France - Math Olympiad Question | An Algebraic Expression | You should be able to solve this!

[@TheEmanoeljr]

Easy. a=2021, c=2020 and b=0.

[@aidan-ator7844]

You missed half the solutions genius

[@TheEmanoeljr]

​@@aidan-ator7844whatever. Its correct! Uhuu!

[@aidan-ator7844]

@@TheEmanoeljr yes but you only have the intuitive solution. There are others.

[@aidan-ator7844]

@ferrel9715 without a doubt. Intuition is only one part of thinking that functions best in conjunction with others.

[@bharati1923]

Good observation, bcz that’s what matter. As computers are there to solve/ evaluate if required…

[@ShawnPitman]

Look at 4:03. This only works if youre working with integers. The single-step assumption that xy = 1 only has two answers is only valid in integers. Counter-example: x = 1/2 and y = 2. This is also 1. (In this case x = (1-b) and y = (a-c)).

[@ShawnPitman]

I'm an idiot. The problem statement says "integer solutions".

[@atrib_]

The video did not stress this point, which is crucial. 3 variables, 2 equations leads to infinite(?) solutions. Even with the additional constraint of integer values, we got 2 solutions

[@muhammadabujabal9387]

​@ShawnPitman But each part was the combination of 2 numbers , so you will get only integers if you add or subtract integers, I think it's the only step that seems need looking the rest is simple algebra could be done in 1 min

[@hickiwawa]

Same here. The thumbnail left that out. I was going to solve before watching the video, only to quickly notice there are infinite solutions.

[@lipers87]

yes, if u name "ab + c = 2020" as X and "a + bc = 2021" as Y, them we can do Y - X we get: (b - 1) • (c - a) = 1. So b can be 0 or 2, but c - a can be anything like (2, 1); (3, 2); (-1, 0); (-2, -3); ...

[@brianbutton6346]

I liked the fact that someone with a nice voice and clear handwriting can provide audio-visuals for an instructional video. I have neither.

[@satvikakshintala8030]

Bro but this isn’t a valid question and neither her answer a valid one Since there are 3 variable you can have multiple answers for eg A = 2021 B = 0 C = 2020

[@eliatgnu]

This is a special case of a much more general problem: ab + c = A (1) a + bc = A + 1 (2) (2)-(1) gives b=1-1/(a-c). Let's replace the variable c by c'=a-c and work with c' from now on (still have 3 independent variables, but more convenient, c can be recovered by c=a-c'.). So b=1-1/c' (3) Substituting this into (1) gives a(1-1/c')+a-c'=A, which yields a=(A+c')c'/(2c'-1) (4). Given an arbitrary A, one therefore only needs to specify c' to get a general solution of a, b and c. For integer solutions, c' must also be an integer, and from (3), b can only be an integer when c'=1 or -1. For c'=1: it follows that a=A+1, c=A, b=0, all integer as long as A is integer. For c'=-1: it follows that a=(A-1)/3, b=2, c=a-c'=(A+2)/3. In order for a and c to be integer for c'=-1, A has to be a multiple of 3 plus 1, i.e., A=3n+1 with arbitrary integer n, which then gives a=n, and c=n+1. The original problem is when n=673. Obviously, there are infinitely many 'problems' with the same condition provided that A=3n+1 with arbitrary integer n. Without restricting to integers, (3) and (4) constitute the general solution for arbitrary A.

[@dorgamahmad6033]

Also 2×0.5=1 and 0.5×2=1

[@andrewcheung7538]

The real answer should be a curve in 3 D diagram... [n, f(n), f2(n)], should you think that you haven't given out a proper answer..

[@hoagy_ytfc]

@@dorgamahmad6033 Except that the problem stated in the first few seconds of the video was "find the integer solutions".

[@adivoma7]

Basically (1-b)(a-c) = 1 is true only when both (1-b) and (a-c) are 1 or both are -1. These are the only integral solutions. The system having infinite solution is being eliminated by the fact a, b, c are integers. This has only 2 sets of solutions.

[@ctsirkass]

@@dorgamahmad6033 we are looking only for INTEGER solution. Read the question 3 times before solving (classic teacher's advice from the elementary school)

[@muskyoxes]

Somewhere, someone's discovered an amazing problem featuring the number 2374 and is just waiting to reach that year to release it

[@donelkorantengbrown9439]

I don't think that they would still be alive

[@mikesmovingimages]

A lot of commenters are boss stating the simple 0 solution or observing that there are infinite solutions, until they finally read the instructions! Integers only! And there is more than one solution.

