IMO 2024 Problem 1 - Neat little problem to start things off!
ฝัง
- เผยแพร่เมื่อ 13 ต.ค. 2024
- #mathematics #olympiad #math
International Mathematical Olympiad (IMO) 2024 Day 1
Solutions and discussion of problem 1
65th International Mathematical Olympiad Bath UK
Problem 1 - Algebra / Number Theory
This took me like 5 mins to solve. (α=k+β where 0
the same thing i did my friend it felt like i am the one writing the comment
My solution: Let α = k+e where e in [0,1). n = 3 shows e in [0,1/3) or (1/2, 2/3), n = 5 shows e in [0,1/5) or (4/5, 1). Combining the 2, we have e in [0, 1/5). We now prove that e in [0, 1/(2m+1)) for all m (*) by induction on m. Hence e = 0, which implies α is an even integer.
The minimality argument is clever. I checked n = 2, 3, wasted a lot of time trying to obtain asymptotic bounds on S = floor(e) + ... + floor(ne) directly to show there must be a contradiction eventually, couldn't find anything strong enough, and tried n = 5 before noticing (*). In my investigations on S I realized there would be a messy solution where you prove frac(e) + ... + frac(ne) ~ n/2 for irrational r and then separately dispose of the case e = a/b by picking n appropriate to b, but didn't start this solution thinking this is too complicated for problem 1.
The first problem took me 1h10min and now im trying the second problem☠️
Me too bro me too 🗿
if u spend 1h10min to 1P, so i think u solved P2 in 2days💀
@@delaxer1828 longer☠3 days
Bro please tell me how I can solve such problems
I have a keen aspiration to go to IMO
@@ghostrider3922what standard are you in?
If your solution is particularly windy, do the graders take off points. To show floor of kf = k-1 in that case, I worked out five base cases for induction. Not needed, but also not incorrect.
As long as the solution is valid, you will get a 7. It's just that longer solutions with many cases runs higher risks of having a hole somewhere
my solution during IMO was a horrible casebash involving 7 cases in the end iirc, because I just quickly noticed that it does work and because of timepressure under IMO I couldn't be bothered to actually find the nice solution. My leaders hated me for it because it was horrible to coordinate (and the writeup was also horrible) and they said it's a mirracle I got 7, but in the end I did
We can say that for a fixed set of integers [1,..,n], the valid real numbers are 2k + [-1/n, 1/n), but since n can be arbitrary large, the solution is 2k.
Ah that’s very smart to write as b-g when odd. I instead did it by doing induction on n=2,3,4… to show that we must have floor(nf)=n-1 for all n which leads to a contradiction. Much longer process than your clean solution.
I'm pretty sure you can just ignore the floor function overall because no matter what you plug in the floor function makes it an integer. My solution was along the lines of the first solution, where I deduced that since an(n+1)/2 is divisible by n, and so a(n+1)/2 is an integer, a(n+1) and subsequently an+a must be even. And since the product of even numbers with any integer is always even, and adding an even number makes it stay even, all even integers satisfy the original problem. The final answer I got was along the lines of all values of a which return an even integer for floor(a)
Nice ! looking forward to questions 2 to 6!
That a lot of questions
i do not think there is 720 question !!
Since i failed P4 during Mock, this returned my confidence 😅
[a] + [2a] + .......[na] = [I+f] + [2I + 2f] +......[nI + nf] = I +2I +3I+......nI + [f] + [2f] + .......[nf] = n(n+1)/2I + [f] + [2f] + .......[nf]
Case 1: n is odd
if n is odd, n(n+1)/2I will be a multiple of n; so we need to make fractional part a multiple of n
if n = 1, [f] = 0 which is multiple of 1
if n = 3 , [f] + [2f] + [3f] should be multiple of 0 or 3 => [2f] + [3f] should be multiple of 0 or 3 i.e f2/3
if n = 5, => [2f] + [3f] + [4f] + [5f] be a multiple of 0,5 or 10 => f 4/5 (after already imposing condition that f2/3)
so generically,
f 1/n for all n till infinity => f =0
scenario 2: n is even
since f is anyway restricted to 0,
if n is even I has to be even
so the set is all even Integers
I don't get 08:42 .. I have m terms that are even summed up ... + m-1 terms of -1 summed up= (1-m)
I do NOT know if m is even or odd ... and then a single -2
1) sum of m parts of even -> still even
2) 1-m -> can be odd or even dependent on m
3) -2 ... only when 2) is odd this leads to the case of not divisible by n
So in the end it does work for some n ... but not any n..
