I ranked 747th at a national math contest

[Artic_counter]

I finally received my results for the Canadian math contest I participated in and god I'm happy. I ranked 747th out of 8716 participants (It was an open contest) and my score is 58/80. I can't even describe how happy I am right now My only regret is that I didn't ranked high enough for the Olympiad but fuck that I'm happy xD

Here's one of the hardest question we got. I would love to see anyone solve it (rubix/leonid/aperson/iironic/remedy ?). Calculator isn't allowed.

[One Winged Angel]

hey, grats bro

I'll take a look at this more in-depth later, should be able to get it but eh, kinda rusty on my math now that I've gone three years without it
(nabbed first in Canada on a couple of the Waterloo national math contests back in high school)

[iironiic]

:O Congrats! The only thing I know about my national rankings is that I am the top 5% of participants in the nation! However, your ranking is much more impressive than mine because you have a harder exam based on the questions posted in your threads haha!

I'm waiting for my Putnam Exam scores and I'm hoping for at least 10 points.

And I"ll be working on that question so I'll brb

Again congrats to you!

[25thhour]

Grats man, I did one of these in grade 8 and I did realllllly bad. ;P

[Artic_counter]

Thanks all xD Well for the difficulty of the questions I'm not sure. It varies alot. There were piss easy questions but also really hard ones.

Wow, you guys are great xD I just started doing contests and stuff. It's actually the first time I did great lol

[remedy1502]

Was this the Euclid? I'm going to get a pizza then I'll go work on your problem. =D

[iironiic]

I think I have an answer to the first 1.5 questions. Still working on the second half of b and c.

EDIT: Got the second half of b I'll post my solutions right now:

a) Determine all x>0 such that f(x)=x.

This implies that f(x) - x = 0 or

x + 1/x - [x + 1/x] - x = 0. With algebraic manipulation, you get that 1/x = [x+1/x].

Note that the above equality is satisfied when 1/x is an integer because [] is the greatest integer function. So suppose x = 1/a for some integer a. This implies that

a = [a+1/a] if and only if 1/a < 1 or a > 1.

Therefore, x = 1/a for all integers a > 1.

Ok posting solution b right now. Just gonna organize my thoughts haha:

EDIT: b)

From part a, if x = f(x), this implies that 1/x = [x + 1/x]

If x = a/(a+1) for a > 1, then:

1+1/a = (a+1)/a = [a/(a+1) - (a+1)/a] -epsilon aka a member of- Z+ (the set of positive integers)

Contradiction.

If f(f(x)) = f(x), then f(x) must be in a form of 1/ something.

We need to prove that f(a/(a+1)) = 1/ something.

So apply the definition of the function:

f(a/(a+1)) = a/(a+1) + (a+1)/a - [a/(a+1) + (a+1)/a] = 1 - 1/(a+1) + 1 + 1/a - [1 - 1/(a+1) + 1 + 1/a]

= 2+1/(a(a+1)) - [2 + 1/(a(a+1))] = 1/(a(a+1))

Because f(a/(a+1)) = 1/(a(a+1)), then x = a/(a+1) with integers a>1 is the solution of f(f(x)) = f(x).

Working on c now.

[Artic_counter]

Originally Posted by remedy1502

Was this the Euclid? I'm going to get a pizza then I'll go work on your problem. =D

Nope it wasn't. Wont' say for now which one it is since the solutions are on the site. I want you guys to work :3

Originally Posted by iironiic

I think I have an answer to the first 1.5 questions. Still working on the second half of b and c.

I only got the first answer right. I was close to b (well, I had the idea going) and I was dumbstruck at c xD
Good luck xD

[Artic_counter]

There is a slight error in your solution for a)

at the end, it's not for a>1, it's : for a (superior or equal) to 2

a>=2 ?

God job though

[dag12]

I remember I did USAMO in high school. Haven't taken the putnam, though it's quite difficult from the past questions I've seen...

[iironiic]

Originally Posted by Artic_counter

There is a slight error in your solution for a)

at the end it's not for a>1, it's : for a (superior or equal) to 2

a>=2 ?

God job though

a = 2 works :P f(1/2) = 1/2 ;D

[Artic_counter]

yeah but f(2/3) =/= 2/3 xD

[iironiic]

Originally Posted by Artic_counter

yeah but f(2/3) =/= 2/3 xD

Haha x = 1/a for positive integers a>1 xD

[warriormag17]

I took the AIME back in 11th grade. Don't recall how I did. Meh i suppose. Was ranked second in Texas in Math/Science in state competitions for a while.

[Artic_counter]

Originally Posted by iironiic

Haha x = 1/a for positive integers a>1 xD

Oups xD

Gosh that's what happens when you dont read lol
I swear, I don't know how many points I lost because of that xD

Originally Posted by warriormag17

I took the AIME back in 11th grade. Don't recall how I did. Meh i suppose. Was ranked second in Texas in Math/Science in state competitions for a while.

I wish I could enter in american contest

Good job !

[iironiic]

Originally Posted by Artic_counter

Oups xD

Gosh that's what happens when you dont read lol
I swear, I don't know how many points I lost because of that xD

That's ok xD I tend to do that very often

For c) The only thing you need to show is that there exist a rational number u such that f(u) = something / (something + 1). I can't find the u .

[qqwref]

Nice job I was pretty good at competition math back in high school, I think the best I ever did was 30th~ in the USAMO. There were certain classes of problems I always had trouble with, though, like geometry and inequalities. I've taken the Putnam twice so far, but I don't really practice much anymore.

[Artic_counter]

Here are the answers to all of the questions in the test. Just scroll for the #4 in part B

http://www.math.ca/Competitions/COMC...tionse2010.pdf