[@masterthnag105]

I solved stuff like this back in highschool. Man life has dragged me down. I need to take math classes again.

[@kathrynstemler6331]

Right? I remember a time when I could try to get my head around this but I’ve forgotten so much.

[@Crom1980]

Interesting that many people can't read the first four words in red.

[@Mr_AbdulRehman]

Indeed. It's painful, too painful reading comments.

[@arteffectsshivam3660]

If u see , there are infinie solutions As 1 can be written 2* 1/2 and infinitely many more 5:41

[@yennhinguyen6746]

The calculation in the video is missing a solution. At 3:52 we cannot assume that a x b = 1 then a=b=+-1 like in the video, there are infinite cases that two numbers multiply each other can equal to 1, for instance a x 1/a = 1 as well. According to my calculations there are two set of solutions, one just like in the video: a = 2021 ; b = 0 ; c = 2020 And the other set is: a = 673 ; b = 2 ; c = 674 You can test my solutions by replacing it to the given equation in the beginning of the video. Ty!

[@the-mathwizard]

Watch the question my brother, it states only integer solutions are allowed

[@felixiduh5286]

a = 2021 b=0 c= 2020

[@yovtobe]

I think I got it!!! Once I opened the video and saw the find all integer solutions I got two answers that both work. I still haven't watched but feel a nice sense of accomplishment!

[@_captain_yt]

2:42 signs work in multiplication? How?

[@Dauti_]

I don't follow what you're doing at 3:21. If you ignore the a(1-b) term, it seems to me you're saying that +c(b-1) is equivalent to -c(1-b), which seems wrong to me, coz if I assign random values to c and b, say c=7 and b=3, then +c(b-1) becomes +7(3-1), which is equal to +14, whereas -c(1-b) becomes -7(1-3), which is equal to +14. OK, you are right, but I don't follow.😭😭😭

[@AhirZamanSairi]

What is the brand of the pen, I love how thin the lines are.

[@Moharidy]

The RHS equals 1 can be expanded as you did,but also can be expanded as multiplication of I and 1/I, which gives infinite number of solutions

[@Mike-rx5uu]

If a, b, and c are all integers (given in the problem statement), there is no way to generate a fraction of the form you're suggesting with 1-b or a-c.

[@lamttl]

Nice use of factorization concepts, well done

[@AYTM1200]

I'm so proud I was able to do this in my head. I even used a different method for the beginning part. Step 1: ab + c = 2020 Step 2: a + bc = 2021 a + bc = 2020 + 1 a + bc - 1 = 2020 You equate the 2 to get. Step 3: ab + c = a + bc - 1 (ab - a) + (c - bc) = 1 a(b - 1) - c(b - 1) = - 1 (a - c)(b - 1) = - 1 Then you solve accordingly like the rest of the video.

[@balkansenjoyer]

In the last step, the two term just need to be reciprocals of eachother and if you get an integer for all values for example (1-b) = 1/2 it is a solution

[@Abhilash26596]

In 4:02, why do you say there is only 2 solutions (i.e., the (1-b) and (a-c) are only 1 or negative 1).. There is a third solution of - (1-b) = 1/(a-c) as well

[@EhsanZia-Academi]

Thanks for the solution and a great explanation.😊

[@yogeshchaure3386]

2 equation and 3 variable so put 1 variable 0. So only one way we can easily satisfy equations is put b=0 then a=2021 and c=2020

[@cm5754]

Easy. c=0, a = 2021, b=2020/2021. With two equations and three variables, we are going to have an infinite number of solutions. We can usually pick one variable to be anything we want. (Edit: I didn’t realize, until after I wrote this that the thumbnail is different from the actual problem, which is misleading. The thumbnail does not say integer solutions.)

[@richardleveson6467]

Thanks! Delightful presentation of a clever little problem.

