Can Calculus Help Solve IMO Problems? | International Mathematical Olympiad 1976 Problem 4

#IMO #Math #MathOlympiad

Here is the solution to IMO 1976 Problem 1!!

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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.

[eduard]

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).

[austinconner]

This is not a proof. Its a tool to guess the maximum, but you still have to prove the claim over the integers.

[sevinchbekbaxtiyorov]

We are all waiting for this.this will be great such as others

[borntolive]

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

[cr]

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?

[santiagoarce]

I heard that there are only a few cases where using calculus to solve a problem isn't severely loathed by IMO graders.

It should be highlighted in the questions that the series of the a have to be all positive integers

[SambhunathMaitra]

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<a1.Thus I have determined the value of n and then applying the rule of AM-GM inequality the value of up to the nth term) comes to approximately (247/2) ^16.

[VSN]

Great explanation & the flex on Euler no!

[soumitrapharikal]

Nice Video, learnt a new approach to them. Thanks, Keep Uploading more such Quality Content

[yurenchu]

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 .

[CreativeMathProblems]

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.

[yoav]

Great video! I liked the introduction in the start it helps keep it up in your next videos please

[ManuelRuiz-xibt]

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.".

[rijubhatt]

Nice. Thank You.This problem was at the back of my mind for quite some time.

[evangozali]

This video makes me realize again that e is really a naturally occuring number.

[tonyha]

I am enjoy this video very much, Thanks!

[dr.merlot]

doesn't it mean that the convex combination of 2s and 3s should be close to e? and not have max number of 3s?

[TusharGupta]

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?

[alainrogez]

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.