France - Math Olympiad Question | An Algebraic Expression | You should be able to solve this!
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- čas přidán 9. 06. 2023
- Maths Olympiads are held all around the world to recognise students who excel in maths. The test is offered at many grade levels and provides them with numerous possibilities to win certifications, awards, and even scholarships for higher studies.
Easy. a=2021, c=2020 and b=0.
You missed half the solutions genius
@@aidan-ator7844whatever. Its correct! Uhuu!
@@TheEmanoeljr yes but you only have the intuitive solution. There are others.
@@aidan-ator7844Yeah... So you can admit he has good intuitions. Lol...
@@ferrel9715 without a doubt. Intuition is only one part of thinking that functions best in conjunction with others.
I liked the fact that someone with a nice voice and clear handwriting can provide audio-visuals for an instructional video. I have neither.
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
Somewhere, someone's discovered an amazing problem featuring the number 2374 and is just waiting to reach that year to release it
I don't think that they would still be alive
Interesting that many people can't read the first four words in red.
Indeed. It's painful, too painful reading comments.
I solved stuff like this back in highschool. Man life has dragged me down. I need to take math classes again.
Right? I remember a time when I could try to get my head around this but I’ve forgotten so much.
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.
Also 2×0.5=1 and 0.5×2=1
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..
@@dorgamahmad6033 Except that the problem stated in the first few seconds of the video was "find the integer solutions".
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.
@@dorgamahmad6033 we are looking only for INTEGER solution. Read the question 3 times before solving (classic teacher's advice from the elementary school)
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)).
I'm an idiot. The problem statement says "integer solutions".
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
@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
Same here. The thumbnail left that out. I was going to solve before watching the video, only to quickly notice there are infinite solutions.
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); ...
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.
Nice use of factorization concepts, well done
a = 2021 b=0 c= 2020
Wonderful. this is how simple things can be best seen.. Thanks a mill
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!
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.
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
Thanks! Delightful presentation of a clever little problem.
Love all your explanations. They are so clear and easy to understand
Not to me.
@@QUABLEDISTOCFICKLEPO 😂
Not clear.... she says a into b when she means a x b.
a into b is b/a
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.)
@@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!
What is the brand of the pen, I love how thin the lines are.
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.
What is the use of these equations...??? Where they are used??? What is the practical application????Please answer
Hm, what is the use of, say, anime pictures? Whom do they depict? What is their application? Please, answer?
@@mig_21bisonif you want to consider science, engineering, physics, absolutely all over the place
Thank you very much, your solution is clear and simple.
It's a clear as mud to me.
@@QUABLEDISTOCFICKLEPOYou would prefer it even slower? Or do you need an explanation of an idea of replacing numbers with letters, so called algebra?
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.
@@QUABLEDISTOCFICKLEPO If you hear a dumb sound when a book hits your head, it is not necessary the book's failure.
Fortunately, I didn't have any "teachers" like that when I was in school. If I had, I never would have learned fractions..
Impressionnant !!
I have seen this classic Olympiad problem so it is easy for me.
Indeed, math is easy if we already seen it before
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)
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
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.
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
How come this video appear in my suggestion, it looks like magic to me
Thanks for the solution and a great explanation.😊
That's nowhere near to a maths Olympiad question
There is a third solution too. (1-b)=1/(a-c)
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.
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.
Ask Grok.
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.
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
What grade is this math problem for? In my country, I encountered this problem when I was in 9th grade
So great , smooth voice , quiet and logic I enjoy
Thank you. Elegant.
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!
The answer will be the same, the extra solution set provided by your method involves functions.
you get the same answers, but it just takes longer. You can set b equal to all the potential values and have 2 solutions 2 equations that you solve normally, but for the solution values of b = -450, -10, -4, -2, 8, 448, you will get non-integer solutions for a and c. The only values of b that give you integer solutions for a and c are, b = 0 and 2.
I solved for natural integers (b+1)(a+c) = 4041 = 3•3•449 (excluding factor 1, which would yield b = 0). I have got 4 equations for
4041=(8+1)(a+c) = (448+1)(a+c) = (2+1) (a+c) = (1346+1) (a+c)
with b>0, of which only the third, with b = 2 gave integer a and c (673 and 674).
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.
This is what I thought just glancing at it. Seems to make sense to me that b = 0, a = 2021 and c = 2020.
In case II, we already have c=a+1. Putting in either equation yields 3a + 1 = 2020. Good job otherwise. 👍
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.
That was really amazing.
Beautiful.
I tried solving this with substitution before watching the video and got stumped. Nice use of factorials.
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.
