GCD of Polynomials | Number Theory | Korea Junior Math Olympiad 2012 Q4
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- Опубліковано 10 лют 2025
- #NumberTheory #Polynomials #MathOlympiad
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.
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Seeing 43 = -7 makes seeing 42² = 49 = -1 easier
I'm obsessed with your channel and videos , plz upload more of em including some "floor" function related videos , that would be great !
Will see what can be done!
Hello!
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FALL IN LOVE WITH UR CHANNEL!! thank for your material which is really helpful for everyone!!! HOPE YOU WILL BE 4EVA here) wish u success !
Awesome video and explanation
Blanka Kövér Thank you very much!!
How do you know there are only 4 cases
Found it easier to do Mod(5) both side . it will concluded that n can be only 1,6,11,16,21 or n=5x+1 form. put it back in equation will get 5^l * 43^m = (5x) * (25x^2 + 15x + 3) . because the primeness .5^l = 5x and 43^m = 25x^2 + 15x + 3
00:00 a legend was born
Excelente
Awesome explained 👍👍👍👍
Superb explanation 👍.
Thank you!!
Can you kindly explain the line where you took the set of {1,43,-1,-43}mod(25)? What is the process there? Thanks 😊
Those are all the possible values of powers of 43 mod 25
@@letsthinkcritically why only 4 numbers are possible?
@@excited3134 43^2 is -1 mod 25 (as shown in video). multiply by 43 to get -43 mod 25. Then multiply again by 43 to get 43(-43) mod 25 which is -43^2 mod 25 which is -(-1) mod 25 = 1 mod 25. Multiply by 43 again gives us 43 mod 25, which is the beginning of the cycle.
So we have a cycle of 43,-1,-43,1 mod 25.
Nice
Very nicely explained
One small correction this is problem 3 not 4, i think.
Yes indeed, thank you for pointing that out!
HK? Yessir.
Hi Letsthinkcritically, I am back with my half baked solution :) Can you please guide me as to how move forward with this approach to provide a well rounded solution? By inspection we notice that the parity of n is even. This means n^3-1 divides 215.(this is because l and m are naturals,so the min value of LHS is 215). We notice that n = 6 holds for l=1,m=1. How do I proceed from here and consider other cases and prove that this is the only solution? (using the parity approach)? Please guide
I think this technique would quickly follow the one done in the video
Are there specifications somewhere for the IMO that state 0 is not in N? Or if you asked a proctor, would they be able to tell you? Most of the time I myself and classes I’ve been in have taken 0 to be a member of N, so I was wondering where the IMO people clarify that.
Natural numbers does not include 0
@@averyinterestingpineapple6038 This is a topic of general disagreement in the mathematical community, and any context in which N comes up should specify which convention is being used.
@@Deathranger999 I know, but I’m fairly confident they exclude 0 for the IMO, at least that’s what I read
@@averyinterestingpineapple6038 Ah, I see. Wonder where that’s stated. Possibly on the cover of the problems.
Please share your story lifestyle
(1,1,6)
Wo else thought it was 5^l * 4 * 3^m + 1 = n^3?
MY EYES HAHAHHA
here
i am always disappointed when a glaringly obvious solution turns out to be the *only* solution.
I dont inderstand why de have contradiction in the end of this vidéo
i think it's because from the last equation in the left that n²+n+1=43^m, and then from contradiction part, n²+n+1=43^m must be 3 mod 25, but we have in the end that 43^m are 1,43,-1, or -43 mod 25, so there's a contradiction and l never be greater or equals to 2
??
Nice