i can see why this problem defeated so many olympiad solvers, couldn't imagine coming up with the solution myself. you explained it very well though, enjoyed solving this with you
What i like about your channel is that it opens my mind to think differently literally! i pause the vid to try and solve it myself but I find you solve it in another way!
Can you do a video on analytic continuation? Especially with the zeta function in the complex plain. I can’t wrap my head around, nor find any good videos on how the analytically-continued zeta function is derived
very clear answer, and this is also one of my favourite questions as it mainly involves prime factors and regular factors of positive integers! but I have a question that I dont understand. when you proved that f(a^n * b) is divisible by N, but dont we also have to prove that f(a1 ^ k1 * a2 ^ k2 * ... as ^ ks) is divisible by N in order to answer the question?
@@niom-nx7kb Hey! All we have to do is show that there exists an n such that f(n) will be a multiple of N for all positive integers N and we've already achieved this with just the case of f(a^k b). Choose any integer you like - let's do 30 for example but it works with any greater than 2. if N = 30, i'll choose n= a^28 * b f(a^28 * b) = (28+2)*2^27 = 30 * 2^27 And so it's a multiple of 30. Choosing n= a^(N-2)*b will yield a multiple of N for any possible case larger than N=2 (which we already dealt with) and so it's been proven!
I actually attempted this problem in the BMO! I believe I got 9/10. My solution was pretty similar to yours (I have it but it's difficult to share on TH-cam as it doesn't allow links to images). I can share the feedback I got for this kind of solution (which I believe is essentially what you did, but I made some arithmetic errors): Nice solution. Check for N=2 case carefully: you’re right to pay attention, but f(6)=3. f(4) is a better way to deal with it.
At the end I mentioned it, f(2^2) is the best way to go about the special case! Thanks so much the comment and that's great that you sat it in the real thing! Incredible to have gotten 9/10 under those conditions - which other questions did you attempt?
@@OscgrMaths I attempted questions 1, 2, 3 and 5. (Also if you're sitting any olympiads/STEP in the future etc: good luck! This olympiad was basically the only exam I ever had a good performance in lol)
@@expectus4673 I think that's the hardest thing about this question - finding a good starting place or plan of attack since it seems so bleak at first. Glad you enjoyed and thanks for the comment!
This problem is similar to the problem of finding the positive integers n such that n | φ(n) where φ(n) Euler's totient function, which are of the form: n= (2^m)*(3^l). Instead, in this problem we are counting all the lists that we can create from the set of all the divisors of n. So aren't we counting all the permutations we can have for the prime factors of n? So if the canonical prime factorization of n = (p_{1}^a_{1})*...*(p_{k}^a_{k}) Then what we are counting is a how many lists L are there such that: L = {p_{i}^a_{i} divides p_{i+1}^a_{i+1} | i in I} for I subset of I' = {1,2,...,k} Or am I completely wrong?
I just want to ask about at 10:20 using his way of swapping the upper row to the lower row of numbers, wouldn’t it led to double counting of the same sequence if the generalise it by 2^k-1 for swapping each time
That's great why do you have put ? There be confident in your answer . If you are not confident in your answer that means you it's the answer is not correct as per you . My maths always say be confident about your answer
@@Lagrange-x4u Hey, thanks for asking! I was really sad to but it was just because there was one part where I felt my explanation could have been better and I didn't want to confuse any viewers since that's the opposite of what I want to do with this channel!
the obvious recurrence for f(n) is on its face a trainwreck for trying to factor in any way, so almost right away we are reduced to trying to brute force a formula for f(n) in special cases, imo a disappointing line to have to take. all prime factors identical? too simple. all prime factors distinct? too complicated. FA til you FO, i guess. at first reading the problem is one of those olympiad immortals, but it kind of falls flat.
@@ytkerfuffles6429 That's fair! I thought it was very tricky - particularly question 6. Have you sat any other BMO papers? If so do you have any favourites? I'd love to go through another question on the channel.
