02-6-2007, 09:22 PM | #1 |
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So, who took the AMC?
AMC 12A 2007 discussion
I took it today. Which ones couldn't you get? How did you do things? How do you do 25? I keep getting answers in the 2000s when the choices are in the 120s Overall, I thought it was of average difficulty (Though I've only taken 2 before)
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02-6-2007, 10:19 PM | #2 |
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Re: So, who took the AMC?
SNOW DAY!
I should have taken the AMC 10 though. |
02-6-2007, 10:24 PM | #3 |
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Re: So, who took the AMC?
I took the AMC 10. Missed qualifying for the AIME becuase of one silly little sign error.
6/9 = -2/3 Doh! Oh well, there's always the AMC 10B... I felt that this year was slightly harder than usual. I never got around to problem 25, but the one concerning paintbrush width (#19?) was very, very, long. Does anyone know an easier method of solving it other than the nitty-gritty method of coming up with a huge-ass equation?
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02-6-2007, 10:57 PM | #4 |
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Re: So, who took the AMC?
ok solving number 25 -
For those who didn't take it and don't know the question: Call a set of integers spacy if it contains no more than one out of any three consecutive odd integers. How many subsets of{1,2,3,...,12}, including the empty set, are spacy? (A) 121 (B) 123 (C) 125 (D) 127 (E) 129 First divide it into 5 cases: empty set, set of 1, 2, 3, 4; imposible for more than 4. Empty set is easy - just 1. set of 1; also easy enough - 12 for the rest this is like the "unfriendly customers" problem. Instead of seats customers sit in, you have numbers being choosen. With two numbers being choosen match one of the "customers" with two previous numbers so 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 Where the bold numbers are selected and the underline represents the groups. As long as the first group never intersects with the second group, and they stay in that order, you can shift the underlined selection (with bold at the end) without having two numbers less than 2 apart. This leaves us with 10 groups, 8 idenical not selected, and 2 which can't be swaped so: 10!/8!2! = 45 Likewise with three, there are now 8 groups, 6 idenical, 3 which can't be interchanged: 8!/3!5! = 56 and with 4: 6!/4!2! = 15 1+12+45+56+15=129 The answer is E. Last edited by RubikRevolution; 02-6-2007 at 11:00 PM.. |
02-7-2007, 04:08 PM | #5 |
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Re: So, who took the AMC?
I got 25 today (I didn't have enough time on the test)
I did it recursively though take a set of {1,2,..... n} we want to find spacy(n) in each subset: Either n is in or n is out If n is out: The problem reduces to spacy(n-1) If n is in: For the subset to be spacy, n-1 and n-2 must be out Then the problem reduces to spacy(n-3) I made the recursive function S(n) = S(n-1) + S(n-3) threw in some trivial seeds and got the answer 129 The reason I had been messing up was because I read the question wrong I read "No more than 1 out of any three consecutive integers" as "No three consecutive integers" silly me
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02-7-2007, 04:18 PM | #6 |
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Re: So, who took the AMC?
I took the AMC 10A yesterday... I thought it was harder than usual and only answered 18/25
I hope I qualified. |
02-7-2007, 04:57 PM | #7 |
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Re: So, who took the AMC?
I got the answer key today
I qualified =)
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02-7-2007, 04:58 PM | #8 |
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Re: So, who took the AMC?
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02-7-2007, 05:07 PM | #9 |
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Re: So, who took the AMC?
I'm pretty sure I said E for 25! =)
I only knew the answer or could make a well educated guess for nine of them. In the beginning I thought the questions were a joke but the test really got harder after the first ten or so. I didn't want to be penalized for guessing and end up with no points. But then again I suck at math for the most part and I kind of got screwed because there were some calculus related problems and I'm only a junior in precalc. =/ |
02-7-2007, 06:11 PM | #10 |
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Re: So, who took the AMC?
I should have signed up for it, but I don't recall ever getting a letter for it. I must have gotten it and shoved it aside without knowing what it was.
I would have AAA'd it, given my past history with discrete mathematics tests for minors and my current history of being a huge math nerd even in my spare time 8) For the people who took 10A, could you tell me what subject is the most prominent on the exam? Like combinatorics, probability, geometry, etc.?
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02-7-2007, 06:13 PM | #11 |
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Re: So, who took the AMC?
I took the AMC 12.
Did anybody get that sequence problem with n=2 and 2007? |
02-7-2007, 08:05 PM | #12 | |
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Re: So, who took the AMC?
Quote:
Just know basic geometry rules, and basic algebra. That will get you through at least problem 15. The last 10 are all experience and finding different ways of thinking about things.
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02-7-2007, 08:07 PM | #13 | |
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Re: So, who took the AMC?
Quote:
There are two circles, one bigger than the other. Two tangents are drawn connecting the circles, and then another tangent to the bigger circle to create a triangle. They gave you the radius of both circles and you had to find the are of the whole triangle. |
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02-8-2007, 08:20 PM | #14 | |
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Re: So, who took the AMC?
Quote:
@Shrimpy I didn't get it on the test, but I figured it out later Basically, between 0 and pi, the sine wave will repeat n times. Each mini sinewave will intersect the regular sin(x) wave twice (once on the up and once on the down) unless the two waves become tangent to each other at a peak or trough. You add up all the intersections with simple arithmetic series and then subtract the tangent exceptions. The two can potentially become tangent at pi/2 and 3pi/2 sin(nx) = 1 and x = pi/2 only when n = 1 + 4r sin(nx) = -1 and x = 3pi/2 only when n = 1 + 4r so whenever n = 1 + 4r, you have to subtract a pair from the big sum and you should get the answer
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02-8-2007, 08:37 PM | #16 |
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Re: So, who took the AMC?
oh
getting over 100 that's easy I thought AAA meant all correct That's generally what the DDR metaphor means where I'm from
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02-8-2007, 09:07 PM | #17 |
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Re: So, who took the AMC?
no u.
Although I would venture a guess that I could at least get over 130, given my past record with this kind-of stuff (22/25 on Math Olympiad in middle school, got a 2 on one test because I was seriously unfocused, hehe, I multiplied 4 and 4 and got 12 on one problem, for example. Also got 19/25 in elementary school which was the highest in the school).
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02-8-2007, 09:09 PM | #18 |
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Re: So, who took the AMC?
Take the test then corbo. The test is extremely hard if your not have not taken geometry and algebra since it is meant to be hard for those taking algebra 2.:
AMC 10B/12B -- Wed., Feb. 24, 2009 |
02-8-2007, 09:38 PM | #20 |
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Re: So, who took the AMC?
Learning is srs business,
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