The American Math Competition

For the discussion of math. Duh.

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DoomyDoom
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The American Math Competition

Postby DoomyDoom » Wed Feb 11, 2009 12:58 am UTC

Anyone else do this today?

Hard test is hard. I managed to get 13 questions answered in 75 minutes.

Censored
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Re: The American Math Competition

Postby Censored » Wed Feb 11, 2009 1:43 am UTC

Managed to do 19.5 points worse than last year :X

132 on the AMC12 went down to a 112.5. I'm getting too old for these multiple choice-type tests.

tonygoaliegreco
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Re: The American Math Competition

Postby tonygoaliegreco » Wed Feb 11, 2009 2:12 am UTC

Yea i did the AMC 12 this year as a senior. last year i got an 88.5 but this year i answered 18 q's with a high degree of certainty. Hopefully I will get through

DivideByZero
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Re: The American Math Competition

Postby DivideByZero » Wed Feb 11, 2009 2:49 am UTC

I answered 16, 15 of which I am certain.

Puffin
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Re: The American Math Competition

Postby Puffin » Wed Feb 11, 2009 3:11 am UTC

Mine got delayed, but I've looked through a practice test. I'm glad I'm not the only one who's struggling. Most of them just have some relationship that you'd never think to see.

Matterwave1
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Re: The American Math Competition

Postby Matterwave1 » Wed Feb 11, 2009 3:25 am UTC

How many questions were there total?
Last edited by Matterwave1 on Wed Feb 11, 2009 3:29 am UTC, edited 1 time in total.

Paiev
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Re: The American Math Competition

Postby Paiev » Wed Feb 11, 2009 3:28 am UTC

Matterwave1 wrote:How many questions were the total?


25.

I'm taking the 10B in fifteen days, looking for AIME qualification. I have fun with these competitions.

bizkut
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Re: The American Math Competition

Postby bizkut » Wed Feb 11, 2009 9:32 pm UTC

I was supposed to take it. I COMPLETLY forgot. >.>

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TheQntty
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Re: The American Math Competition

Postby TheQntty » Wed Feb 11, 2009 9:39 pm UTC

Well since feburary 10th is over, we can discuss the problems. Did anyone get number 24?

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Generic Protoplasm
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Re: The American Math Competition

Postby Generic Protoplasm » Wed Feb 11, 2009 11:36 pm UTC

I answered 14, got at least 1 wrong. It pissed me off.

What was #24?
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DoomyDoom
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Re: The American Math Competition

Postby DoomyDoom » Thu Feb 12, 2009 3:46 am UTC

It was one of the last problems, so it was one of the ones with 2009 in it. I remember one with a bunch of logarithms in it. I'm supposed to get my booklet back Soon™ so I'll see then.

tonygoaliegreco
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Re: The American Math Competition

Postby tonygoaliegreco » Thu Feb 12, 2009 4:17 am UTC

hey i have my booklet ill type out a question for you if you want but its in school. I can get to it soon probably. Tomorrow?

Matterwave1
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Re: The American Math Competition

Postby Matterwave1 » Thu Feb 12, 2009 5:19 am UTC

What's the average on these things? Are they super hard? or moderate? O.O

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qinwamascot
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Re: The American Math Competition

Postby qinwamascot » Thu Feb 12, 2009 7:00 am UTC

IIRC a score of 100 on the AMC 12 roughly corresponds to the top 5% of the 11-12th graders. A score of 120 is about the top 1% of 9th and 10th graders. In order to make the AIME, you have to pass either these scores or percentages.

Personally, I always found the AMC to be boringly easy. I qualified for the AIME all 4 years (scores of 131, 136, 125, 129 i think). The AIME is harder; my best score was 9/15 (barely made the USAMO, but couldn't take it on the given day :cry:) Personally, I never found the problems to be at all interesting or hard on the AMC; the people with the highest scores just memorized lots of "competition theorems" in euclidean geometry and number theory that proved useless everywhere else.
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Matterwave1
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Re: The American Math Competition

Postby Matterwave1 » Thu Feb 12, 2009 7:51 am UTC

So I'm assuming the max is 250? (10 points per question?)