Is this solution not rigorous enough;
the given expression can be written as:
a(n)(n+1)/2 + {a}+{2a}+…+{na}
The sum of fractional parts is ‘k’ for 0
I think you are incorrectly evaluating the expression of the fractional part. It should be [f] + [2f] + [3f] and so on. Assuming that alpha = a + f, where f is fractional of alpha
@@mithilleua7730 ive represented a number x=[x]+{x}, where [x] denotes the greatest integer function or the floor function and {x} denotes it's fractional part...all i did was write [x]+[2x]+...=x+2x+... -({x}+{2x}+...)
Lol am I the only one to notice the hint toward problem 6? The alpha as written at 2:04 is a function of the type defined in P6
Probably a coincidence!
Wow. Never done any olympiads in my life (almost) but were able to solve this. I have practically the same solution as shown in the video but with very poor and scuffed execution.
What of when n>m (here m>1 is the smallest integer such that mg>1)... then it remains to check whether the floors of the fractional part can equal n...in which case won't we have the divisibility condition being met? I got stuck here. And the video doesn't consider this case.
The question asks for alpha that works for all values of n. So to disqualify an alpha, you only need 1 example of n that fails for that alpha. There is no need to consider n>m since n=m is already the failure case for that alpha
I thought that any particular n must divide that expression. I didn't realise that the question is saying for all n. So i came up with another answer. Though that became more complicated. Successfully failed.
Sir, in the part "a is odd integer + fractional part", how is [2a]= 2b-1 when 2g =< 1 ?? It is because if 2g =< 1, then [2a] will be an even integer and hence will not be of the form 2b-1. Same problem goes to the second case. Sir please explain this. Also, how did you write [a] + [2a] + ... + [ma] = b + 2b + ... + mb - 1 - 1 - ... - 1 - 2
g is basically what your subtracting from b to have a. An example is when b = 3 and g = .4, a will be 2.6. [2 * 2.6] = 5 and 2 * 3 - 1 = 5. Thus 2 * B - 1 = [2a] when 2g
It helps to use an example! For example a = 1.7 (g=0.3 here). 2a = 3.4 and [2a]=3, which is not even
@@abhaymangla5678 brother, but it was taken that b is an even positive integer, not odd. That's what I am saying...
If I still have something which I didn't understand, please feel free
how about the floor definition for negative? [-0.c] = -1
?
Nice solution
is 2k
Alpha is to be fixed regardless of n. However, your range can lead to the same solution by observing that if you intersect all the ranges across all values of n, you get alpha = even integers
@@dedekindcuts3589 Understood, thank you sir.
For alpha being an odd integer case: we showed for just 1 case that it does not divide n, but what about all of the other cases?
The question asks for alpha that works for all n, which means to disqualify that alpha, you only need to show one n that fails
thank you Great explanation
You are welcome!
Hey just wanted to ask,I'm kinda newbie in competion level math problems I really want to get there,no hurry but I don't really get your videos,I want to understand them someday. Can you suggest me some roadmap or strategy to implement,like where should I start. I havent done any such maths since I have graduated from high school decade ago almost. So please,tell me some kind of rough roadmap@@dedekindcuts3589
thank you sir
Most welcome
How did you find problem 1? Were you able to solve it?
Exactly the same solution. Shows how straightforward this problem is. Ideal for a P1.
@@quite_unknown_1 Agree!!
Took me 3 hours to solve it (I am not satisfied)
@@xquadsw2448 Solving it is a good start! Will get better next time!
Solved it in 1hr 5 mins, it is the first IMO problem I have managed to solve 🎉
Another solutions is when a = + or -n.
yes i think you need to include alpha = +/-n as part of the answer, not just even integers
@@leeriri no, the single fixed value alpha as to work for all positive integers n, you can't have alpha equal to something that its not fixed
As cralts pointed out, the alpha needs to work for all values of n
@@dedekindcuts3589 thank you for your clarification. i wonder how many students will get this question right. a lot of the countries have chinese contestants. for example, in canada, almost always all 6 contestants are chinese. and half of them will go to school in the US.
@@leeriri This question should have many 7s, given that it is problem 1
Lovely problem 1, I'm glad the IMO chose it!
Hell no in my solution i used induction and limits. Dont ask me how cuz it was a pure insanity. At least im glad i used fractional part
xD whatever works still gets a 7!
Nice
Tell me App u use ?????
I use powerpoint
took me 40 to 45 mins, showed that kf = k-1 by induction :)😊
easy one honestly
It's problem 1 after all!
@@dedekindcuts3589 Easier than average for a problem 1 though