Math

Useless information for AIME

Posted November 28th, 2008 by Mgccl
in

For AIME, there are always questions ask m+n or mn where m and n and relatively prime integers. It's easy to infer the only way to m or n = 0 is the other one is 1. The answer of AIME are always integers from 0 to 999.
Suppose m and n are distributed evenly as long as m+n<1000 and mn<1000. How is the distribution of m+n and mn? Because if m and n can't be found easily, an educated guess can increase the chance.

Case 1: For the m+n case, the amount of (m,n) pairs for x are \varphi (x)
Case 2: For the mn case, the amount of (m,n) pairs for x are 2^{p(x)}
Where \varphi (x) is the Euler's totient function,p(x) is the number of unique prime factors for x, I can't find any named function with that definition.

So for m+n one, it's better to chose a larger number, even a prime number.
For mn case, it's better to chose numbers with a lot of unique prime factors.

Case 3: It happens sometimes when m and n doesn't have any other requirement except m+n<1000. Then we have a linear relationship. The amount of times m+n = x is x-1.

Case 4: Sometimes there are Squarefree requirement for n, then it is still approximately linear.

A nice plot. Green = Case 4, Brown = Case 3, Purple = Case 2, Blue = Case 1 * 50
Then one just have to have a random number generator, and then convert it into a pick in the plot.

Small summation problem

Posted November 20th, 2008 by Mgccl
in

\sum _{i=1}^{n-2}\sum _{j=1}^{n-1-i} (n-i-j) = 2\sum _{i=a}^{b}\sum _{j=c}^{d} (n-i-j)
n is even.
Find a,b,c,d, where a,b,c,d are all linear combination of n/2, i and 1.

I stumbled though this problem because I was trying to solve a even harder one
2\sum _{i=a}^{b}\sum _{j=c}^{d} (n-i-j)f(i,j) = \sum _{i=1}^{n-2}\sum _{j=1}^{ n-1-i} (n-i-j)|f(i,j)| where f(i,j) = -f(j,i)

I don't know the difficulty of this question, but I don't have any technique for solving problem like this. So I did exactly what the question asked. Decompose the variables into linear combinations of n/2, i and 1 and brute force all the solutions.
2\sum _{i=k_1\frac{n}{2}+c_1}^{k_2\frac{n}{2}+c_2}\sum _{j=k_3\frac{n}{2}+m_1i+ c_3}^{k_4\frac{n}{2}+m_2i+c_4} (n-i-j) = \sum _{i=1}^{n-2}\sum _{j=1}^{ n-1-i} (n-i-j)

There are a lot of solutions. I put an restriction on all the coefficients. But I found a particular nice one:
2\sum _{i=1}^{\frac{n}{2}}\sum _{j=1}^{\frac{n}{2}-i} (n-i-j) = \sum _{i=1}^{n-2}\sum _{j=1}^{ n-1-i} (n-i-j)

And... I don't like my current MimeTex system. The output equations looks ugly.
I need to find other servers. Currently MathBin is nice and fast, but it is created to be a pastebin. Ahh. I want a delicate server. Or leech from CodeCogs LaTeX Equation editor. Wordpress one is good. but blocked in China.

Mathematica 7 released

Posted November 19th, 2008 by Mgccl
in

As a Mathematica user(MUHAHAHA, that's what's so good about having a parent work at a research institution. MUHAHAHAHA!!!), the release of Mathematica 7 is an exciting news.
Mathematica has becoming Photoshop for math. In the future I will edit photos by typing equations instead of mouse gestures. Love this thing.
Can't wait to get my hands on it...
While it is good. I can't help to notice Mathematica 7 is way bloated and a released too soon. Consider Mathematica 6 was released only a year ago. Does anyone really need speech output system for Mathematica? lol.

I can get my own copy if I gold USAMTS. The last problem for this round is quite hard consider I use the naive and obvious sum up all area and divide method. Still searching for nice identities.

Best xkcd comic not shown on xkcd

Posted November 8th, 2008 by Mgccl
in

This was drawn for The New Yorker's Cartoon-Off
I loved this one. In fact this is the best one I have seen recently(like the past 6 months).

Higher dimensional string as a 1 dimensional string in higher dimensional space?
That's what I thought. Correct me if I'm wrong.
2 dimensional manifolds can knot in 4 dimensional space right?

16th place on the iTest

Posted November 6th, 2008 by Mgccl
in

Today I stumble upon iTest website again, they have posted the results. Nice.
It's a major shock that 83 made my team into No. 16 overall and No.2 in NY.(If there is no more changes to the result due to some teams failed to submit their stuff correctly)
Wat...We are that good?
Sadly. Only top 10 in US get recognized and the first in each state get to be noticed.
If only I didn't just give up in the last hour and stop checking the answers. Since I had enough with 18 hour per day for 4 days and sleep deprived.
Yep. Since I wasn't in school I was working on it NON-STOP(except sleep)
I'm such a beast.

It gave me some hope and now I know where to place myself in the math community. Finally it happened I'm actually pretty good.

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