I ranked 747th at a national math contest

Re: I ranked 747th at a national math contest

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)

Re: I ranked 747th at a national math contest

: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!

Re: I ranked 747th at a national math contest

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

Re: I ranked 747th at a national math contest

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

Re: I ranked 747th at a national math contest

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

Re: I ranked 747th at a national math contest

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.

Re: I ranked 747th at a national math contest

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

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

Re: I ranked 747th at a national math contest

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

Re: I ranked 747th at a national math contest

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...

Re: I ranked 747th at a national math contest

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

Re: I ranked 747th at a national math contest

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

Re: I ranked 747th at a national math contest

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

Re: I ranked 747th at a national math contest

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.

Re: I ranked 747th at a national math contest

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

I wish I could enter in american contest

Good job !