[@rnseby]

Side note: I gave both Bing Chat and Google Bard this problem. While Bing Chat gave a great step by step, it got the wrong answers. Google Bard got the same answers as the video. Bing Chat: a = 2019.49 or -1919.49 b = 0.51 or 3940.49 c = 2019.98 or 22.02 The first set of answers seemed to be a rounding error. But the second set was completely off. My comment has nothing to do with the video, I just find it interesting how far off these AI are still. Google Bard got this one right but I've had times where Bing Chat gets it right and Google Bard gets it wrong as well. I've found if I ask both the same question, I'll either get a good answer or funny one.

[@xxxBradTxxx]

I can wait until chat bots can do math properly. I kinda wish OpenAI and Microsoft would block those questions for now until they figure out how to make them accurate.

[@santiagoferrari1973]

Ask Grok.

[@fongwinson5017]

So far they are Language Models with some math capability. "Intelligent" in some areas only. SImilarily there should be very powerful math AI that could not have a nice "chat" with you.

[@andrewcrayton2424]

I watched just to make sure that my answer was right. I saw the problem while scrolling here on youtube and took only a few moments before I saw the solution. All of the extra steps were entirely unnecessary, but probably can help folks who aren't able to easily see the answers to math problems like this one.

[@huynhaibac2020]

What grade is this math problem for? In my country, I encountered this problem when I was in 9th grade

[@LightWaveLtd]

How come this video appear in my suggestion, it looks like magic to me

[@jan.kowalski]

Well, you do not need any calculations for a logic solution: just observe that b cancels a in first equation and also cancels c in second, so if b equals 0 then c is 2020 and a is 2021. 5 second solution.

[@avalagum7957]

My generic solution for problems with integer solutions: convert the problem to A * B = a small constant. As A, B are integer, we can find all the possible values of (A, B). So, this problem gives a(1 - b) + c(b - 1) = 2021 - 2020 = 1, so (a - c = 1 and 1 - b = 1) or (a - c = -1 and 1 - b = -1)

[@proman9297]

That's nowhere near to a maths Olympiad question

[@coolfreaks68]

ab + c = 2020 and a + bc = 2021 => *(c-a)(b-1) = 1.* Since, a, b and c are natural numbers, so the only way the above product( written in *bold font* ) can be 1, is when b-1=1 and c-a=1, which implies b = 2 and c = a+1. Putting b = 2 and c = a+1 in ab + c = 2020, we get a = 673 and c = 674.

[@eddie31415]

not natural numbers, but integers

[@ctsirkass]

We are talking about integers, not natural numbers, so you can get a product of 1 by (-1)*(-1) so you missed one solution.

[@danvalean2217]

An easier step at the 2a+c/a+2c part would be just adding them. 3a+3c=4041 a+c=1337 Then you take the a-c=-1 2a=1336 a=673, then c=674 But that I guess it depends on the education system or on your mood. Nice solution.

[@gardenjoy5223]

? Where did you leave b?

[@xyntas]

@@gardenjoy5223tell me you didn't watch the video without saying you didn't watch the video

[@tcz1757]

What about the a = 2021, b = 0, c = 2022 solution?

[@gardenjoy5223]

@@xyntas Actually I did watch it. What happened to the b? Was I distracted at that bit? Can you give me the time, where one can leave b out to find the answers?

[@danieltatar7575]

@@gardenjoy5223 6:32

[@mauriciogerhardt3209]

Why are you only using integer solutions?

[@slavasavitsky]

A pretty clear explanation except that it is not proven at 3:45 that 1 * 1 and (-1) * (-1) are indeed the only products of two integers that yield 1.

[@toveirenestrand3547]

b(c - a) - (c - a) = 1 = (c - a)(b - 1): 1) b - 1 = 1 and c - a = 1, so b = 2 and a = 673 and c = 674; or 2), b - 1 = -1 and c - a = -1, so b = 0, and a = 2021 and c = 2020.

[@crcurran]

This is what I thought just glancing at it. Seems to make sense to me that b = 0, a = 2021 and c = 2020.

[@deathstarresident]

Just like in video, instead subtracting on equation from the other - you can also add one equation to other and get a simplified product for (b+1)(a+c) =4041. It’s pretty much a very straightforward problem

[@GauravSingh-gd1yj]

This equation is not at all straightforward,the easiest soln was given in the video

[@dudedujmovic6562]

@@GauravSingh-gd1yj Agree, the easiest is shown. It is a nice problem for basic algebra astute.

[@geralynpinto5971]

Love all your explanations. They are so clear and easy to understand

[@QUABLEDISTOCFICKLEPO]

Not to me.