The question asks: given any N, find n such that N | f(n) I think u only showed this for N of the form b*a^k where a,b are primes. This does not cover all integers! E.g. N=2*3*5
@@alexs3243 It does! In your example when N = 2*3*5 = 30, I would choose k = N-2 which is 28 and so n = a^28 b f(a^28 b) = (28+2)2^27 (by our conjecture that we proved) = 30*2^27 this is clearly divisible by 30. It works this way for any integer! This means whatever you choose N as there exists an n (n= a^N-2 b) that satisfies the condition.
There are lots of approaches to this question so feel free to share here!
Just commenting to say I'm actually the one that came up with this problem and I'm really touched by all the comments about it!
@@rhyslewis9057 Wow! It's an absolutely incredible problem and I've loved thinking about it - thank you. I hope I did it justice.
Just commenting to say How do we know you are telling the truth?
I honestly don't watch this kind of math but your teaching style encourages me to watch right till the end. Excellent work once again.
Bro thanks so much! I really appreciate that.
Maths 505 is here! Love your vids❤
@@Player_is_I thanks mate
Commenting to help you with the algorithm. This was a good one and I hope one day you have as many subscribers as Michael Penn; keep it up!
@@josephbrisendine2422 Thanks so much! Very kind of you.
i can see why this problem defeated so many olympiad solvers, couldn't imagine coming up with the solution myself. you explained it very well though, enjoyed solving this with you
@@riot1133 Thanks so much! It is very tough but a really interesting question all the same.
What i like about your channel is that it opens my mind to think differently literally! i pause the vid to try and solve it myself but I find you solve it in another way!
@@anas.aldadi Thanks so much! This is especially true for olympiad problems which could have lots of different constructions to reach the same result.
Interesting problem and nice and systematic explanation!
Thank you very much for making such a good video ❤
@@حسينالقطري-ب8ص Thank you!
Try some "JEE advanced"Questions, specially that probability stuff from there
as a person who just finished my first year of ChemE, I have no idea what 90% of that was but I still watched it
Can you do a video on analytic continuation? Especially with the zeta function in the complex plain. I can’t wrap my head around, nor find any good videos on how the analytically-continued zeta function is derived
very clear answer, and this is also one of my favourite questions as it mainly involves prime factors and regular factors of positive integers! but I have a question that I dont understand. when you proved that f(a^n * b) is divisible by N, but dont we also have to prove that f(a1 ^ k1 * a2 ^ k2 * ... as ^ ks) is divisible by N in order to answer the question?
@@niom-nx7kb Hey! All we have to do is show that there exists an n such that f(n) will be a multiple of N for all positive integers N and we've already achieved this with just the case of f(a^k b).
Choose any integer you like - let's do 30 for example but it works with any greater than 2. if N = 30, i'll choose n= a^28 * b
f(a^28 * b) = (28+2)*2^27
= 30 * 2^27
And so it's a multiple of 30. Choosing n= a^(N-2)*b
will yield a multiple of N for any possible case larger than N=2 (which we already dealt with) and so it's been proven!
@@OscgrMaths oh now I get it
Did you compete in the BMO?
I actually attempted this problem in the BMO! I believe I got 9/10. My solution was pretty similar to yours (I have it but it's difficult to share on TH-cam as it doesn't allow links to images).
I can share the feedback I got for this kind of solution (which I believe is essentially what you did, but I made some arithmetic errors): Nice solution. Check for N=2 case carefully: you’re right to pay attention, but f(6)=3. f(4) is a better way to deal with it.
At the end I mentioned it, f(2^2) is the best way to go about the special case! Thanks so much the comment and that's great that you sat it in the real thing! Incredible to have gotten 9/10 under those conditions - which other questions did you attempt?
@@OscgrMaths I attempted questions 1, 2, 3 and 5. (Also if you're sitting any olympiads/STEP in the future etc: good luck! This olympiad was basically the only exam I ever had a good performance in lol)
@@trueboxguy5421 I might do a video on question 6... I've not seen an olympiad problem like that before, I'm sure it confused most people in the exam!
