offline though.

[@OrenLikes]

@@KF888888 I almost fell asleep anyways - this is the longest video of his that I watched. The simple stuff in detail, the complex stuff rushed and skipped. I like most of his other videos.

[@OrenLikes]

at 8:00 why write "A (mod 10)" and not, simply, "A", as A is a (single) digit? You probably meant to write "cx (mod 10)" and "cy (mod 10)"... Or, does the "(mod 10)" somehow pertains to the whole expression/line?

[@user-gd9vc3wq2h]

Yes, the (mod10) belongs to the entire line, so something like (173=223)mod10 would be a much clearer notation, or (173)mod10 = (223)mod10. Even clearer is [173]_10 = [223]_10. Here the square brackets together with the subscript 10 indicate that actually the equality is between the images of the ring homomorphism which maps the integers Z to the Ring Z_10 of integers modulo 10.

[@charlesglidden557]

Neat first time I saw his identical twin

[@cdenn016]

Jesus......😮😢

[@OrenLikes]

all the modulo is too confusing to me... do you have a video "modulo for dummies"? :)

[@eartphoze]

Who does your hair in winter

[@omargaber3122]

You are a genius, Michael. I wish you a life full of psychological peace, and it is good board.

[@donaldhoot7741]

Huh?

[@Igorious92]

The solution is OK. But I don't like the task itself. Instead of a single elegant solution, we have many-many cases that we need to check - it's annoying.

[@OrenLikes]

this is the longest, most confusing, with errors and "magic", video of yours that I've seen. all others are clear, coherent, and logical. summary of my notices: --- at 8:00 why write "A (mod 10)" and not, simply, "A", as A is a (single) digit? You probably meant to write "cx (mod 10)" and "cy (mod 10)"... Or, does the "(mod 10)" somehow pertains to the whole expression/line? --- ~ 9:20 yes, 10, 20, 30, or 40 - but you didn't explain why. --- ~ 11:15 why can't A=0? You didn't explain... --- ~ 12:40 if c=2, "x=7. y=2" is not allowed: c=y... --- ~ 18:00 SIX cases! each possible value of c eliminates one ordered pair of (x,y) !!! --- ~ 22:00 why did you skip c=4 and c=8? ---------------- see my brut force(?) javascript code and long division calculator confirmation in another comment.

[@mihaleben6051]

...? This makes no sense. Its not... Its not even a long division problem. There is an A this is multiplying three numbers to me.

[@ToanPham-wr7xe]

😮

[@MH-sf6jz]

Why can’t the 4 digit number at the start be a 3 digit number instead? Circle can be 0 after all.

[@CM63_France]

I think circles can be 0s unless they are at the beginning.

[@theartisticactuary]

At 21:40 you show case 3 works but don't investigate cases 4 to 9. How do we know none of those six cases will get us to a solution?

[@KF888888]

It's not case "3". To avoid putting the audience to sleep (consider that he already solved the modular inverses offline anyway), he checked all the other cases offline and explained that only one of them has a single-digit solution, hence it being case (question mark) so as to not specify the order in which he checked the cases.

[@clownhands]

Clicked for a Feynman story but too much blatherskite in the first two minutes

[@varmituofm]

Great video, but as a warning, do not use colors alone to differentiate variables. You have a pink box and a green box, but to colorblind viewers, you have two boxes.

[@phitsf5475]

I don't like long division

[@just.a.viewer]

did when you made a ads exactly after sponsership video, we pause you and unsubscribe and close this window. bye chip

[@roberttelarket4934]

I'm not exactly a fan of Feynman but this is a really really really great problem!

[@SupGaillac]

I'm curious to know why? Because of a specific thing? (Genuine question)

[@roberttelarket4934]

@@SupGaillac: All this hype for Einstein, him, Tesla, Descartes . . . Cult status. I don't buy it! Gauss, Von Neumann, Turing, Ramanujan yes.

[@theupson]

lots of working by cases, lots of tedious calculation. im not faulting your solution, but your problem selection is a little questionable here.

[@kylecoleman6221]

This is now the second time he's made assumptions, that though true, are not explained. This we his constant looking to some sheet below the camera gives me no credibility that he actually knows the answer and isn't just copying with good editing and presentation. The assumptions are that c cannot be equal to A, or that y cannot be equal to A. Not to mention that earlier he seems afraid to explain what modulo 10 or congruency actually is while he painstakingly slows down to write factors and other basic maths. I'm sure there will be more if I continue the video. This is a pretty, yet frustrating explaination.

[@kylecoleman6221]

I've just realized part of the rules of the problem states that all A's a listed, therefore none of the other variables are listed. I feel dumb not only because of the criticism I made, but also because I spent so long proving that was the case in my own solution and thought he was ignoring the difficulties for convinience... I'm sorry Michael.

[@sumdumbmick]

I mostly have questions about how you're writing it out. why do you have an entirely different notation for every time you're ruling something out? why aren't you just writing paired options as ordered lists in the first place, when you later do write them as ordered lists? and why are you completely rewriting it every single time until finally giving up and just using a vertical ellipsis? clearly you're reading off a script, which means you pre-planned this disaster of a presentation.

[@patricius6378]

If it got you this mad, you should get yourself checked, man

[@nonamenoname6921]

Yer, you should do a better vid on it. Can’t wait.

[@phasm42]

I brute-forced this, and got 26,236 distinct solutions, e.g 135240 ÷ 120 = 1127, and 9277809 ÷ 989 = 9381

[@alexsokolov8009]

Therefore there are 2 conditions: no leading zeros anywhere and each dot is NOT equal to A. However, your examples violate both of them, since first is obviously 6-digit number and second violates first step of division: 989 * 9 = 8901 which doesn't end in 88

[@MichaelPohoreski]

Feynman's original problem _also_ stated that **the dots can NOT be equal to A.** Brute forcing it shows there is only ONE solution: 484 * 7289 = 3527876

[@phasm42]

Ah, I misunderstood the problem.

[@user-fu1iz6ne5n]

เวลาเปลี่ยนได้ ถ้าตำหนวดแก้เวลาอีกนี่โป๊ะแตกแน่555555