[@popliceanumihai9653]

@@QUABLEDISTOCFICKLEPO 😂

[@PreservationEnthusiast]

Not clear.... she says a into b when she means a x b. a into b is b/a

[@christianherrmann]

She explained every little step, but not the most crucial one, why none of the factors can be a fraction and hence only can be equal to +/- 1. (because a, b, c are to be integer, the factors (1-b) and also (a-c) are always integer.)

[@gardenjoy5223]

@@christianherrmann For someone without foreknowledge, she forgot several steps! She did not explain why -(ab + c) is the same as -ab -c. Then she fails to explain why a - ab is the same as a(1-b). Explain those steps in between, and we are good to go. Without: nothing goes. To me this constitutes a bad teacher!

[@themathiasP]

I got to 3:37 myself but I had a different train of thought. I thought to myself what do I multiply by each other to get one. I realised that x * 1/x always is one. I got stuck on that and watched the video. I forgot that integer means that it could not be a fraction. English is not my native language although I should have realised it since in a programming I know an integer is a round number from- 2^15 and + 2^15.

[@Dannychii]

Thanks for the explanation, I was confused at that too :D German here 😄

[@mark2talk2u]

I’m confused. At the 3:49 mark, there are many products that equal one. How about 7 and1/7 for factor values? Is there a requirement that a, b, and c are integers?

[@BurstWalkthroughs]

it says at the top of the paper "integer solutions"

[@夢月天林]

The question only requires integral solutions.

[@dalchandagrawal7312]

Assumptions that they are integers.

[@fuzzblightyear145]

as my old teacher once said "Read the question. Then read it again"

[@ctsirkass]

@@fuzzblightyear145 haha this is what my teachers said and I found it not to be enough. What I say to my son now is to read it 3 times, slowly, while focusing on each word separately and understanding what it means to be there.

[@kumarchandrahas1462]

Good answer...it is factual... Based on multiplication principle- 0 multplied by any number is 0

[@phoenixiiita]

Great solution. Is this really a math Olympiad question ? Looks quite simple

[@ceansonnery5937]

I had problems falling asleep. This video was my cure.

[@alanklajnsek4400]

The easiest is you choose one variable to be zero like a = 0 Then you eliminate quite a lot. c = 2020 and b = 2021/2020. So multiple solutions Not a fair Math problem but resourcefull one.

[@tawfikahmed.2526]

This is decimal solution not integer solution for b

[@alanklajnsek4400]

Ab+c = (a + bc) - 1 Now solve it... There are probably more solutons... or more easy c =2020 - ab Now replace this c in second equation with (2020 - ab). a + (b(2020-ab))= 2021 a + (2020b -ab^2)=2021 Now we get: 2020b-ab^2 = 2021-a and c = 2020-ab Here on we choose one variable and multiple solutions present ourselves... One of them: b = 1 we get 2020-a=2021-a 2020=(2021-a)a a^2-2021a+2020=0 a=1 c=2019

[@tawfikahmed.2526]

Wrong

[@CellarDoor-rt8tt]

