The American Math Competition
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The American Math Competition
Anyone else do this today?
Hard test is hard. I managed to get 13 questions answered in 75 minutes.
Hard test is hard. I managed to get 13 questions answered in 75 minutes.
Re: The American Math Competition
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 choicetype tests.
132 on the AMC12 went down to a 112.5. I'm getting too old for these multiple choicetype tests.

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Re: The American Math Competition
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

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Re: The American Math Competition
I answered 16, 15 of which I am certain.
Re: The American Math Competition
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.

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Re: The American Math Competition
How many questions were there total?
Last edited by Matterwave1 on Wed Feb 11, 2009 3:29 am UTC, edited 1 time in total.
Re: The American Math Competition
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.
Re: The American Math Competition
I was supposed to take it. I COMPLETLY forgot. >.>
Re: The American Math Competition
Well since feburary 10th is over, we can discuss the problems. Did anyone get number 24?
 Generic Protoplasm
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Re: The American Math Competition
I answered 14, got at least 1 wrong. It pissed me off.
What was #24?
What was #24?
Life is like a rock candy, you never know which attribute you're gonna increase next.
Re: The American Math Competition
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.

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Re: The American Math Competition
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?

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Re: The American Math Competition
What's the average on these things? Are they super hard? or moderate? O.O
 qinwamascot
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Re: The American Math Competition
IIRC a score of 100 on the AMC 12 roughly corresponds to the top 5% of the 1112th 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 ) 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.
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 ) 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.
Quiznos>Subway

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Re: The American Math Competition
So I'm assuming the max is 250? (10 points per question?)
Sounds hard based off the top scores
Sounds hard based off the top scores
 qinwamascot
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Re: The American Math Competition
Matterwave1 wrote:So I'm assuming the max is 250? (10 points per question?)
Sounds hard based off the top scores
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.
Quiznos>Subway

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Re: The American Math Competition
Oh, then it doesn't sound nearly as hard...based off these scores XD
 Generic Protoplasm
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Re: The American Math Competition
I remember a question. This one had a big number in it. It was
Define a function T(n) = 2^T(n1) 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.
Define a function T(n) = 2^T(n1) 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.
Life is like a rock candy, you never know which attribute you're gonna increase next.
Re: The American Math Competition
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.
EDIT: Wait, that's for the repeated log to be an integer. For a real number, it's T(2008)+5.
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.
 Generic Protoplasm
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Re: The American Math Competition
Ah, but all the answer choices were something you could write down, like in the range 20004000. It was nothing that huge.
Life is like a rock candy, you never know which attribute you're gonna increase next.
Re: The American Math Competition
Remove the T?
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.
Re: The American Math Competition
Generic Protoplasm wrote:I remember a question. This one had a big number in it. It was
Define a function T(n) = 2^T(n1) 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:
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?
Re: The American Math Competition
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. ;;
Re: The American Math Competition
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.

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Re: The American Math Competition
If you use an abacus, you may want to get in touch with 21st century technology.
"It is common sense to take a method and try it. If it fails, admit it frankly and try another. But above all, try something."
 Franklin D. Roosevelt
 Franklin D. Roosevelt
Re: The American Math Competition
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 yvalues as the adjacent vertices), otherwise there were only 85 possibilities.
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 yvalues as the adjacent vertices), otherwise there were only 85 possibilities.
Re: The American Math Competition
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]}.lingomaniac88 wrote:If you use an abacus, you may want to get in touch with 21st century technology. :P
[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?
Re: The American Math Competition
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(n1) 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:
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...
Re: The American Math Competition
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.
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.
Happy hollandaise!
"The universe is a figment of its own imagination" Douglas Adams
"The universe is a figment of its own imagination" Douglas Adams
Re: The American Math Competition
lingomaniac88 wrote:If you use an abacus, you may want to get in touch with 21st century technology.
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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Re: The American Math Competition
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 510 minutes on a problem (that I could easily do in 13) just thinking about how I only have 510 minutes to do it.
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