Solving the hardest question of a British Mathematical Olympiad

Thanks to Nahian for the suggestion! This is a difficult factorial problem.
British Mathematical Olympiad 2002-2003 problem 5
bmos.ukmt.org.uk/home/bmo1-20...
A Mathematical Olympiad Primer Paperback - 1 Aug. 2011
www.amazon.co.uk/Mathematical...
Math Forum post
mathforum.org/library/drmath/v...
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[@MindYourDecisions]

I am always excited and nervous when I post a video like this! One the one hand, I am thrilled to share a challenging mathematical proof. On the flip side, proofs require perfection and it is very challenging to make a video with no errors. You guys have great attention to detail, so if you see any mistakes, let me know! For major mistakes I will repost a corrected video; for minor mistakes/typos I will leave a note in a comment. Hope you enjoyed this problem!

[@hardestmaths6975]

Hy sir i wana talk to you

[@tentic]

ily! and ur videos!!, they help me learn so much! (I have completed pre calc course and I'm in middle school lol)

[@ilias-4252]

You wanted solutions on the positive integers so you didnt have to check a=0 at the start of the video. Very minor mistake.

[@replicaacliper]

More challenging stuff like this please!

[@jameswan2963]

13a=13+a please help me for the solution

[@sounak5853]

"Pause the video if you would like to give this problem a try" Thanks for your concern, I'll just skip 10 seconds instead.

[@Azimsajid1729]

kzbin.info/www/bejne/omm2YaNjbpabbaM

[@tfdtfdtfd]

No need to pause.....the short interval provided was enough to put together the proof

[@RosidinAli]

😅😅😅

[@wushu1984]

He should have said turn off your device for a week as you attempt to give this problem a try...

[@mydkarthikmecharena9010]

😂

[@samuelking4723]

I understand the steps. What I don’t understand is how *anyone* was able to figure them out in the first place.

[@user-jn4sw3iw4h]

This was indeed more of a "keep watching the video for the answer" than a "keep watching the video to figure out how to solve this" video

[@josmin5384]

Exactly, and imagine if someone was able to solve this type of question in the exam. These are the sort of questions that are said to be "easy" for Math Olympians which just doesn't make any sense.

[@hydrochloricacid2146]

I'd agree that this feels more like randomly stumbling into the solution and less like reasoned mathematical logic...

[@daominhdung5347]

I did participate in some of the same competition, and actually, you can use some kinds of ideas for such problems; but as this is a high-level competition, you must be very creative. With the kind of problem like the one above, the idea is you will try to EXPERIMENT with properties related to integers (for example, divisibility), and with each experiment, you will try to NARROW DOWN the range of value for each number. It is not easy; some experiments will only narrow down very little, others can help a lot. As a result, the competition often takes a lot of time, and the students should be very creative

[@josmin5384]

@@daominhdung5347 Interesting. I will take note of that.

[@maanasagarwal7518]

Many brain cells were lost trying to solve this problem.

[@derekhudgin6660]

Many were lost watching this video

[@jongyon7192p]

yeah people become dumber after trying to solve math problems

[@dey9293]

Brain cells get activated if one is interested in the subject!😀

[@joekerr5418]

F

[@MagruderSpoots]

The carnage was awful, brain cells laying everywhere. Some where still alive and trying to crawl away, others were gasping their last breaths. The lucky ones died quickly.

[@Zavstar]

Contradiction is such a powerful tool

[@fburton8]

No it isn't! (Sorry. :P)

[@prithujsarkar2010]

@@fburton8 xd

[@Trockenshampooleopard]

@@fburton8 Yes, it is!

[@ramone.chacon5084]

@@fburton8 yes, it is x2

[@jongyon7192p]

Assume contradiction is a weak tool. We have a counterexample showing contradiction isn't weak. Thus by contradiction, contradiction is a strong tool.