Before watching here’s my method. We’re going to do this in a way where we are going to find every integer solution and prove that we have found all of them First we consider any fixed b and solve as a system in terms of a and c. I did so using Cramer’s rule which gets the job done and obtain that a = (2020b - 2021) / (b^2 - 1) and c = (2021b - 2020) / (b^2 - 1) First we must check explicitly what happens at b = 1 and at b = -1 as cramer’s rule will fail if our system is linearly dependent, but our system being linearly dependent does not preclude the possibility of there being solutions so we must check. This is easy as for b = 1, a + c = 2020 and a + c = 2021 so 2020 = 2021 -> 0=1 which is a contradiction. If b = -1 then -a + c = 2020 and a - c = 2021 so -2020 = 2021 so 0 = 4041 -> 0=1 which is a contradiction. So, now we have that a = (2020b - 2021) / (b^2 - 1) c = (2021b - 2020) / (b^2 - 1) We have every solution, but now we need to isolate the integer ones. Use partial fraction decomposition. We get that a = 4021/(2b+2) - 1/(2b-2) c = 4021/(2b+2) + 1/(2b-2) Now assume a and c are integers and b is an integer that’s not -1 or 1, then 2ab + 2a = 4041 - (b+1)/(b-1) 2cb + 2c = 4041 + (b+1)/(b-1) In both instances, the left hand sides are necessarily integers. On the right hand side, 4041 is an integer so both right hand sides will result in integers if and only if (b+1)/(b-1) is an integer. So b-1 is a factor of b+1. In the case that b < -1, (b+1)/(b-1) will always fail to be an integer because in that case abs(b+1) < abs(b-1) -> abs( (b+1) / (b-1) ) < 1 and (b+1)/(b-1) = 0 only at b = -1 If b > 0 then (b+1)/(b-1) is not an integer if b > 3. This is because of prime factorization. Since every natural number larger than is uniquely the product of primes, that means that the largest integer factor of any natural number larger than 2 is itself and the next largest factor is at most the size of that natural number divided by the smallest prime which is 2. But, b+1 =/= b-1 ever and b > 3 -> 2b > b + 3 -> 2b - 2 > b + 1 -> b -1 > (b+1)/2 So, b must be between -1 and 3. We already showed b = -1 and b = 1 don’t work. If b = 0 then a = 2021 and c = 2020. if b = 2, then a = 673 and c = 674. If b = 3, then a = 4039/8 which isn’t an integer. So the only 2 solutions are (a, b, c) = (2021, 0, 2020) and (a, b, c) = (673, 2, 674).

[@BALLYRM]

Question: what is your audience- Initial 4 minutes could be reduced to less than 60 secods!

[@noyanozyakali5281]

Yes but missing. Basically you have 3 unknowns and 2 eqnS, so it can be said “infinitely many solutions” to be exact. Also consider “negative integers” as well.

[@ZIN24031980]

Thank you very much, your solution is clear and simple.

[@QUABLEDISTOCFICKLEPO]

It's a clear as mud to me.

[@sobolzeev]

​@@QUABLEDISTOCFICKLEPOYou would prefer it even slower? Or do you need an explanation of an idea of replacing numbers with letters, so called algebra?

[@QUABLEDISTOCFICKLEPO]

If I couldn't understand it, the fault is not mine. I won't waste time by looking at it again.If I said that it was unclear, it was.

[@sobolzeev]

@@QUABLEDISTOCFICKLEPO If you hear a dumb sound when a book hits your head, it is not necessary the book's failure.

[@QUABLEDISTOCFICKLEPO]

Fortunately, I didn't have any "teachers" like that when I was in school. If I had, I never would have learned fractions..

[@Nevyn515]

I’d have said a = 2000, b = 1 c = 20… Because I’m dumb and just pop out the first answer that comes to mind, give it zero thought, then never think about it again, because I’m not a maths professor or in a maths class, so it’s not like I need to do any equations ever, just like 99.999% of everyone else in the world. The remainder probably do physics or work at NASA or something so they need more maths skills than basic addition, subtraction and potentially basic multiplication or division… But we all have phones with calculators on them for a reason, pretty much for doing DIY, maybe doing refunds at a customer service job, working out how much each person needs to pay at a restaurant, and pretty much nothing else.

[@perweryoussef6947]

Three unknowns cannot be deduced from two equations... There are an infinite number of solutions

[@jimbrooks1452]

You are correct if there are no restrictions. But, as I tell my students, "read the problem." The problem restricts the solutions to *integers.*

[@rzbonilla]

Is assuming b = 0 possible?

[@misomiso8228]

Beautiful.

[@pandu2129]

So, 2 variables and 2 equations give you only 1 possible case answer whereas 3 variables and 2 equations give you 2 possible case answers.

[@setsunaes]

(Me proud because I solved for the 2 results of this problem): Hehe, I still got it, here I am solving a math Olympics probl... (Someone:) Yeah this is a third-grade student problem (me)... well, gotta train to help my daughter then

[@avidelahi5224]

So interesting, thanks 👌✨️

[@desmondaubery9621]

Thank you. Elegant.

[@GineAce]

(1)(1) = 1 (-1)(-1) = 1 (x)(1/x) = 1， so this expression has infinitely many solutions

[@mayihelpyou5557]

1/x 不是整數

[@夢月天林]

But the question only requires integer solutions, which means the third case should not be considered or otherwise at least one of the unknowns has to be a function.

