did anyone take the AMC math test today?

[MikeMike]

i was stumped. had NO clue wtf was going on. answered like 13 questions out of 25. here is a question... (you were allowed to keep the test sheets)

The polynomial X^3 - 2004x^2 + mx + n has integer coefficients and three distincet positive zeros. Exactly one of these is an integer and it is the sum of the other two. How many values of n are posible?

250000
250250
250500
250750
251000

i had no idea.

MIKE

 

[BigJ]

Had no idea on that question. I wound up getting a 99.5 though. Highest in my school, considering we didn't prepare for the AMC and really didn't care about lines tangent to a circle when we need to review for an AP test in May.

 

[DrPizza]

Originally posted by: nourdmrolNMT1
i was stumped. had NO clue wtf was going on. answered like 13 questions out of 25. here is a question... (you were allowed to keep the test sheets)

The polynomial X^3 - 2400x^2 + mx + n has integer coefficients and three distincet positive zeros. Exactly one of these is an integer and it is the sum of the other two. How many values of n are posible?

250000
250250
250500
250750
251000

i had no idea.

MIKE

My first impression is that the logic is almost, but not quite complicated... grabbing a pencil now. brb.

I have gone in 3 different directions on that problem... but have to think each one through a little more.
Grr.!

 

[DrPizza]

nope.
yup.
Darn it, changed my mind again. I keep forgetting simple logical things.
Since all 3 are positive, call the roots A,B,C. A<B<C. And, A+B=C
Now, if one of the roots is an integer, the other two roots have to occur as a conjugate pair, such as 5 + sqrt(3) and 5 - sqrt(3).

It's necessary for the conjugate pair to be added together to cancel out the irrational part in order to sum up to an integer. Thus, A and B need to be irrational roots and C is the integer root.

Factors would be
(x-A)(x-B)(x-C) where A+B = C, but A&B are irrational. (and the first two factors would have to be factored over the irrational numbers)

multiply it out
x^3 - Ax^2 - Bx^2 + ABx - Cx^2 + ACx +BCx - ABC which must equal
x^3 - 2400x^2 + mx +n

So, - A - B - C = - 2400 therefore A+B+C = 2400
-ABC = n
and
AB + BC + AC = m

 

[DrPizza]

more: and this led to a duhhhh moment. if A + B + C = 2400 and A + B = C, then C + C = 2400 and C = 1200.

And, if A + B are conjugate pairs as c + sqrt(d) and c - sqrt(d) then c + c = 1200
Therefore, A and B are in the form 600 +/- sqrt(d)


uh, before I go on, how much time should I have been able to solve this problem within? I believe I am on the right track now, but I may be going from NY to Boston via Los Angeles...

 

[DrPizza]

anyway, continuing, the revelation A and B are in the form of 600 + sqrt(d) and 600 - sqrt(d)
leads to the complete factorization:
(x - 600 + sqrt(d)) (x - 600 - sqrt(d)) (x - 1200) = x^3 - 2400x^2 + mx + n
First two are conjugate pairs (x-600) +/- sqrt(d)
so
((x-600)^2 - d) (x - 1200)
x^3 - 1200 x^2 + 360000x - dx - 1200x^2 + 1440000x - 4320000000 + 1200d

1800000 - d = m
and
1200d - 4320000000 = n

Neither of these restricts the value of d, however, for every value of d, there exists one value of n

 

[DrPizza]

So, lastly, since the roots are all positive,
600 - sqrt(d) must be positive (and d must be positive to have 2 other zeroes rather than imaginary roots.)
600 - sqrt(d) > = 0 (I know it can't be equal, I'll take care of that restriction in a minute)
So,
sqrt(d) < = 600
d < = 360000

that leaves 360,000 possibilities

However, d can't be a perfect square either. that eliminates 1^2 and 2^2 up to 600^2... 600 of those 360,000 are eliminated.
So, final count is 359,400.
.
.
.
which isn't one of the choices.
dammit.

 

[Noirish]

argh!!!
so much brain power coming from DrPizza.

 

[LordMorpheus]

I got a 109.5 on the Junior/Senior test. Enough to have to take one mour huge test. Tied for third in school (of course, two of my classmates are in Calc 3 over at Washington University . . . ) We never bother to prepare. The classes I've had are Algebra II, Adv. Geometry, PreCalc and I'm in AP Calc BC. about 10 more of my classmates got >100 on the test.

