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The Ultimate Maths Competition Thread

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Original post by Maths465Man
No. I've not made it to any camps yet but next year I'm hoping to go to the Hungary camp and maybe even try for the Trinity camp (not too sure if that will happen though). I'm going into year 11 next year so I'm hoping to make the national summer school then.

Have you been to any? What are your Olympiad experiences and what year are you going into?


Really? I'm surprised. I'm in the year above you and went to the national mathematics summer school last year, and this year I'm going to the olympiad initial training camp. I'm not even that good, I got 30 in BMO1, 3 in BMO2 and a merit in Maclaurin lol.

Yeah if you keep on going like this I'm confident you'll go really far, plus you have loads of years left. Me on the other hand, I've been too lazy to do any proper olympiad practice this summer:biggrin: Probably should start soon so I don't look like an idiot at camp...
Original post by Aldrstream
Really? I'm surprised. I'm in the year above you and went to the national mathematics summer school last year, and this year I'm going to the olympiad initial training camp. I'm not even that good, I got 30 in BMO1, 3 in BMO2 and a merit in Maclaurin lol.

Yeah if you keep on going like this I'm confident you'll go really far, plus you have loads of years left. Me on the other hand, I've been too lazy to do any proper olympiad practice this summer:biggrin: Probably should start soon so I don't look like an idiot at camp...


Congrats on going to so many camps already :smile:
I'll just keep working on my olympiad maths this year and hopefully the camps will follow. Last year (this past year), I got to BMO2 so this year I want to make it to Trinity. I just want to make one stage every year. So year 11 I'll make trinity, year 12 I'll make tonbridge (I think it's called that) and year 13 I'll make the IMO. Oh well, I can dream :smile:
Definitely start practicing now. It's what this thread is designed for :smile:
It's quite easy but I thought I'd share it nonetheless. From the Oxford MAT:

Original post by Maths465Man
Congrats on going to so many camps already :smile:
I'll just keep working on my olympiad maths this year and hopefully the camps will follow. Last year (this past year), I got to BMO2 so this year I want to make it to Trinity. I just want to make one stage every year. So year 11 I'll make trinity, year 12 I'll make tonbridge (I think it's called that) and year 13 I'll make the IMO. Oh well, I can dream :smile:
Definitely start practicing now. It's what this thread is designed for :smile:


Wow, that's dedication, I salute you :smile: I know IMO is waay out of my league so my dream is EGMO, but then you get genius kids like that girl who also got on BBC Young Musician wth :mad: So yeah I'm just in it for the fun I guess
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Could anyone help with
Problem 1.1.12(Poland 2004).Find all natural n >1 for which value of the sum 2^2+ 3^2+......+n^2 equals p^k where p is prime and k is natural.

I got n=2,3,4 so far but need help with proving if p=-1 mod 2^m
(edited 7 years ago)
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Original post by 11234
Could anyone help with
Problem 1.1.12(Poland 2004).Find all natural n >1 for which value of the sum 2^2+ 3^2+......+n^2 equals p^k where p is prime and k is natural.

I got n=2,3,4 so far but need help with proving if p=-1 mod 2^m

No wait I think my method is wrong could someone help to solve the whole lot
Original post by Aldrstream
Wow, that's dedication, I salute you :smile: I know IMO is waay out of my league so my dream is EGMO, but then you get genius kids like that girl who also got on BBC Young Musician wth :mad: So yeah I'm just in it for the fun I guess

you should always aim higher than what you think is possible

Original post by 11234
Could anyone help with
Problem 1.1.12(Poland 2004).Find all natural n >1 for which value of the sum 2^2+ 3^2+......+n^2 equals p^k where p is prime and k is natural.

I got n=2,3,4 so far but need help with proving if p=-1 mod 2^m


what is the sum of consecutive squares as a product? then factorise
Original post by Aldrstream
Wow, that's dedication, I salute you :smile: I know IMO is waay out of my league so my dream is EGMO, but then you get genius kids like that girl who also got on BBC Young Musician wth :mad: So yeah I'm just in it for the fun I guess

Who tomoka.
Oh


Posted from TSR Mobile
Original post by gasfxekl
you should always aim higher than what you think is possible



I'm going for Fields medal by november.
Mark my words.


Posted from TSR Mobile
Original post by physicsmaths
I'm going for Fields medal by november.
Mark my words.


Posted from TSR Mobile


feasible
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Original post by gasfxekl
you should always aim higher than what you think is possible



what is the sum of consecutive squares as a product? then factorise


I did but then I just get stuck on p=-1 mod 2^k because one of them has to be even
Original post by 11234
I did but then I just get stuck on p=-1 mod 2^k because one of them has to be even

what? if you factorise you get two factors whose only gcd can be 13 so...
(edited 7 years ago)
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Original post by gasfxekl
what? if you factorise you get two factors whose only gcd can be 13 so...


sorry im so confused could you show working
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What score do i generally need to qualify to round 2
Original post by 11234
What score do i generally need to qualify to round 2


Mid 30s.


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Original post by 11234
sorry im so confused could you show working

∑i=2ni2=n(n+1)(2n+1)−66=(n−1)(2n2+5n+6)6\sum_{i=2}^{n} i^2 = \frac{n(n+1)(2n+1)-6}{6} =\frac{(n-1)(2n^2+5n+6)}{6}
and the gcd of these two factors divides 13, so is at most 13, so can this be a prime power for n sufficiently large?
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Original post by gasfxekl
∑i=2ni2=n(n+1)(2n+1)−66=(n−1)(2n2+5n+6)6\sum_{i=2}^{n} i^2 = \frac{n(n+1)(2n+1)-6}{6} =\frac{(n-1)(2n^2+5n+6)}{6}
and the gcd of these two factors divides 13, so is at most 13, so can this be a prime power for n sufficiently large?


THanks very much could you give me a hint on bmo1 2010/2011 q3
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Original post by 11234
THanks very much could you give me a hint on bmo1 2010/2011 q3


Never mind didnt read the question sorry...
Reply 178
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could anyone help with bmo1 1991 q6
Reply 179
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could someone check my answer for bmo2 1996 q1 I got x,y,z = 4,2,5 and 0,1,2 3,0,3 as the only solutions

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