[@JSSTyger]

After 10 minutes of fuming, I proved that my desk is really, really strong.

[@RocketRaju2212]

@xd hello he was banging his desk all the time

[@jokerjackass8461]

😮mine too

[@Chadicle7]

This. THIS is what I’m talking about! Keep this content up Presh!

[@carterqualls5266]

He ripped off this exact problem from Michael Penn, posted two weeks ago. kzbin.info/www/bejne/gp7HpoaYfbKBjdk

[@hybmnzz2658]

@@carterqualls5266 this is an olympiad problem lmfao there is no trademark. That being said y'all should check out Michael Penn's channel if you want to see many math problems like these!

[@bhaswatibhattacharya4201]

Imagine someone writing "easy" in the comments for this one.

[@rustemtehmezov9494]

Easy!

[@daddahamza6831]

@@rustemtehmezov9494 HAHA

[@justingolden21]

Easy. Just find the upper limit because factorials grow exponentially, and multiplying beyond a certain point will always be greater than adding, and then experiment and realize it's impossible with a or b as 1 or 2, and that 4 is an upper bound, and then that a and b must be 3, and then solve for c. (Except this doesn't work because you don't know if there could be a c that's exactly large enough to compensate for the difference between multiplying and adding a! and b!)

[@tusharkumar3998]

We're simpletons it is not easy for us 😂

[@giuseppebassi7406]

@@justingolden21 i used a different method but with the same trick, the product of two factorials had to be greater than the sum of three at some point

[@tarunrathitra1158]

It's not easy to even understand this proof I can't even start to imagine solving this on a timed test These Olympians are probably superhumans

[@advaykumar9726]

IMO aur inmo Ka level dekhe ho?

[@marvinmartinez3818]

Not so much, it's matter of practice and interest in the subject

[@tosca1883]

not really, after solving a few problems you start to learn new methods and ways to understand math, this problem was really fun, and it wasn't difficult at least for me, but the solution presh used is actually more complicated that the one i had, so its just practice

[@advaykumar9726]

@@akshaj7011 what?

[@sweetpotato_35]

Yeah if you had in mind divisibility this would have been an easy problem. It is just different kind of maths that you aren't used to.

[@GravitationalRealism]

Great solution! One small enhancement I noticed - for the section starting at <a href="#" class="seekto" data-time="281">4:41</a> to prove a = b, it is quicker to divide the equation by b! (rather than a!) and notice a! / b! is an integer iff a = b (and everything else are integers).

[@schungx]

Yes, symmetry arguments are so powerful. a!/b! is an integer iff a >= b. However, since the equation is symmetric wrt swapping of a and b, the only solution is a=b. Once you get that, it is a no-brainer. Take n = a!. n^2 = 2n + c!. n^2 - 2n - c! = 0. Use the quadratic formula to find that n = a! = sqrt(c! + 1) + 1.

[@thexoxob9448]

Your way still works even if you divide by a!.

[@alexeyrb1807]

Nice problem! Some alternative ways that I found when solved it first: Rewrite the equation as (a! - 1)(b! - 1) = c! + 1, check cases like 0 and 1 and then look at (a! - 1) = (c! + 1) / (b! - 1). This allows to establish that a >= 3 and c > b. For case a = b: solve a quadratic equation in terms of a!. It gives a! = 1 + sqrt(1 + c!). This then gives a = 3, c = 4, but also an upper bound on c, since if c can't be cubic or higher expression in terms of a (eliminating some low number cases first). But it's sloppy, your proof on c

[@Bry10022]

If you need the not equal to, less than or equal to, or greater than or equal to symbols, Here they are: ≠ ≤ ≥

[@json6573]

this is what i thought too.

[@romanshevtsiv3526]

Yeah did it more or less the same way. Looks faster and cleaner for me than the original one.