I did better last year, but it doesn't really count for anything because unless you can score a 140 or better, you aren't going to any of the higher level competitions.

 

[DrPizza]

sh!t sh!t sh!t

Can anyone find where I made a mistake in logic? What'd I forget? This is seriously pissing me off.

I even went back to the point where I had -ABC = n,
so, -AB = n/C
and -AB = n/1200
and -1200(AB) = n
so -1200(600 + sqrt(d)) (600 - sqrt(d)) = n
= -1200(360000 - d) = n, which agrees with late in my solution...
Hmmm.
SOB what did I do wrong?

 

[DrPizza]

If the OP put down the wrong multiple choices, I'm driving to Ohio and beating him into a pulp.

 

[PowerMacG5]

I took the AMC 12 competition almost 3 or 4 weeks ago now. We did, however, get our scores back today. I got a 101, second highest in my school for that grade level. The highest was a 110, and he is in my calc class. Surprisingly, he is number 3 in the grade, and I am number 72 (as of the beginning of the year, but I believe I am closer to 40-50 now). I was shocked when I broke 100 because last year I don't think I got above a 70. I was also shocked, because like people 1-10 in rank are in my calc class, and I beat all of them, but number 3. The reason my rank sucks so much is because I sucked at history and english the previous three years, but this year I am much better because of the teachers I have. Good luck, and let us know how you score. The next test I take in relation to this is the AIME, and I hear it's brutal. Also, does anyone know if this test will help me get into college because I am already a senior, or am I doing it just for fun? Because it would be nice if the colleges get a list of people who move onto the next level, etc...

 

[Kyteland]

DrPizza, you are assuming that D must be an integer. What if D is .5?

n = 1200*(600*600-.5)

This is sstill an integer

Edit: nm.

m = 1800000 - d, which would not be an integer

 

[DrPizza]

d has to be an integer, because
1800000 - d = m
and m is an integer.

lol, edit, I didn't read your edit before I instantly went to the quick reply box. Never mind this post

 

[DrPizza]

bumpity, 'cause I know you ATOTers need something to get those synapses firing on the weekend
With the number of .99999repeating = 1 people out there, I figured we'd have an answer by now.

(and with the number of .99999repeating doesn't equal 1 people out there, I also figured we'd have plenty of other answers to sort through such as "uhhhhh 4.2" )

 

[sash1]

I took it last week, the AMC-12. Possibly the easiest math test ever...

`K

 

[BigJ]

Originally posted by: Kauru
I took it last week, the AMC-12. Possibly the easiest math test ever...

`K

Not a very good joke. Probably a 2/10.

 

[dds14u]

Is this AMC a high school test or a college test...I'm in high school and I've never heard of it...I just know about AP tests lol.

 

[BigJ]

Originally posted by: dds14u
Is this AMC a high school test or a college test...I'm in high school and I've never heard of it...I just know about AP tests lol.

It's a HS test usually reserved for students in advanced mathematics classes. Only problem is, it doesn't really test Calculus concepts. I really don't give a damn about finding the length of a circle's tangent line(and other topics that we haven't covered since 9th grade) when I have a quiz on finding Volumes of Cross-sectional Areas.

 

[dds14u]

I can relate to you on the area/volume of cross sectionals hehe, just took a test on that...weird though I'm in AP calc and I've seriuosly never heard of it...

 

[amoeba]

I got like 114 on the AMC back when I was in high school. Then I took the long 3 hour 10 question test after that and that just blew my mind. The 10 question test is not multiple choice. all answers are integers between 0 and 9999. I think I got 1 out of 10. I think the test was called the Aimee?

Supposedly if you do well enough on the 10 question test, you get on the American Math Olympic team.

 

[amoeba]

The trick on AMC is to not answer everything. I answered like 15 and left the rest blank.

 

[PowerMacG5]

Originally posted by: amoeba
The trick on AMC is to not answer everything. I answered like 15 and left the rest blank.

Exactly, I think I answered like 13 or 14 questions and got a 102. Also, the next step is the AIME, and then I am told its the ASME after that. I have never taken the AIMI but heard it's a crazy test.

 

[MikeMike]

i answered 13, and damn, i never knew this thread got any responses.

but i do not know my score yet, highest i could have gotten was a 103.

MIKE

 

[DrPizza]

nourdmrolNMT1,
Not only did the thread get responses, but it became a topic in HT

btw, if your multiple choices are incorrectly typed, I'm driving to Ohio and smacking you upside the head with a calculator.