Can Calculus Help Solve IMO Problems? | International Mathematical Olympiad 1976 Problem 4
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- Опубліковано 16 жов 2024
- #IMO #Math #MathOlympiad
Here is the solution to IMO 1976 Problem 1!!
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This is not a proof. Its a tool to guess the maximum, but you still have to prove the claim over the integers.
Official solution:
Answer: 2·(3^658).
There cannot be any integers larger than 4 in the maximal product, because for n > 4, we can replace n by 3 and n - 3 to get a larger product. There cannot be any 1s, because there must be an integer r > 1 (otherwise the product would be 1) and r + 1 > 1.r. We can also replace any 4s by two 2s leaving the product unchanged. Finally, there cannot be more than two 2s, because we can replace three 2s by two 3s to get a larger product. Thus the product must consist of 3s, and either zero, one or two 2s. The number of 2s is determined by the remainder on dividing the number 1976 by 3.
1976 = 3·658 + 2, so there must be just one 2, giving the product 2·(3^658).
This is kinda easy considering the toughness of imo
great
@@vijaybalajin3259 Almost all math tests have gotten easier over the years. If you look at ~1960 Putnam, IMO, etc., they are easy in comparison to today's questions. I'm sure 40 years from now, kids will be saying how much easier 2020 Putnam, IMO were lol
True
Thank you. I convinced myself of this answer, but I knew my approach was not rigorous. You gave me a firm foundation. Your answer is more complete than given in the video.
We are all waiting for this.this will be great such as others
9:26
Flexing *e* digits...
Yeah I too thought the same.
XD
2.7 double year of birth of Leo Tolstoy and angles in icosceles right triangle. Ez
Say a to a(n-1) are made of positive and its equal negative counterpart. So that each number cancels out leaving, a(n) which is +1976. So if we replace a to a(n-1) with infinity and its negative counterpart, we get the maximum value of a*......a(n)
That is a great point.
I don't think the way it's described in the video is anywhere close to being rigorous. I understand the intuition though but how would one write it down rigorously?
I heard that there are only a few cases where using calculus to solve a problem isn't severely loathed by IMO graders.
There are 0 instances of calculus being helpful at the IMO in recent years. There is perhaps some intuition but it can usually knock you off the wrong path because questions are chosen so that no one who knows more theory has a significant advantage.
Great explanation & the flex on Euler no!
It should be highlighted in the questions that the series of the a have to be all positive integers
Nice Video, learnt a new approach to them. Thanks, Keep Uploading more such Quality Content
I have solved the problem by assuming the given sum as the sum of a1, a2, a3....
upto nth terms which are in A. P series with common difference b=1 such that b
Great video! I liked the introduction in the start it helps keep it up in your next videos please
There is a simpler way to get intuition to use only 3s.
2^3 < 3^2
They sum up to the same thing, yet 3^2 is greater.
If the a_n are allowed to be postive reals, then the maximum product would be
(1976/727)^727
(where n = 727 , and a1 = a2 = ... = a_n = 1976/727 )
and not (e^726)*(1976 - 726*e) . This can be easily shown by defining f(x) = (x^726)*(1976 - 726*x) , and showing that f(x) has its maximum at x = 1976/727 instead of at x=e .
Nice. Thank You.This problem was at the back of my mind for quite some time.
I am enjoy this video very much, Thanks!
The question here is not nicely phrased. The official question is "Determine, with proof, the largest number which is the product of positive
integers whose sum is 1976.".
doesn't it mean that the convex combination of 2s and 3s should be close to e? and not have max number of 3s?
If we use the derivative for 10 instead of 1976, the maxima is attained at 10/e => we should 3s i.e. 3^3 * 1 but the maximum is obtained using five 2s i.e. 2^5. What’s the mistake?
3^2 2^2 = 36
This video makes me realize again that e is really a naturally occuring number.
Can't we solve without calculus or the AMH GH inequality? I don't think anyone would think of this. I hope you can PLEASE answer when you can. Thanks very much.
Really wouldn't they? I looked and immediately saw that it would need to be mostly 3s and some 2s, because it clearly relates to e
I don't understand. I thought we have AM-GM equality if all the an are equal. Isn't the result which we're looking for ? The maximum value of the product of the an is when all the an are equal to each other.
We need all the a_n to be integers, that is why you cannot apply the AM-GM result directly since the true maximum requires non integer partitions of 1976
@@rocky171986 but, I don't understand the answer : why is there a "2" times a power of 3?
@@alainrogez8485 the best result is if they are all e. But we need them to be integers not e, so instead we use the nearest integer which is 3, and 1976 = 3(658) + 2, so we have 658 terms equal to 3 and one equal to 2, i.e. n = 659. After that we take their product. But wait, what if we had 1976 = 2(988) so 2^988 instead? Well, that is less than 3^658 * 2, it is not really shown here but it should happen that choosing 3 is better as 3 is close to e.
And if you want to really prove that 3 is the best, maybe the justification to choose all 3s instead can be justified by studying the equation
n^floor(A/n)* (A%n)
and proving that it has maximum for n close to 3
@@moonlightcocktail ok, thanks. English isn't my mother tongue and I didn't understand all of these in the video.