[@heysqualito]

@@romanshevtsiv3526 I also thought of the same

[@anshumanpanigrahi7323]

Correct eqn will be (a! - 1)(b! - 1)=c! + 2

[@mohdhassan2677]

Now I know why exclamation mark is used for factorial!

[@dyosorkan]

As a french viewer,your videos make me practice both english and maths... And how to think outside the box. Genius

[@Harriett2423]

Oui, moi aussi

[@WOLLEY104]

I'm a GCSE student and these kinds of videos are always fun to watch. I love maths but never understood how people even know where to start with these questions. I lost track of what was going on like a minute into the video.

[@vontabi]

The last time I was this early, the Gougu theorem was still called the Pythagoras theorem

[@ericzhu6620]

you mean Gougu? It's Chinese way to call it, but I believe the Europeans still call it Pythagoras theorem, btw I live in Portugal.

[@EXOPLANETnews]

Hey i have an interesting channel about space science and mysteries if ur curious about it do visit my channel once pls 🙏 🙏🙏.

[@Grizzly01]

@@ericzhu6620 It's a joke, Eric. Varun was making a joke. Watch a fair selection of Presh's previous geometry-based videos, read the comments sections of those videos, and hopefully it should become clear.

[@yosuakrisnando]

@@Grizzly01 , he might know it before you "advise"...

[@chanlokhim3182]

I am from Hong Kong, we call that Pythagoras theorem too tho, but we also know the name of GouGu Theorem

[@matthewcarlyon5133]

Wow, great video Presh. I especially loved the section of proof at <a href="#" class="seekto" data-time="300">5:00</a>. Brilliant maths.

[@VinaySingh-jm8iw]

Great proof, but how would one know they needed to prove that b

[@typingcat]

Probably with lots of previous experiences with similar "proof" questions that used this kind of boundary checking/contradiction tests.

[@bigboboluager5003]

You don't know that. The way he presents this is not the way you start thinking about the problem. I started out, trying a few easy values like 1 and 2 for a, b and c. You immediatly get a contradiction so you can keep that in the back of your mind. Sometimes this will come in helpfull later on and sometimes it wont. If you have some practice you will learn more tricks, that you can just kind of throw at the problem and see what happens.

[@thunderstorm8653]

Experience

[@logannasty3240]

Constructing these problems, that’s the truly magnificent feat! I imagine it takes weeks or months to come up with such a beauty. Then this is only one of many problems in an Olympiad. So much intellect invested into competitive maths...

[@mj9765]

This video is a gem! Thanks Presh

[@pushkarshukla1229]

Wonderful!!!, it's a treat to see you go through the proof. Keep it up !!

[@danyd1583]

You are just amazing. You must have spent days on this video... Thanks for this awesome explanation. Wondering how people solved the exercise in minutes during the olympiade lol

[@pokmaster4475]

This is one of the best math proofs, I have seen

[@JuanMataCFC]

actually amazing proof, i think u covered every possibility. i don't think i've ever seen contradiction used better than this before!

[@eytansuchard8640]

The method of negation is seen over and over again in integer equations. Thank you for this enlightening video. While performing other tasks I watched your video and it reminded me that sometimes in order to think we need to stop doing what we are used to do.

[@dey9293]

Once you eliminate the impossible, whatever remains must be the truth.- Sherlock Holmes😀 This is the only theorem that solves the above problem.

[@honyi1995]

When you said "as always" at the end, I immediately said in my head: "Stay awesome, bros!"

[@achaemenid]

I luv ur videos! Keep up the great work!

[@ilmionomenonloso]

Wow, this was really awesome! Great job!

[@ibrahimmalki150]

I never would’ve thought combining elementary concepts such as divisibility with basic middle school stuff like factorial and algebra would be so complicated

[@marvinmartinez3818]

That's math, going from the simple to the complex

[@meenashetty9782]

Halfway through i stopped listening and started reading comments.

