IMO 2024 #1

This is my thinking on this 2024 Olympiad problem.

[@tanxuanyuanmoe]

Just a suggestion: Please explain other questions asked in the IMO 2024, thanks.

[@pijanV2]

you can see he intends on doing it since he put #1 after the end

[@PrimeNewtons]

Lol 😆

[@PrimeNewtons]

I am flattered that you think I can solve them. No promises.

[@poghuman8676]

@pijanV2no… thats because its the first problem bro 💀

[@Nerdiest_nerd]

I would also like to see more of imo questions

[@fluffybear46]

Immediately clicked cuz I thought I recognized that question. Thank you for providing a solution. I would like for more content on competition maths questions. Maybe you could do all the questions on last year's IMO too.

[@klausg1843]

Tak!

[@qwertyman123]

Excellent job! Please do more IMO problems😊

[@deibis4522]

Your videos are amazing!!! Very well explained and your voice is so relaxing 😂😂😂😂

[@aleksandr_le_gens4164]

Great video and exercice ! At 7:25 shouldn't we precise that a and b are real 'positive' numbers ? Thank you for all you have done and have a good day sir.

[@ChronoQuote]

10:31 I think the nα on the right should instead be ⌊nα⌋. Also, by claiming all the terms in the sum before ⌊nα⌋ are 0, I think you're assuming n is the _smallest_ natural number such that nα>1, but the archimedean property only states n is _a_ natural number such that nα>1. But this is easy to resolve with the well-ordering principle.

[@kavimahajan892]

yeah it should be floor nalpha but its fine cause itll be 1 since hes assuming smallest n, and 1 still wont be a multiple

[@LorenzoFabbri-g4l]

Thank you. I'm trying to get into this competition and I'll definitely follow all of these videos.

[@GreenMeansGOF]

I think your Archimedian argument is essentially correct but let’s make it rigorous. For any α in the interval [1/n,1/(n-1)), where n>1, F(α)+F(2α)+…+F(nα)=1 which is not a multiple of n. Since every number in (0,1) falls into one of these categories, then the interval (0,1) is excluded from the solution set. It follows that only integers can possibly have a chance of being solutions, just as you argued.

[@PrimeNewtons]

This is so precise. I wish I thought to clearly.

[@klausg1843]

@@GreenMeansGOF hmmm. (n-1)•alpha is in [1-1/n, 2-2/n) and you dont know if the integer value is 0 or 1. ???? So it’s not correct.

[@GreenMeansGOF]

@ I believe I have fixed it?

[@klausg1843]

@@GreenMeansGOFyes sorry I did not see your last and correct version. 🙏

[@pizza8725]

Couldn't tehnically alpha be odd and integral part of b be non zero such that if they are added than we get a multiple of n? Or it doesn't work

[@StudyOnly-nn1xb]

Yeah i am wondering that too

[@klausg1843]

No see my remark on that.

[@auztenz]

b is less than 1 greater than 0 so there is no non zero integral part of b

[@StudyOnly-nn1xb]

@@auztenz but the multiples of b like 2b,3b can be >1

[@francoferreira5378]

Same question i have

[@a.hardin620]

I saw a good problem. It’s not that hard but interesting: Prove that all Prime numbers greater than two can be represented as a difference of squares. For example: 2^2- 1^1 = 3.

[@Aman_iitbh]

P+1/2=x, p-1/2=y

[@erikberg7891]

Take the difference between the square of a integer n and the next-higher integer (n+1) and you get (n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 This is just the set of odd integers. Since all primes greater than 2 are odd, this equation will generate all of these prime numbers (and many odd numbers which are not primes).

[@erikberg7891]

An even-number generator is 1/2 ( (n+1)^2 - (n-1)^2 ) for n is an integer.

[@glorrin]

Great video. I had doubt at first but I finaly get it case alpha = a + b when a is odd. S = a*n*(n+1)/2 + s0 If n is even, then n does not divide a*n*(n+1)/2, For n to divide S b must not be 0 because n must not divide s0. Assume there is an a and b so that for any even n, n divide S. when n is odd, n divide a*n*(n+1)/2, b is not 0 so n does not divide s0 so n does not divide S if n is odd, then n divide, a*n*(n+1)/2, for n to divide S it has to divide s0 so b = 0. so alpha = a. but then when n is even, n do not divide a*n*(n+1)/2 so it does not divide S. Therefore a cannot be odd.

[@nicolasb11]

Good job Prime Newton...! So I have a question why should you take ''b'' as a fraction between o and 1 ?

[@PrimeNewtons]

Because every number has a fractional part. It is either 0 or between zero and 1.

[@nicolasb11]

@ ok I got you !

[@kateknowles8055]

Whow ! I am going to watch this .I know thst some triangular numbers n(n+1)/2 are involved, by how does the induction carry to infinity for all the solution set? Post video: Oh at the end we have found at least one n that causes it to fail iff alpha is not an even integer. You are really good! Right 0:40.. I'm listening. 12:50 every RATIONAL number has an integer part and a fractional part and most REAL numbers have an irrational particle or a transcendental particle additionally. Splitting hairs. But I am listening again...What you intend that we understand at 15:55 is that there is a certain value of n for which all the terms with floor(kb) sum to 1. I am getting there. Here you explain something quite complicated quite well. I will be here for another quite complicated one.❄Thank you.

[@Dominus_Potatus]

I think your necklace is scrubing against microphone. I love this kind of math. The one that is really in the competition.

[@blackdye2420]

Why the floor function is the same for negative too?

