France - Math Olympiad Question | An Algebraic Expression | You should be able to solve this!
Вставка
- Опубліковано 9 чер 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
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)
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
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.
Nice use of factorization concepts, well done
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.
Interesting that many people can't read the first four words in red.
Indeed. It's painful, too painful reading comments.
Thanks! Delightful presentation of a clever little problem.
Wonderful. this is how simple things can be best seen.. Thanks a mill
a = 2021 b=0 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.
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
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!
Impressionnant !!
Thanks for the solution and a great explanation.😊
Beautiful.
Thank you. Elegant.
So great , smooth voice , quiet and logic I enjoy
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)
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.
That was really amazing.
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.
I have seen this classic Olympiad problem so it is easy for me.
Indeed, math is easy if we already seen it before
There is a third solution too. (1-b)=1/(a-c)
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..
Thank you
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.
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
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)
This reminds me linear algebra
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
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
Clear solution
How come this video appear in my suggestion, it looks like magic to me
Wow very impressive.
As a 7th grade asian we can be sure with you that this math equation is a piece of cake
This was great!
So interesting, thanks 👌✨️
it was exactly my answer spending just 30s thinking
love it.
I tried solving this with substitution before watching the video and got stumped. Nice use of factorials.
love these
Good answer...it is factual...
Based on multiplication principle- 0 multplied by any number is 0
Diofantic problem. Very interesting.
That's nowhere near to a maths Olympiad question
awesome !
Side note, neat idea with the name on a pen
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.
What grade is this math problem for? In my country, I encountered this problem when I was in 9th grade
This is so simple.
Nice solution
If u see , there are infinie solutions
As 1 can be written 2* 1/2 and infinitely many more
5:41
In case II, we already have c=a+1. Putting in either equation yields 3a + 1 = 2020. Good job otherwise. 👍
Thank you very much
I loved this.
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.
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
Great solution. Is this really a math Olympiad question ? Looks quite simple
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.
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.
i have to count to 20 by taking my socks off and i was able to do solution 1 in my head :)
2:42 signs work in multiplication? How?
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
In my augustus opinion, add up all the stuff and put its variable b in evidence.
I had problems falling asleep. This video was my cure.
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).
Why are you only using integer solutions?
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 😄
I was not able to follow the solution. But the first one was obvious to me looking at it. 😂
Thank you.. i am watch for 3 minute and subscribe later
abc (nailed it)
Solving systems of equations is so satisfying. Plug and chug, baby
I need to reteach myself math fr.
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.
Through inspection a = 2021, b = 0, c = 2020 lol
Nice, now tell me where do I need it in real life.
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.
Man… it’s been too long since college.
0.5*2 equals to 1 as well
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.
not natural numbers, but integers
We are talking about integers, not natural numbers, so you can get a product of 1 by (-1)*(-1) so you missed one solution.
1-b=1/2, a-c=2.
Use it in a realistic world problem
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
Wrong
easier to solve using a system of equations. we express A, and then substitute it into the first equation..., and then it’s obvious
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
My solution also seems to work a=0 not b, b=2021/2020, c=2020. Waht is wrong with that ?
I loved hear tuani(twenty)
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.)
Interesting
He took 8 mins to explain this
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.
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
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).
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
There is an additional constraint that a, b, c are integers
So simple