[@targetiitbcse1761]

same😂

[@justanonverifiedyoutubechannel]

Sugoi!! That was awesome 👌 I had a lot of fun watching this one too .....keep up the good work

[@rogersledz6793]

Thank you so much for uploading this video. It is helping me to get through the pandemic!

[@helo3827]

You are my favorite youtuber, I never miss any of your videos

[@niejocelyn1635]

Cool

[@advaykumar9726]

Me too as I have turned on the bell

[@NaHBrO733]

nice!! its good to have some really challenging problema once in a while👍🏻

[@DavidAult]

Very nicely done! Thank you for that!

[@chunli1225]

Nice video Presh. I enjoyed it so much. Please share more videos like this about challenging problems.

[@billy.7113]

It took me longer than 11 mins to understand the whole proof.

[@itsjustanapple5452]

For all or us.

[@harris6498]

😂😂😂

[@arianrolex6456]

Bill Y. Dude I got no idea why I’m watching this I’m 13 what the heck is he talkin about . Why am I watching this I have math paper 2 tomorrow...

[@sadkritx6200]

@@arianrolex6456 don't worry bro. Come back to this video in a few years, maybe it will become easier as you will probably know more about mathematics then than you do now. I noticed this many times. When I was 13/14 many math videos seemed too hard to even understand. But now when I'm 17 the impossible to understand videos have gone down but not become 0 lol😅

[@timnauwelaers6876]

I still don't get it :(

[@yhchong4270]

I have watched more than 3 times to follow the logic. Thanks for brushing my mind.

[@user-wp1uw8fv6y]

Highly Impressive Solution!

[@kalpanarms9597]

Some people like these videos before even watching them, that is how much we love your videos! :) nice video btw....

[@gaganupadhyay4978]

Nope that's just how KZbin algorithm works.

[@ahasabamirraiyan8031]

Yeah

[@laavanayaggarwal6671]

‌‌

[@gordoncharles741]

No, that's just blind stupidity.

[@EXOPLANETnews]

Hey i have an interesting channel about space science and mysteries if ur curious about it do visit my channel once pls 🙏 🙏🙏.

[@frankie391]

Wow, it's super duper rare to encounter a Factorial problem like this. Thank you Presh Tall Walker, what an amazing solution. And also, can someone tells me how can we think of the ideas at the beginning of the solution? Like how can Presh know we need to prove b=a or c

[@GeorgeFoot]

Olympiad problems tend to be open like this, you have to creatively think of ways to narrow the field. Rather than trying to find the different numbers, maybe we can find three numbers all the same? Or prove that it's impossible. What about a b or c being really low, like 0, 1, 2? You just have to explore things like this and watch out for any good leads. Also know that where factorials are concerned it's almost always something to do with divisibility or primeness, as the fact that all factorials divide each other is very powerful.

[@leonais1]

I'd consider that to be a summary rather than insight. Factorials can be hard to deal with, however you can work with values such as n!/m! so cases where n>m, n=m are immediately interesting.

[@kevinm1317]

A lot of it boils down to experience and intuition, which only comes from practice - its often difficult to explain why exactly you try something. But I'll try my best to explain here. It's very natural to assume a=3, you might realize that 2 and 3 must divide b!, so 2 and 3 must divide the RHS as well. If b and c are much larger than a, its instantly clear this is impossible, as the RHS would be odd+even+even. As such, exactly one of b or c must be equal to a or a+1, from which we get c>=a+2 (as c>b). It immediately follows that b=a or a+1, and from there its quite natural to try to see which one it is/can be. If you try b=a, then of course you'll get the solution mentioned, and if you try b=a+1 then you'll get the same equation as in the video.