[@Jake-p8s]

Try it for some values. You will see.

[@arknight2024]

it is difficult how xould you solve

[@joelklein3501]

I'll restate the details of the proof, and by doing so I think it will make some detail missing from the proof clearer Any real number r can be written as r = a + b Where a is an integer, and 0

[@davibz2243]

You're right that there's a hole in the given proof for odd values of a. The trick seems to be to consider different cases for values of b. Namely b in (0, 1/2) then b in [1/2, 2/3) then [2/3, 3/4)... If you go through a couple of examples you can see what value of n goes with each case. Hope this is enough of a hint

[@ccdsah]

7:45 your assumption is only valid if both real numbers are positive. if a is negative and b is positive there's no natural number such as n*a>b

[@PrimeNewtons]

Yes. However, if i was doing it for negative numbers, I'd simply reverse the sign.

[@magnusPurblind]

Suppose alpha = 1.999. Pause the video at 18:05 or so. Can you see how the given explanation tells us that alpha should be rejected? If you check the first few hundred values of n, it looks OK. Basically I think the argument needs to be changed for odd values of a so you're looking at an even number minus a fractional part. Apart from that the video is very good

[@rashiartandcraft485]

Sir I was doing some mathematics and accidentally I made this equation ( x^7=7× x^x) . So I want to ask that is this equation can be solved algebraically. I have been stuck in this for a day. So I request you to give some time to this equation. Thankyou 🙏🙏🙏🙏

[@francoferreira5378]

You probably need the lambert W function for that as the variable is both on the exponent and on the base.

[@mayukhintesarislam306]

damn thanx a lot, i never thought i could actually understand any imo problem lol

[@surendrakverma555]

Thanks Sir

[@diegorosa7517]

Im not sure, but I tought that you may need to prove that an(n+1)/2 + M is never a multiple of all n belonging to the natural numbers, to make sure that there is never a fractional number such that an(n+1)/2 + M (being the fractional part that is never a multiple) sums up to being a multiple of n Obviously if an(n+1)/2 is a multiple of n then it wont sum up to a multiple of n, but what if it isnt a multiple of n, then the sum wont be as certain, because numbers that are not multiples of a natural number may add up to a multiple of this natural number (14+4=2×9, 14 and 4 arent multiples of 9)

[@DrPhunsukhWangdu]

Thanks sir ❤

[@Aminh7s]

What if [b] + [2b] + ... + [nb] is a multiple of n/2 for all even values of n and odd values of a?

[@PrimeNewtons]

Give an example

[@auztenz]

That sum is actualy 1 (let 1/n≤b

[@Aminh7s]

@PrimeNewtons My apologies. I made a compulsive reasoning. There might be such an M, but it does not hold true for odd values of n.

[@Aminh7s]

​@@auztenzif b=0.9, it is true for n=2,4,6,8,10,12. So n=2 is not the only option.

[@Aminh7s]

​@@auztenzfor b=0.3, n=6: sum=3=n/2 so n=2 is not the only option.

[@nedmerrill5705]

Alpha must be an even integer. I agree.

[@American_Language]

watching this guy is always the greatest part of my day :] never stop learning.

[@iljas275]

You did not prove that odd numbers with fractional parts are not the solutions!

[@iispacedustii]

thank you!

[@glennberry4829]

The proof for negative alpha will be different, since the floor behaves differently with negative non-integers than it does with positive non-integers. If x is a positive number that is not an integer, floor(-x) = -1 - floor(x).

[@PrimeNewtons]

It's the same idea

[@R097-p2y]

Wauw

[@StudyOnly-nn1xb]

I am not able to understand some parts of the solution, the only thing of material is the a+b solution part, i dont understand the relevance of the former half of the solution. Is it to get familiar with what the question is asking? Besides that the solution assumes if M is not divisible by n, then the other term must be a multiple n but thats is not true

[@ZDTF]

Buf arent there infinite numbers What if theres infinite number

[@alexandrezeddam7817]

The question asks for alpha a real number. It does not matter that there are infinitely many reals - all reals are finite.

[@ZDTF]

@alexandrezeddam7817 Woaht

[@ZDTF]

@alexandrezeddam7817 can you rephrase that for a 13 year old pls

[@alexandrezeddam7817]

@@ZDTF It means that there is no such thing as an infinite number.

[@ZDTF]

@@alexandrezeddam7817 Damn that's crazy

[@VictorTeachesMath]

13:23 huh

[@mrpringles4479]

Putnam please mr newtons

[@spectra1037]

I just did a(n(n+1))/2 = kn a(n(n+1)) = 2kn a(n+1) = 2k therefore a(n+1) must be even, and if a is odd, its not true when n is even.

[@deltalima6703]

A floor? Deep in the weeds here...

[@seasickskipper]

Come on, zero is NOT the nicest number in mathematics. 🙂

[@Dominus_Potatus]

😏

[@klausg1843]

Great explanation. Nothing essential to fix. You shall prove that the sum equal to a/2•n•(n+1) +1 is not divisible by n. This is obvious for a even. For a uneven you split into cases n even/ n uneven. Or just see that a/2 • (n+1) + 1/n can never be an integer except for n=2. I will follow you and learn a lot. Thx.

[@solmonhue]

IMO 2026

[@subhajitpaul-es8rz]

First comment

[@pijanV2]

alpha cannot be zero since how can zero multiply with any other number to get a nonzero number

[@ChronoQuote]

If α=0, the sum given in the question is 0, and 0 is a multiple of all positive integers, so the claim is true.