? Where did you leave b?
@@gardenjoy5223tell me you didn't watch the video without saying you didn't watch the video
What about the a = 2021, b = 0, c = 2022 solution?
@@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?
@@gardenjoy5223 6:32
This reminds me linear algebra
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
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)
Wow very impressive.
Good answer...it is factual...
Based on multiplication principle- 0 multplied by any number is 0
Thank you
As a 7th grade asian we can be sure with you that this math equation is a piece of cake
If u see , there are infinie solutions
As 1 can be written 2* 1/2 and infinitely many more
5:41
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.
Three unknowns cannot be deduced from two equations...
There are an infinite number of solutions
You are correct if there are no restrictions. But, as I tell my students, "read the problem." The problem restricts the solutions to *integers.*
awesome !
love it.
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.
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
This equation is not at all straightforward,the easiest soln was given in the video
@@GauravSingh-gd1yj Agree, the easiest is shown. It is a nice problem for basic algebra astute.
2:42 signs work in multiplication? How?
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.
Thanks for the explanation, I was confused at that too :D German here 😄
Why are you only using integer solutions?
In my augustus opinion, add up all the stuff and put its variable b in evidence.
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.
So interesting, thanks 👌✨️
Thank you.. i am watch for 3 minute and subscribe later
Through inspection a = 2021, b = 0, c = 2020 lol
Diofantic problem. Very interesting.
This was great!
Thank you very much
awesome
Great solution. Is this really a math Olympiad question ? Looks quite simple
Can you please make a video solution on this-:
x-5/3=x-3/5
x is impossibile
subtracting different values from same x and equalling them is solutionless, there is no solution for that equation.. :) those x's disappear when we subtract x from both sides and -5/3=-3/5 is left and they are not equal.. ;) Sincerely..
Clear solution
Side note, neat idea with the name on a pen
it was exactly my answer spending just 30s thinking
Nice solution
i have to count to 20 by taking my socks off and i was able to do solution 1 in my head :)
easier to solve using a system of equations. we express A, and then substitute it into the first equation..., and then it’s obvious
love these
My solution also seems to work a=0 not b, b=2021/2020, c=2020. Waht is wrong with that ?
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.
I was not able to follow the solution. But the first one was obvious to me looking at it. 😂
0.5*2 equals to 1 as well
1-b=1/2, a-c=2.
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
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).
I had problems falling asleep. This video was my cure.
Man… it’s been too long since college.
He took 8 mins to explain this
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 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.
This is decimal solution not integer solution for b
I am surprised because I was taught that you need always the same number of equations as variables.
You were taught correctly. The reason for that is because if you have less equations than variables then there is an infinite set of solutions. This video only covers 2 integer solutions (not sure if more exist or not, too lazy to check). Vectors and matrices cover the full infinite set of solutions, but I'm assuming the Math Olympiad this video covers is based on less advanced topics.
You need 3 restrictions. Third equation is obviously the integer restriction itself. So the fast way to say is it solvable or not is not relying on 3 different equations but to rely on 3 different restrictions on variables.
Are there any results where the solutions are complex?
There are an infinite number of solutions even for reals a, b, c, because for most real values of b (namely, ones where determinant b^2-1 != 0) the resulting linear system can always be solved for a, c.
i dont get it, wouldnt there be infinite solutions? like the product of 1-b and a-c could 3 and 1/3, 2 and 1/2, wouldnt it?
But we have to find only integer values.
Me trying to learn, I'm already lost at why "a -ab = a (1-b)"
a×1/a = 1 ? Test ?
This is so simple.
but n*1/n=1 how you can say it is 1 or -1?
I do the first 6 steps in 1 step, but then I'm lost
Hello, you said that if we multiply two numbers one by one or if we multiply negative one by negative one, the result will be one, while multiplying one number by its inverse will also be one. How did you not say this? Thank you.
at the start it says ‘find the integer solutions’, which gets rid of the case where you have some number times its inverse.
otherwise there are definitely infinite solutions
looking at the comments it seems like at some point the thumbnail didnt show the integer part (but it does now), so that might be part of the confusion
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 directions state integer solutions only.
I am happy because my student join math olympiad
From which country ? Congrats !
Como se resolvería si la condicion en los resultados de las ecuaciones originales fuera 2020 y 2022 respectivamente? (Simplemente con que la diferencia entre ambos numeros sea mayor que 1)
Tienes que encontrar todos los números enteros que den como resultado 2 (la diferencia entre 2022 y 2020), después, encontrar los posibles resultados de a, b y c para que (-b+1)(a-c) = 2
I need to reteach myself math fr.