[@loglnlg]

He knew it because he solved it before making a video When I was solving this problem at that moment of progression (getting frormula at 8:06) I was stuck on this general problem, so I divided them into subproblems where c=a, c=a+1, c=a+2, etc. and then I firstly thought about divisibility by 4 and proved that if c>=a+4 then there is no solutions, so I got only 4 little subproblems (instead of infinity of them) and each of them could be solved more easily than the whole problem. For me it is helpful to divide problems into cases (if we could do it, sometimes it is just could't be done), not even to solve whole problem by them, but to see particular examples and work with something little at the beginning and maybe see something more general in them

[@PrashantSharma-nw6gc]

In such types of problem you must have to make some assumptions and be dare to do so because if they are wrong and your method is right, the great genius mathematics itself told you their errors. May be there some contradictions occurs by which you are able to say that the assumption is wrong.

[@suyashgupta4019]

Your best problem in a long time. Keep it up, presh :)

[@earthbind83]

This was a beautiful proof. I had fun seeing all the pieces fall into place.

[@benheideveld4617]

And without using the Gougu theorem...

[@Macieks300]

Nice puzzle. I solved it kind of similarly but when I got to a

[@rorschachkovacs7867]

I did the same way that you do, and eventually you can get the conclusion that n=0 and m=1, just aplying that n! is even if n is equal or greater than 2

[@latenter1428571]

There is an easier way to show a contradiction. Once a < b < c, then b! divides a!*b!, b! divides b! and b! divides c!. But b! cannot divide a! Then c < b (that's not our case) or a = b.

[@moinfahad]

Love ur videos bro, keep up the good work.

[@egillandersson1780]

One of your best proof ! Thank you

[@Timotheeee1]

how does anyone even come up with this

[@typingcat]

Drugs. Lots of drugs.

[@nelser1160]

For real, how would someone know what to prove at the start?

[@user-qs3ih3ll5f]

miracle, I guess.

[@rustemtehmezov9494]

@@nelser1160 So, when you solve so much of questions like these, it's easy to check these metods and get bounds.

[@nelser1160]

@@rustemtehmezov9494 I wonder how many I have to solve to get that lol

[@gfasoulas]

Whenever I watch a video from this KZbin channel my brain is twisted.

[@laavanayaggarwal6671]

small brain small pp

[@EXOPLANETnews]

Hey i have an interesting channel about space science and mysteries if ur curious about it do visit my channel once pls 🙏 🙏🙏.

[@Mrpopperer]

nah smol brain usually results in big pp xD

[@ABHISHEK-xm5mt]

@@Mrpopperer Power of belief, Dare to believe 👇 kzbin.info/www/bejne/r5jGXnWVhK2tb9k

[@piman9280]

Whenever *I* watch, my brain wants to Go to goo!

[@pruthvipatel8720]

Thank you for this video! Really enjoyed watching it :-)

[@otakurocklee]

This is one of the coolest problems posted on the channel.

[@zapzordx5440]

3rd time I solved any of your questions... I'm a 6th grader(your education system might treat 6th grade differently) and solved it and it made my day as I got it right :) Well I just used trial and error method...I know they asked for a real explanation but I'm still happy to get the answer 3,3,4. 😁

[@vanditseksaria5897]

the main crisp of the problem is to show that there are no other solutions

[@afbanales]

Your definition of wonderful is WAY different than my definition of wonderful!

[@ABHISHEK-xm5mt]

Power of belief, Dare to believe 👇 kzbin.info/www/bejne/r5jGXnWVhK2tb9k

[@alekosbantias3962]

I didn't even try. I had to pause the video several times to digest the information. Wonderful indeed

[@GioFaber]

Bravo Presh!

[@williamleinonen4888]

yeah, right. Glad I gave up after a couple minutes.

[@shaiavraham2910]

"Find all solutions..." There is only one solution. lol 🤣

[@piman9280]

.... and *the* solution can be found after a few seconds of "guess and check!" 0! = 1, 1! = 1, 2! = 2, 3! = 6, 4! = 24. Then 6 x 6 = 36 = 6 + 6 + 24.

[@RGP_Maths]

So? You don't know there's only 1 solution until you prove this. There was no need for the question to tip you off as to how many solutions you are looking for.