@@OscgrMaths I'll subscribe to you (on my other account) so I can watch it if it comes out!
@@trueboxguy5421 Okay great!
Dont know how I ended up here but glad I did, I somewhat understood the proof but noway I could have thought of it lol
@@expectus4673 I think that's the hardest thing about this question - finding a good starting place or plan of attack since it seems so bleak at first. Glad you enjoyed and thanks for the comment!
Great vid❤
@@Player_is_I Thank you so much! I really appreciate it.
This problem is similar to the problem of finding the positive integers n such that n | φ(n) where φ(n) Euler's totient function, which are of the form: n= (2^m)*(3^l).
Instead, in this problem we are counting all the lists that we can create from the set of all the divisors of n.
So aren't we counting all the permutations we can have for the prime factors of n? So if the canonical prime factorization of n = (p_{1}^a_{1})*...*(p_{k}^a_{k})
Then what we are counting is a how many lists L are there such that: L = {p_{i}^a_{i} divides p_{i+1}^a_{i+1} | i in I} for I subset of I' = {1,2,...,k}
Or am I completely wrong?
yeah youre wrong mate
I just want to ask about at 10:20 using his way of swapping the upper row to the lower row of numbers, wouldn’t it led to double counting of the same sequence if the generalise it by 2^k-1 for swapping each time
PLEASE DO MORE BMO Qs!!!!!!
@@wjeksababakqabzzhzaab873 Will do!! Glad you enjoyed it!!
I dont get the question? isnt that always true?
N = 1 will always divide f(n) lol
@@uggupuggu Ah I see what you mean! You have to show that for any N, there will be a value of f(n) that N divides. Hope this helps!!
Let me ask you a very easy question if sec A + tan A= z then what will sec A - tan A = ??
1/z? since sec^2A - tan^2 A = 1 = (secA+tanA)(secA-tanA)
That's great why do you have put ? There be confident in your answer . If you are not confident in your answer that means you it's the answer is not correct as per you . My maths always say be confident about your answer
Let me tell you to PAY A TUTOR rather than ask an off topic question.
Why you deleted the jee video??
@@Lagrange-x4u Hey, thanks for asking! I was really sad to but it was just because there was one part where I felt my explanation could have been better and I didn't want to confuse any viewers since that's the opposite of what I want to do with this channel!
the obvious recurrence for f(n) is on its face a trainwreck for trying to factor in any way, so almost right away we are reduced to trying to brute force a formula for f(n) in special cases, imo a disappointing line to have to take. all prime factors identical? too simple. all prime factors distinct? too complicated. FA til you FO, i guess.
at first reading the problem is one of those olympiad immortals, but it kind of falls flat.
i proved this by induction on n=3x2^k when i sat this paper (:
@@ytkerfuffles6429 Nice! How did you find the other questions?
@@OscgrMaths i did question 1 because it wasnt that hard but the rest of them i did after because i didnt have time time
@@ytkerfuffles6429 That's fair! I thought it was very tricky - particularly question 6. Have you sat any other BMO papers? If so do you have any favourites? I'd love to go through another question on the channel.
@@OscgrMaths i sat this years one and i got 47/60 and 20/40 in the bmo2
@@ytkerfuffles6429 WOAH! Incredible. Huge congrats! Which was your favourite from this year? (including BMO 2)
The question asks: given any N, find n such that N | f(n)
I think u only showed this for N of the form b*a^k where a,b are primes. This does not cover all integers! E.g. N=2*3*5
@@alexs3243 It does! In your example when N = 2*3*5 = 30, I would choose k = N-2 which is 28 and so n = a^28 b
f(a^28 b) = (28+2)2^27
(by our conjecture that we proved)
= 30*2^27
this is clearly divisible by 30. It works this way for any integer! This means whatever you choose N as there exists an n (n= a^N-2 b) that satisfies the condition.
@@OscgrMaths My mistake, thanks for clarifying :)
@@alexs3243 No problem! Thanks for commenting with questions, I'm happy to have helped.
ASNWER=1(SIN-COS)(SIN+COS) ISIT