Sounds hard based off the top scores :P

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qinwamascot
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Re: The American Math Competition

Postby qinwamascot » Thu Feb 12, 2009 8:06 am UTC

Matterwave1 wrote:So I'm assuming the max is 250? (10 points per question?)

Sounds hard based off the top scores :P


150. 6 Points per question. There's some partial credit if you don't answer; I think it might be 1.5 points, but I'm not sure.
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Matterwave1
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Re: The American Math Competition

Postby Matterwave1 » Thu Feb 12, 2009 8:32 am UTC

Oh, then it doesn't sound nearly as hard...based off these scores XD

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Generic Protoplasm
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Re: The American Math Competition

Postby Generic Protoplasm » Thu Feb 12, 2009 12:43 pm UTC

I remember a question. This one had a big number in it. It was

Define a function T(n) = 2^T(n-1) for n > 0. So it's basically a power tower of 2s that is n units high. Then define a number A such that A = T(2009)^T(2009). Then let B = A^T(2009).

Find the maximum m such that [imath]log_2 log_2 log_2 log_2 ... (m times) ... B[/imath] is a real number.
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Token
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Re: The American Math Competition

Postby Token » Thu Feb 12, 2009 1:24 pm UTC

m = T(2008)+1?

EDIT: Wait, that's for the repeated log to be an integer. For a real number, it's T(2008)+5.
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Generic Protoplasm
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Re: The American Math Competition

Postby Generic Protoplasm » Thu Feb 12, 2009 10:06 pm UTC

Ah, but all the answer choices were something you could write down, like in the range 2000-4000. It was nothing that huge.
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Re: The American Math Competition

Postby Token » Fri Feb 13, 2009 12:38 am UTC

Remove the T?
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antonfire
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Re: The American Math Competition

Postby antonfire » Fri Feb 13, 2009 2:02 am UTC

Generic Protoplasm wrote:I remember a question. This one had a big number in it. It was

Define a function T(n) = 2^T(n-1) for n > 0. So it's basically a power tower of 2s that is n units high. Then define a number A such that A = T(2009)^T(2009). Then let B = A^T(2009).

Find the maximum m such that [imath]log_2 log_2 log_2 log_2 ... (m times) ... B[/imath] is a real number.
Spoiler:
lg means log2. lgk x means lg applied to x k times, so that, e.g., lg4 x = lg lg lg lg x.

Note that lg T(n) = T(n-1). Also, I hate numbers and like generality, so take N=2009.

lg B = lg AT(N)= T(N) lg A = T(N) lg T(N)T(N) = T(N)2 lg T(N) = T(N)2 T(N-1)
lg lg B = 2 lg T(N) + lg T(N-1) = 2 T(N-1) + T(N-2)

This is strictly between T(N-1) and T(N) for N big enough (2009 is definitely big enough, work this out if you must), so taking N-1 more lgs, lgN+1 B is strictly between T(0) and T(1), i.e. 1 and 2. So one more lg makes it between 0 and 1, and another one makes it negative, at which point you can't take any more and have the thing be real. So, m=N+3=2012.


As for it staying an integer, lg lg B = 2 T(N-1) + T(N-2) = 21+T(N-2)+2T(N-3) is not a power of 2 unless 1+T(N-2)=T(N-3), which is certainly not the case for N=2009, so even lg lg lg B is not an integer.
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VDOgamez
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Re: The American Math Competition

Postby VDOgamez » Wed Feb 25, 2009 3:02 pm UTC

Why did they remove calculators? Every year, I got best in school, until they removed calculators. Then I only got very high scores on it. ;-;

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Re: The American Math Competition

Postby t0rajir0u » Wed Feb 25, 2009 3:15 pm UTC

If you have to use a calculator on any AMC problem, you're doing it wrong. None of the best solutions to any of those problems require the use of a calculator.

lingomaniac88
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Re: The American Math Competition

Postby lingomaniac88 » Thu Feb 26, 2009 3:23 am UTC

If you use an abacus, you may want to get in touch with 21st century technology. :P
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Poohblah
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Re: The American Math Competition

Postby Poohblah » Thu Feb 26, 2009 3:58 am UTC

Took it today. I qualified for AIME with a score of 103.5

My first year doing the AMC, too. The trickiest question was the last one, #25, you had to take into account that you can also use rotated squares (vertices not having the same x- or y-values as the adjacent vertices), otherwise there were only 85 possibilities.

