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

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

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

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.

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)).

a = 2021 b=0 c= 2020

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.

Easy eq1>eq2 by 1 so 2nd equation has 1 extra b c is therefore a+1 b is 2 (one extra plus 1 in row 2) 2020/3 =673 Remainder is 674

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

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

Nice use of factorization concepts, well done

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

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

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)

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

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.

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.

Wonderful. this is how simple things can be best seen.. Thanks a mill

Thanks! Delightful presentation of a clever little problem.

There is a third solution too. (1-b)=1/(a-c)

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

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!

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

Impressionnant !!

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.

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.

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

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

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

Hi, Did you look at the variation where you add the two equations and get (1+b)*(a+c) = 1*3*3*449. So you could possibly get multiple solutions in addition to what you provided. Curious what kind of solutions are generally acceptable? Thanks!

Thanks for the solution and a great explanation.😊

Thank you. Elegant.

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.

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

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

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

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.

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

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.

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

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

So great , smooth voice , quiet and logic I enjoy

Through inspection a = 2021, b = 0, c = 2020 lol

That's nowhere near to a maths Olympiad question

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

Thank you

awesome !

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.

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.

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

This reminds me linear algebra

Why are you only using integer solutions?

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

Wow very impressive.

love it.

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

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.

In case II, we already have c=a+1. Putting in either equation yields 3a + 1 = 2020. Good job otherwise. 👍

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.

Beautiful.

awesome

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

Clear solution

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

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.

That was really amazing.

Поскольку ни француским, ни ангийским не владею, то нюанс про целые числа я упустил. Только когда он начал разбираться с "а" и "с" я понял в чем дело. А потом еще увидел и "Іnteger" в условии.

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

So interesting, thanks 👌✨️

Nice solution

Diofantic problem. Very interesting.

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.)

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!!

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!

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

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

He took 8 mins to explain this

This was great!

Side note, neat idea with the name on a pen

In my augustus opinion, add up all the stuff and put its variable b in evidence.

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).

Thank you very much

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).

(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

Solving systems of equations is so satisfying. Plug and chug, baby

lol i can't do all of that weird stuff but I did solve it in my head by just making b 0 cause it was the only one multiplying in both and working from there lol.

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

love these

I was not able to follow the solution. But the first one was obvious to me looking at it. 😂

Que linda 😊

Mam 1/2*2=1 also .

You know it’ll be a good math video when the person has an Indian accent.

Interesting

i have to count to 20 by taking my socks off and i was able to do solution 1 in my head :)

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 ?

Is assuming b = 0 possible?

Hello from Spain. One question: why 1-b=1 on (1-b)(a-c)=1

2:42 signs work in multiplication? How?

Can you please make a video solution on this-: x-5/3=x-3/5

I need to reteach myself math fr.