[@pdpgrgn]

@piman that doesn't prove that there aren't other solutions

[@piman9280]

@@pdpgrgn - who said it did? I merely commented on the fact that the only solution is very simple to find. The very fact that the "problem" is so shamelessly contrived speaks volumes about the "contriver."

[@GRBtutorials]

But you also need to prove there's only one.

[@manswind3417]

That was a treat to solve and watch! Indeed an Olympiad Number-Theory problem (though not as insane as the present ones by virtue of its age...)

[@Below10IQ]

It's amazing how people can think of these solutions. And it's even more amazing that a solution like this was to exist.

[@williamadams137]

I have another number theory problem that I’ve been stuck on. Find all integers u such that u³ + 2u + 1 is a perfect square. (You must have a complete solution.)

[@ananthchopra106]

Only one integer that is 1

[@williamadams137]

Ananth Chopra nope.

[@williamadams137]

Actually it’s much harder than it looks.

[@yurenchu]

u = 0, u = 1 and u = 8 are solutions, but I don't know if those are all solutions. There are no negative integer values of u that are solutions (because it can be easily shown that for any u ≤-1, f(u) = u³+2u+1 ≤ -2 which is negative and hence cannot be a perfect square). There are also no other positive integer values of u ≤ 32 that are solutions (confirmed simply by hand). However, I haven't been able to prove that there are no solutions where u > 32 . I can prove though that (u+1) cannot be a multiple of 4, nor a multiple of any of the prime numbers 5, 7, 13, 23, 29, 31, 37, 47, 53, 61, 71, 79, ... (the rest of the list of prime numbers probably goes on infinitely but is apparently unpredictable though).

[@justinmay2295]

I was so close to the answer

[@harshitvarma7867]

I bow down to you my lord.

[@harish6787]

Good

[@TechToppers]

How close?

[@carlosmspk]

What did you do?

[@drsonaligupta75]

Good content after a really long long time

[@indrashisbhattacharyya7854]

This made me cry

[@nigglefiggle]

It’s not incredible. It’s *magnificent*

[@advaykumar9726]

I thought so

[@maiphen4537]

*french spy intensifies*

[@johnjordan3552]

The funny thing is the question it self isn't really hard, just need to keep track of your logic string yet that seems to be the hard part

[@jayatheerthasg8735]

Hey... That was really MINDBLOWING...

[@adityamukhopadhyay6803]

This problem is one of the most interesting ones I have ever seen.

[@QuantumHistorian]

