#mathematics #olympiad #math International Mathematical Olympiad (IMO) 2024 Day 1 Solutions and discussion of problem 2 65th International Mathematical Olympiad Bath UK Problem 2 - Number Theory
Пікірлер: 47
@valentioiverson5992 ай бұрын
This is my problem! Great explanation and motivation :) Hope you enjoy the problem.
@JoPaskalis2 ай бұрын
Kerennnn🔥🔥
@dedekindcuts35892 ай бұрын
Amazing!!
@Linhkinhbrods2 ай бұрын
could u stop capping? everyone knows it's just a cap
@JoPaskalis2 ай бұрын
@@Linhkinhbrods no bro, he isnt lie. in my country, he is a gold medalist and absolute winner in our National Olympiad.
@bamsuth96502 ай бұрын
hes our gold medalist @@Linhkinhbrods
@shreyameswargupta924613 күн бұрын
It is a great and beautiful problem ❤❤✨️✨️
@이선근-n5hАй бұрын
This video was very helpful to turn back my sense of imo problems.
@zmaj123212 ай бұрын
q=ab+1 felt natural to me because of that lcm problem from the Japan Math Olympiad you posted recently (in that video, we set n+f(m)=mf(m)+1 to make gcd(n+f(m), m)=gcd(n+f(m), f(m))=1).
@dedekindcuts35892 ай бұрын
That's great to hear! Yeah doing similar problems before helps!
@PetersLabAviation2 ай бұрын
really clear explanation and solution! looking forward to turbo the snail :D
@dedekindcuts35892 ай бұрын
😂
@dedekindcuts35892 ай бұрын
I think this is a quite an elegant problem! At the same time, it is not impossible. What do you think?
@wesleydeng712 ай бұрын
Can be a little quicker in the end: if q | a-1 then a-1=0 since q>a.
@awkwardhamster85412 ай бұрын
i proved for (a,b) with gcd=1 , setting x=a+b , for every n=T*(ord of b modulo x)+1 , g=(a multiple of x) . And otherwise , g =1 . Then i had to prove for wlog a1 and I divided it into 2 cases being 1)k=a and 2) k
@wintersfan2 ай бұрын
I love Number Theory but failed to solve this in the IMO 😢
@Ephemeral_EuphoriaYT2 ай бұрын
Damn u went to imo?
@mihailgrecu6542 ай бұрын
But what if I LOVE *geometry* and am scared of NT ?
@dedekindcuts35892 ай бұрын
Let's just say this year's problem set isn't very favourable to those strong in geometry! Only a P4 geometry!!
@vukdjokic87982 ай бұрын
@@dedekindcuts3589 im never placing to the imo(got 4 years left) but there is no way id ever get this lucky
@shamilkhusnullin33632 ай бұрын
Drop maths, geometry is for nerds only
@vlv54342 ай бұрын
I found the solution in this way: let q>2 is prime and not divide a,b. We want q | GSD(n). Then a^n=-b mod q, b^n = (-a^n)^n = (-1)^n * a^(n^2) = -a , a^(n^2-1) = a^((n-1)*(n+1)) = (-1)^(n+1) mod q. It is easy to see that n=q-2 is ok, because a^(q-1)=1 mod q. Then b = -a^(q-2) = -1/a mod q. a*b = -1 mod q. We can use nay prime divisor of the a*b+1 as q.
@ayoubaassou43612 ай бұрын
nice solution but at the end q can be equal to 1 in case a = 0 or b = 0 . in reverse, for ( a , 0 ) ( 0 , b ) that works to .
@dedekindcuts35892 ай бұрын
a and b need to be positive. That's why I said verbally in the solution that q can be 1 or 2, but the only solution is q =2 with a=b=1.
@sagnikbiswas32682 ай бұрын
I had all the intuition you did until the 10th minute of the video. I just really struggled with developing a construction. Any insights how to improve on this?
@dedekindcuts35892 ай бұрын
The construction is the hardest part of this problem. Someone pointed out a similar past problem where you want the power to be -1 mod and -1 mod b, so you use ab-1. I think doing more problems can help!
@RSTATHER2 ай бұрын
Tried it for 2 hrs but didn’t quite get there, I am very pleased with how far I got however
@dedekindcuts35892 ай бұрын
As long as progress is made!
@justmath.15332 ай бұрын
I think this problem was too easy for p2. but it was very nice, it took me around 1 hour.
@dedekindcuts35892 ай бұрын
Yeah! I think it's pretty manageable and suitable as a great P2!
@theller2k3752 ай бұрын
My solution: Let’s take the gdc of the smallest value of n: gdc( a^1 + b, b^1 + a) = a + b Logically, a + b is the gdc so it has to divide every value of n in the values of a and b we want. a + b | a^n + b a + b | b^n + a => a + b | a^n + b^n + a + b => a + b | a^n + b^n => a + b | ba^(n-1) + b^n => a + b | b^2a^(n-2) + b^n ••• => a + b | 2b^n Let’s suppose that an and b aren’t equal. Rewriting a as b + x we have: 2b + x | 2b^n For n=1 2b + x | 2b Contradiction. Therefore, a=b. gdc( a^n + a, a^n + a) = g g = a^n + a The only value which is constant for every value of n is 1. So, a = b = 1.
@dedekindcuts35892 ай бұрын
The gcd only has to be *eventually* constant. It need not be constant from n=1. So even though a+b is the gcd when n=1, it does not mean a+b need to divide all the gcd for larger values of n.
@theller2k3752 ай бұрын
@@dedekindcuts3589 indeed. Thank you for pointing the mistake.
@narayansareekuthir53302 ай бұрын
Honestly I am preparing for first stage of imo but I solved problem 1 and 2 because I learnt only number theory
@dedekindcuts35892 ай бұрын
It's a good start!
@berlinisvictorious2 ай бұрын
Can you elaborate how q|a^N +b and q|b^N +a implies q|a^N+1 +b q|b^N+1 +a at 12:42
@il_caos_deterministicoАй бұрын
I suppose for hypothesis since for all n>=N the gcd is g and q divides g.
@tomykill52322 ай бұрын
Good problem
@dedekindcuts35892 ай бұрын
I agree!
@IneidSmz-qz4nq2 ай бұрын
But only a=b is necessary...... Need not be equal to 1 as g don't meant to be same in every case!!