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antonfire
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Re: The American Math Competition

Postby antonfire » Thu Feb 26, 2009 5:10 am UTC

lingomaniac88 wrote:If you use an abacus, you may want to get in touch with 21st century technology. :P
It's mod mental health awareness month here on the xkcd fora. If people seem to say strange things, it might be because of "wordfilters"[1].

[1] "wordfilters" filters to "awesomification"
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

mgcclx
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Re: The American Math Competition

Postby mgcclx » Fri Feb 27, 2009 11:36 pm UTC

antonfire wrote:
Generic Protoplasm wrote:I remember a question. This one had a big number in it. It was

Define a function T(n) = 2^T(n-1) for n > 0. So it's basically a power tower of 2s that is n units high. Then define a number A such that A = T(2009)^T(2009). Then let B = A^T(2009).

Find the maximum m such that [imath]log_2 log_2 log_2 log_2 ... (m times) ... B[/imath] is a real number.
Spoiler:
lg means log2. lgk x means lg applied to x k times, so that, e.g., lg4 x = lg lg lg lg x.

Note that lg T(n) = T(n-1). Also, I hate numbers and like generality, so take N=2009.

lg B = lg AT(N)= T(N) lg A = T(N) lg T(N)T(N) = T(N)2 lg T(N) = T(N)2 T(N-1)
lg lg B = 2 lg T(N) + lg T(N-1) = 2 T(N-1) + T(N-2)

This is strictly between T(N-1) and T(N) for N big enough (2009 is definitely big enough, work this out if you must), so taking N-1 more lgs, lgN+1 B is strictly between T(0) and T(1), i.e. 1 and 2. So one more lg makes it between 0 and 1, and another one makes it negative, at which point you can't take any more and have the thing be real. So, m=N+3=2012.


As for it staying an integer, lg lg B = 2 T(N-1) + T(N-2) = 21+T(N-2)+2T(N-3) is not a power of 2 unless 1+T(N-2)=T(N-3), which is certainly not the case for N=2009, so even lg lg lg B is not an integer.

actually. answer was 2013.

I got 93 on this one... epic failed it... I mean... non of the question are really that hard... If I was in a better condition. maybe like 120 is possible...

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scikidus
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Re: The American Math Competition

Postby scikidus » Sat Feb 28, 2009 3:27 am UTC

Gaah! How did I miss this thread?

I took AMC10A (when's 10B?), and got a 116.5, or 3.5 points under qualification. >_>

It's not my fault! Stupid mistakes...*wanders off muttering*

Then again, my friend and I got the top two scores in the school, which is downright sad.
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Mazuku
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Re: The American Math Competition

Postby Mazuku » Sun Aug 29, 2010 5:31 am UTC

lingomaniac88 wrote:If you use an abacus, you may want to get in touch with 21st century technology. :P


I would not knock the abacus, while they ain't as versatile as modern scientific calculators, the abacus is a great tool for giving your brain a good workout and if you use it enough you won't even need the abacus anymore, you will be able to add/subtract/multiply faster than you could with a calc.
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nash1429
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Re: The American Math Competition

Postby nash1429 » Sun Aug 29, 2010 5:40 am UTC

qinwamascot wrote:the people with the highest scores just memorized lots of "competition theorems" in euclidean geometry and number theory that proved useless everywhere else.


There are very few math competitions that I like for this very reason. My other problem is that I get stressed out by the time limit when I don't need to, often spending 5-10 minutes on a problem (that I could easily do in 1-3) just thinking about how I only have 5-10 minutes to do it.


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