So, who took the AMC?

[chickendude]

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)

[RandomPscho]

Re: So, who took the AMC?

SNOW DAY!

I should have taken the AMC 10 though.

[hi19hi19]

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?

[RubikRevolution]

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.

[chickendude]

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

[DDR_Jerred]

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.

[chickendude]

Re: So, who took the AMC?

I got the answer key today
I qualified =)

[DDR_Jerred]

Re: So, who took the AMC?

I got the answer key today
I qualified =)

Congrats!

[Megmo]

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. =/

[lord_carbo]

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.?

[shrimpy]

Re: So, who took the AMC?

I took the AMC 12.
Did anybody get that sequence problem with n=2 and 2007?

[hi19hi19]

Re: So, who took the AMC?

For the people who took 10A, could you tell me what subject is the most prominent on the exam? Like combinatorics, probability, geometry, etc.?

The AMC actually does a pretty good job covering all topics. This one was a bit light on probability and combinatorics (IIRCjust 2 problems, and both were easy). This one also seemed a bit heavier than usual on "set theory-ish" stuff. I can't really explain it any better; there were a bunch of tough problems involving finding the highest or lowest values a given set of rules will hold. The geometry problems were not particularly advanced, just long, leading to a high probablity of making a minor error.

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.

[RandomPscho]

Re: So, who took the AMC?

For the people who took 10A, could you tell me what subject is the most prominent on the exam? Like combinatorics, probability, geometry, etc.?

I took it last year, still havent taken the AMC 10 yet because of a snow day, but it was mostly geometry mixed with a lot of algebra. First few are always easy, then it gets hard. One question that I didn't know the math for last year was:

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.

[chickendude]

Re: So, who took the AMC?

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)

Little overconfident there? I have always been a huge math nerd as well, but time became an issue for me. I only got to problem 23.


@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

[lord_carbo]

Re: So, who took the AMC?

Little overconfident there? I have always been a huge math nerd as well, but time became an issue for me. I only got to problem 23.

No, I'm pretty big. I'm 100% sure I could have gotten over a 100 on the 10A.