There's a much more elegant solution using modulo arithmetic. Honestly, when dealing with integer problems, modulo arithmetic should be the first port of call. Some of it overlaps with the video a lot so I'll leave the details out, but overall its a much tidier presentation and doesn't require so much guesswork about what to prove next. Really, there are only three steps here. As a prelude, consider x! mod y!. If x >= y, then y! | x! and so x! mod y! = 0. If x < y, then the modulo base is larger than the number, and so x! mod y! = x!. First, assume as before that b >= a. Taking mod b! of both sides, we get: a! b! mod b! = a! mod b! + b! mod b! + c! mod b! 0 mod b! = a! mod b! + 0 + c! mod b! (*) There are two cases to consider. Firstly, a < b. That implies 0 = a! + c! mod b!, and thus that the last term is non-zero, hence that c < b, which in turn implies that c! = b! - a!. Substituting that back into the starting equation we have a! b! = 2 b! which clearly has no solution. Thus, we move on to the second case, where a = b, leading to a! mod b! = 0 and so (*) tells us that c! mod b! = 0 and therefore that c >= b. The case c = b can be discarded by noting that it would lead to the equation a! a! = 3 a!. Thus, we have now restricted ourselves to the cases a = b < c. Going back to the original equation, we have a! a! = 2 a! + c! a! (a! - 2) = c! a! - 2 = (a+1)(a+2) ... (c-1) c (**) Where we have divided out a! in the last line. Taking mod 3 of both sides (and assuming a >= 3, the smaller cases can be checked by hand) we have a! mod 3 + (-2) mod 3 = (a+1)(a+2) ... (c-1) c mod 3 0 + 1 = (a+1)(a+2) ... (c-1) c mod 3 If the RHS contains 3 or more consecutive terms, then one of them will be a multiple of 3 and so the modulus will be 0. If it has 2 consecutive terms, none of them can be a multiple of 3 so we have (3n+1)(3n+2) mod 3 = 2, so this is also impossible. Thus, the RHS can only contain one term which is exactly 1 above a multiple of 3. This tells us that c = a + 1 and a = 3n. Substituting back into (**) we have (3n)! - 2 = 3n + 1 (3n)! - 3n = 3 (3n)! mod (3n) - 3n mod (3n) = 3 mod (3n) 0 = 3 mod (3n) n = 1 Thus (a, b, c) = (3, 3, 4) is the only possible solution. Evaluating it shows that it does indeed work. Note that we don't have to deal with the complexities of n=0 because we ruled out a=0, 1, 2 earlier. I know that this is similar in many details to the video, but a much more streamlined chain of thoughts IMO. Remember, when dealing with integer problems, modulo is your friend. _Especially_ with factorials flying around. As an aside, there's also a proof of this result using Wilson's Theorem, but that's unnecessarily complicated here.

[@Authen194]

Bro what you do? A teacher😀

[@QuantumHistorian]

@@Authen194 haha, no, but I know how to write up proofs in a detailed and understandable way

[@Authen194]

@@QuantumHistorian can you help me

[@ninad7694]

dude you make such good videos, you inspired me to make a youtube channel! keep up the good work!

[@ihaveamathquestion7377]

After you derive at a

[@gavin.d.m]

This channel reinvigorates my interest in maths. Generally I can follow the solution but this one lost me. But still, it's fascinating to watch. 👏😄

[@rubyjha4298]

I'm done for the day!!!!

[@typingcat]

That's what she said.

[@kritika1413]

Thumbnail: solve Me : NO Because I can't 😂😂😂

[@JohnDlugosz]

The numbers are small, so trial and error will take less than ten minutes.

[@kaisarsihaloho975]

Thumbnail : solve Me : Yes Brain : Can't Me : No Presh : HAH! I KNEW IT!

[@leung77]

incredibly elegant proof for this sound simple question

[@manglamshukla8395]

We also like to your video before watching, such a powerful voice and technology

[@pavelavietor1]

hello I give it a try and fail 🙂 🙃. thanks for sharing 👍 saludos

[@abdulkadiralat5197]

Could you add Turkish subtitles if I would like?

[@kienthanhle6230]

You can practice English listening bruh

[@kienthanhle6230]

@Emma Beckman Vietnamese and English

[@cowabunka]

i find it amazing that people can derive this prove out of thin air I have trouble following it even as I am watching the solution and I wonder how can someone take the necessary steps and making the correct deduction without knowing what the solution / direction to the solution is well done for having this problem, made my day

[@goebelmasse]

That's a great and incredible hard problem. I want more of this stuff… 😉

[@hamster8706]

Fun fact: A prime number in russian is "простое число", which can mean simple number too,therefore, a prime number is the opposite of a complex number

[@yurenchu]

_Prime number_ is the "opposite" of _composite number_ . A composite number is a (positive) integer that can be written as the product of two (positive) integers that are both not equal to 1 ; in other words, a composite number is _composed_ as the product of two (non-unit) integers. (For example, 35 is a composite number because 35 = 5 * 7 . Another example: 9 is a composite number because 9 = 3 * 3 .) The word _complex_ also means "composite"/"composed", but in a different way: a complex number is composed as the sum of a real number and an imaginary number. (Since the real number may be 0 and/or the imaginary number may be 0i, a complex number may also have just a real value, or just an imaginary value.)