[@ctsirkass]

...and the 3rd solution allows for only values x=+/-1, in order for fractions themselves to be integers (x: integer & 1/x:integer). Read the question more carefully or you are going to make many mistakes in your exams.

[@induwara2513]

Clear solution

[@rameshsingamsetti9690]

The product of 1 can be obtained by 1/2 × 2 or 1/3 x 3. There are infinite solutions to that problem.. You have 3 unknowns and 2 equations. You can't have a definite solution for that!!

[@andykyllo6856]

The directions state integer solutions only.

[@dussaramaa5392]

a=673,b=2,c=674

[@WannaKnowMyName]

3:52 “1 is a product of 1 times 1 and -1 times -1”. 1 is a product of infinite numbers. 1 is a product of 2 and 1/2, 3 and 1/3, 4 and 1/4, etc. There are infinite solutions. By saying that 1 is only a product of 1*1 and -1*-1 you’re limiting the scope of the solutions here. This problem in fact has infinite solutions

[@syedarbishalishah5140]

You need to read the question more carefully.

[@Laci-ps9xq]

As a 7th grade asian we can be sure with you that this math equation is a piece of cake

[@PraveenKumar-bk7yk]

My way to solve this one: ab + c = 2020 ---(1) a + bc = 2021 ---(2) Now (2) - (1): a -ab + bc -c = 2020 - 2021 => a(1 - b) -c(1 - b) = 1 => (a -c)(1 -b) = 1 ---(3) From equation (3): Both factors could possibly be either -1 or 1 itself. i.e. Case-1: Either (a -c) = -1 and (1 - b) = -1 Or, Case-2: (a -c) = 1 and (1 -b) = 1 If (a -c) = 1 and (1 -b) = 1; => b = 0, Put this value in eq(1) and (2), we get: c = 2020 and a = 2021. If (a -c) = -1 and (1 -b) = -1 Then b = 2; And we have: 2a + c = 2020 ---(4) a +2c = 2021 ---(5) Subtracting them: a -2a +2c -c = 1 =>c = 1 + a ---(6) Substituting, in (4): 2a + 1 +a = 2020 3a = 2019 Answer set is either, a = 673 b = 2 c = 674 Or, a = 2021 b = 0 c = 2020

[@tienphung5871]

a=2021, b=0, c=2020. a=2019/3, b=2, c= 2022/3

[@kagegakurenokuni]

before reading comments i had doubts that an olympiad question could be so simple (probably it's for 7th grade tho), but now i know how many people don't know what an integer is.

[@bonevgm]

I am embarrassed to say I spend 15 min going in circles until I realized what I need to do and it was solved in 5 min.

[@pedagoclown2267]

So great , smooth voice , quiet and logic I enjoy

[@jasonsternburgh8363]

I brute forced 0's and 1's and found a solution quick. But I'm not taking the test under pressure, I wouldn't have been able to do this in school.

[@aurelusentertainment5303]

My solution also seems to work a=0 not b, b=2021/2020, c=2020. Waht is wrong with that ?

[@kenkennio4452]

Sooo... We have few informations. What I found was simple: b belongs to the set of reals, and it can be any real value, with the exception of 0 and 1. For negative b, a>c. For positive b, a

[@ctsirkass]

Yeah, you need to pay attention to the question. My suggestion would be to slowly read 3 times and pay attention to each word separately before solving. (hint: we are looking for integer solutions only)

[@edemshirinskiy324]

0.5*2 equals to 1 as well

[@pastorgarcia4676]

There is an infinite solutions, as you can express any variable as dependent of the other two, as that is a two equation system with 3 incognities

[@jige1225]

There is an additional constraint that a, b, c are integers

[@bimokayoba]

I found b = 0, a = 2021 and c = 2020, is this correct?

[@strangersingapore4121]

It's not correct if we substitute it back. This video is not concluded properly ✌️

[@bimokayoba]

@@strangersingapore4121 oh ok

[@eddie31415]

Subtracting the two equations work here. I wonder how difficult versions of this problem would look like: I am imagining some function of the first equation + some function of the other equation to give some hint in the general case.

[@mig_21bison]

What is the use of these equations...??? Where they are used??? What is the practical application????Please answer

[@sobolzeev]

Hm, what is the use of, say, anime pictures? Whom do they depict? What is their application? Please, answer?