Yeah, e is the best base for processing and storage capacity, this problem is pretty much about the same. Since 3 is closer integer, it's better. This idea was utilised in russian "+,-,0" trinary computers, that weren't developed further for some reasons. So now we're using less effective "0,1" binary computers.
Another great video
Great Video , beautiful solution
A.M. is greater than or equal to G.M.
From the thumbnail:
1976 if n = 1
988*988 if n = 2
unbounded if n ≥ 3
EDIT: Oh, _positive_ , and _integers_ . And we have to find the "optimal" value of n?
The function becomes 1976^n/n^n which maximises at n =1976/e ... So n=727 almost ?
The only thing I can imagine is making:
a(1) = a(2) = ... = a(n) = a
In that case:
a = 1976/n
product = f(n) = a^n = (1976/n)^n
max product -> f'(n) = 0
n = 1976/e
max product = e^(1976/e)
(edit)
Since n is a natural number:
n = ⌊1976/e⌋ = 726
To make sure it is the right value:
e*(n+0.5) > 1976
e*726.5 = 1974.8
So n = 726+1 = 727
Finally:
max product = (1976/727)^727
Ok so the numbers are integers lol all the work for nuttin...
Breaking n up using “3”s maximizes the value since x^(n/x)=(x^(1/x))^n whose global maximum over the reals is at x=e. Checking for the cases x=2 and x=3 let us conclude that the maximum over the integers is x=3.
So in the case n=1976=658*3+2 the maximum is 2*3^658.
Thank you very good
Mention the problem statement clearly.
Wow, that is not a proof
To have the largest product you need each number to be equally as large as possible
What I mean by that is that if you were doing it for 10 and n was 2 the two numbers would have to be 5 and 5, rather than something like 6 and 4 or 3 and 7
So if the numbers didn't have to be integers the solution would just be (1976/n)^n with each number just being 1976/n
But since they do have to be integers and 1976/n isn't guaranteed to be an integer, some have to be greater and some have to be less so that overall they average out to be 1976/n
You find how many of each by doing
1976modn
For the greater ones and
n - 1976modn
For the lesser ones
And so then you get
(Ceil(1976/n))^(1976modn)×(Floor(1976/n))^(n-1976modn)
Thanks, is, waiting for moe videos
thanks it's very 🥰🥰
But sir we can't write in it the proof ,huh ?
I think we can
India?
@@vijaybalajin3259 no I think calculus isn't allowed in maths olympiad
Yes india 🇮🇳
Check the hbcse solutions of previous years I think they themselves have given a solution based on calculus
Not sure which year
Blew me away!
Btw u cant use a calculator for IMO
no calculator needed for this
I got to the each number must be “about e” part, albeit this is similar to other questions I’ve seen. I then assumed that this would mean decomposing into 70% 3’s, and 30% 2’s, to get about 2.7 on average. So solved 1976=3m+2n, with m/n=70/30. This gives 512 3’s and 220 2’s. It looks like I got this wrong but curious as to why I had an “intuition fail”!
Ah I have now seen the Eduard 992 comment, and now get it!
because instead of using 3 2's you can use 2 3's which will will give the product 9
1976/e
You need to round it, then prove which ones gives the maximum, 727 or 726
@@riadsouissi how to decide which one
@@amiyancandol4499 anyway, all that is incorrect since now that the video is playable, it appears that a(n) must be integers.
@@riadsouissi yeah but if it wasn't then
@@amiyancandol4499 late, but this is how I did it because (1976/n)^n was too large for readily accessible calculators:
Assuming that the sum with n terms has a smaller product than the sum with n-1 terms,
(1976/n)^n < (1976/n-1)^n-1
=> 1976/(n^n) - 1/[(n-1)^(n-1)] < 0
=> 1976/n - [1 + 1/(n-1)]^(n-1) < 0
=> 1976 < n*[1 + 1/(n-1)]^(n-1)
Plugging in n = 727 to the formula on RHS gives ~1975, which implies by contradiction that n=727 gives a larger product than n=726.
Probably better ways to go about it, but that was the most intuitive way for me to check.
Why should we not use am gm
I think if it is +ve we can use
If -ve is allowed then maximum value is infinity I believe
Yes i thought same
Sry for the confusion we don't know the value of n so we can't use am gm
May be using trial and error we could find the maximum value and prove that it is monotonically decreasing after it as well as before it
This problem can be generalized: Given (a₁)ⁿ + (a₂)ⁿ + (a₃)ⁿ+…=S for n≥2, max(a₁a₂a₃…) = (2ⁿ)^floor(S/2ⁿ).
Did u notice the problem is just a special case of something u knew or u thought of generalization by yourself? If so, how?
@@omenuawo I thought of the general question myself, as it seemed quite natural.
My question was of course about the result, not just the obvious generalization of the sum. I wrongly assumed u saw both from the original problem
@@omenuawo I indeed saw the result too. It can be shown by considering what happens when, say, a bunch of 2^n s are replaced by a higher nth power to give the same sum: the product decreases. In other words, the same approach given in the official solution works.
Face reveal brother
And is infinite imo
*e*