[@laavanayaggarwal6671]

*Fresh tall walker*

[@jagadishyerragunta8230]

Love your way man

[@sharmashivam1054]

Purely Impressive bro 🔥🔥

[@facilvenir]

It was hard enough to follow your explanation! Thank God I didn't even try to solve it myself!

[@yashwantsingh6703]

before this video: presh's problem are easier nowadays after this video:🤐

[@vacuumcarexpo]

This is literally incredible❗

[@ABHISHEK-xm5mt]

Power of belief, Dare to believe 👇 kzbin.info/www/bejne/r5jGXnWVhK2tb9k

[@matthieu5096]

I cannot believe it! So interesting and logical prove!

[@PSquared-oo7vq]

Great video, but I have never, ever gotten more ads in an 11 minute video than this one. At least seven, two of which were > 3 minutes long, and one less than 15s into the video (and that's after a double pre-roll). If the creator set the video up for this many ads, then boo you. If not, then KZbin is screwing with you (more than usual). I try to support creators, but... adblockers are calling.

[@manswind3417]

Well, I fortunately didn't encounter any ads at all, and in general I don't get as many ads as you say, despite using the regular KZbin version, so evidently the creator hasn't set the vid up for these many ads - it's youtube's doings Btw, are you on aops, coz your the guy with your username there is a 13 yo IMO medallist...

[@bangchui4465]

<a href="#" class="seekto" data-time="482">8:2</a>(1+3) = 16 or 1

[@afuu794]

8:2 (1+3) = 4 (4) = 4 × 4 = 16

[@perishedgalaxy1732]

It will be 16:1 as it is given in ratio format.

[@faridmk3707]

Classic discrete math exam problem. Love it.

[@aryanshrajsaxena6961]

What an amazing evening problem!!

[@armacham]

<a href="#" class="seekto" data-time="108">1:48</a> "suppose A is either zero or one" -- why would I suppose that A could be equal to zero, when the problem clearly states that the values of A, B, and C are limited to positive integers?

[@theqwertyman9309]

For the sake of argument, I guess?

[@yurenchu]

In some languages/cultures/education systems, 0 is considered positive (as well as negative). Presh was just being rigourous.

[@vikaspandey3537]

Got it in first look

[@jonasdaverio9369]

Obviously 😂

[@CraigNull]

Once you establish 3

[@Name-is2bp]

Hello Presh, can you please add in the description of your videos the requirements of what math concepts we need to be familiar with to solve each problem??

[@gojosan1376]

Finally Presh bro is back with his god level stuff.. And yes I missed gogu 😂😂

[@piman9280]

*Nobody* misses "gogu!"

[@Tiqerboy]

It's Gougu and I knew this would be the longest Presh video where that mighty word wasn't spoken without even watching it!

[@jasonzheng5839]

This problems looks like something along the lines of “if Paul bought 10 cookies and gave his friend 2, how many kilos of meth do I have in my pocket”

[@kukulroukul4698]

:)))

[@abysslight2490]

You could’ve proven the fact that a=b much easier if you divided both sides of your equation by b! instead of a! at <a href="#" class="seekto" data-time="293">4:53</a>. This is because we already knew that a is less than or equal to b, so the case where they are not equal would mean aa (as previously shown), c!/a! is an integer, and 1 is obviously an integer, but since b>a, a!/b! is not an integer, thus a! isn’t an integer. And that’s a contradiction. It’s a lot simpler and quicker way to get to the same conclusion.

[@bobjordan5231]

Well done Presh. Not a verbal pause or miscue, which shows you know this material ice cold. Impressive... most impressive...

[@em_the_bee]

Or that he has prepared a script?

[@bobjordan5231]

@@em_the_bee Perhaps, even reading something with no pause or miscue is impressive!