[@neonblack211]

​@@mig_21bisonif you want to consider science, engineering, physics, absolutely all over the place

[@kennethkan3252]

ab+c=2020 a+bc=2021 a is must be odd, then b,c is even.

[@sammail180]

easier to solve using a system of equations. we express A, and then substitute it into the first equation..., and then it’s obvious

[@md.mamunorrashed4924]

There should be an infinite number of solutions for this problem. For 3 variables you should have at least 3 different conditions. Otherwise no specific solutions will be found!

[@szymcio9931]

Use your brain before you write something: ab=3 and assuming a and b are integers (just like in the video), you have 1 equation with 2 variables. Still thinking it has infinite solutions?

[@md.mamunorrashed4924]

not taking 'integer solution' as a condition was my mistake. either way there should have at least equal number of conditions compare to variable numbers to solve any equation.

[@shoutplenty]

not watched the video but the intuition that comes to mind here is that b scales either a or c with very similar results (2020 vs 2021), so a and c must be very close, hence write c in terms of a by substituting d := c - a (so d will be small), and yeah subtracting the top equation from the bottom then gives d(b - 1) = 1, so (c - a)(b - 1) = 1, then it's easy cos they're integers dividing 1

[@advertisingagency5840]

Thank you

[@pakdionmtk]

Thank you.. i am watch for 3 minute and subscribe later

[@lucianofinardi7222]

(1-b) (a-c) = 1 can have an infinite combination of multipliers as solutions and not only 1*1 and -1*-1. It could be 0.5 * 2 or 0.25 * 4, etc, etc.

[@Crom1980]

But they are not integer.

[@ctsirkass]

As my teacher used to say "Read the question 3 times carefully"

[@chriscotton4207]

So I follow the video everything is going well until you get to a point where it's (1-b) [a-c] Why I'm confused is because multiplication and subtraction properties are different. You do them in different order. How are you able to take a cluster inside parentheses and move it out without affecting the order of operation? You're moving it ahead of its pemdas A+b*c does not equal a*b+c. It worked here but can you help me understand why an addition from one side can move to the multiplication of an unsolved other half.

[@zdrastvutye]

how does this solve in this universe? 10 for a=1 to 1999:for b=a+1 to 2000 z1=2020-a*b:z2=(2021-a)/b:if z1=z2 then stop next b:next a

[@Неиначекак]

Why (1-b) equals (a-c)? 1-b=1/(a-c)

[@TuanNguyen12]

a×1/a = 1 ? Test ?

[@alster724]

I have seen this classic Olympiad problem so it is easy for me.

[@the-mathwizard]

Indeed, math is easy if we already seen it before

[@irish3353]

Its good it says integer solution because with two equations and 3 variables you can get infinite real solutions, lol. a = 1125 b = ⅘ c = 1120 For example. Very nice, I forgot to factor that way, and so I trailed and errored to get (a-c)(1-b).

[@ESSServices-x7p]

a= 673; b=2; c=674 and a=2021; b=0; c=2020

[@felixmartin9156]

What language is this?

[@РоманЕфимов-ф2ц]

1-b=1/2, a-c=2.

[@ghostracer_978]

For finding solution for 2a+c=2020 a+2c=2021 Just add the equation Now you get 3a+3c=4041 By factoring, You get, 3(a+c)=4041 Now a+c=4041/3 a+c=1347 Now take any two random numbers that add up to 1347 Lets take a=1000 c=347 Verification: 3a+3c=4041 3x1000+3x347=4041 4041=4041

[@shirinmalik2235]

I miss. High school math 6:09

[@lancelink88]

That was really amazing.

[@abderrahimhilali4092]

Why did she Say that we can get 1 just by multiplying 1x1 or -1x(-1) ? Can’t we use for example 2x(1/2),3x(1/3) ... and so on ?

[@the-mathwizard]

It says only integer, so we cant use fraction

[@JeffreyBue_imtxsmoke]

I tried solving this with substitution before watching the video and got stumped. Nice use of factorials.

[@Shiraori999]

Besides the 0 solution. C=674 A=673 B=2

[@smt4940]

but n*1/n=1 how you can say it is 1 or -1?

[@akinamegu9896]

hhhhh lucky for you that ( a & b & c ) were integers and not floats 😛

[@KingSalah1]

Isnt that just a Matrix which ist diagonal?