AHAHahhashdhsfYAAAAAAAAYYYY

Today, at 11:56 am, by Thunder365

Oh Oh OHHH ITS MAAAGIC!!!
YOU KNOW-OH-OH....
NEVER BELIEVE ITS NOT SO!!!!

AHA W00T! So i can get a max of 30 points, but most likely a 25, which is WAY more than needed to get to the top 100, since last year a 24 on part 1 and 17 on part 2 got you in the top 35...


BUt Im sure that I got #1,3, and 4 correct after comparing with mankill... but since the test graders are weird and take off points for no reason, I will probably get a 9/0/9/7/0, even though my solutions for number 1 and 3 were very solid, and SHOULD earn me 10 points each

*knocks on wood*...
ARGHHHH number 2 was SUCH A COMMON QUESTION YET I NEVER BOTHERED TO LEARN THE STUPID CONCEPT Cursing Furious Censored Wallbash Red

So basically what happened was that I started the test, looked at all the questions, and wet my pants (not really).
Then for the next half an hour I managed to get absolutely nothing done.
But my luck changed at around the 45minute mark, as I finally got 1 problem solved! (number 4).
Then I looked at number 3, tried to do some mod bash but failed miserably, then went to number 1 and was like "hey this is really easy even though it looks tough." So I did 1a), then was like "wow 1b is hard" so I went back to number 3.
After doing some random crap, I realize the solution is extremely easy, so I got 3a), and then 3b) sorta followed easily.
I went back to 1b) and was like "oh hey I think I got it!"
I swore a bit after looking at number 2 and realizing that I had seen the problem so many times but never bothered learning the concept, so I left it blank.
Number 5 was apperently pretty easy, but I ran out of time somehow

A Poem

Yesterday, at 4:35 pm, by Thunder365

Twas the night before MMPC part 2
And all through the house,
THUNDER365 WAS FREAKING OUT
AHHHHHHHHHHHHH



yeah, that doesnt rhyme, w/e

Also, USAMTS questions came out...they look _____ as h3ll

EDIT: lol, I spoke too soon...... yay i wont fail!

ohhhh boyz

Mon Nov 30, 2009 6:15 pm, by the cliu

The Math Counts School Test is on December 9th for my school.

Bah

Mon Nov 30, 2009 5:42 pm, by Thunder365

All 20 vertices of a regular dodecdahedron are labeled with a different integer from 1-20. The numbers on the vertices of each face are then added up and written on the face itself. Show that it is not possible for all the faces to have the same number written on them.

just fyi, a dodecahedron has 12 faces, 30 edges, 20 vertices.
lol

Assume, for the sake of contradiction, that m is the number written on each face. Then 12m is the sum of all 12 faces.
We know that the sum of all the numbers on the vertices is 210. However, in a regular dodecahedron, each vertex is shared by 3 common faces. Hence, the sum of the numbers on all the faces is 210 \cdot 3 = 630. This means that 12m = 630 \implies m = 52.5. However this is a contradiction, since the sum of the five integers of the vertices of the face always sums to an integer.

Hi

Mon Nov 30, 2009 3:37 pm, by Thunder365

Let P(x) be a polynomial with integer coefficients such that P(0) and P(1) are both odd. Show that if c is an integer, P(c)\neq 0.

Blah
Let P(x) = a_nx^n + a_{n - 1}x^{n - 1} + \dots + a_1x + a_0.
Then P(0) = a_0, which means a_0 is odd...(observation 1)
We also have P(1) = a_n + a_{n - 1} + \dots + a_1 + a_0 Since P(1) is odd and a_0 is odd, the set {a_n, a_{n - 1}, \dots ,a_1} must consist of an even number of odd elements and an odd number of even elements, otherwise P(1) will not be odd...(observation 2)
Now we go back to the question at hand. If c is even then P(x) = a_nc^n + a_{n - 1}c^{n - 1} + \dots + a_1c is clearly even, so by observation 1, P(x) = a_nc^n + a_{n - 1}c^{n - 1} + \dots + a_1c + a_0 is odd.
If c is odd, then by observation 2, P(x) = a_nc^n + a_{n - 1}c^{n - 1} + \dots + a_1c contains an even number of odd elements and an odd number of even elements. This implies that P(x) = a_nc^n + a_{n - 1}c^{n - 1} + \dots + a_1c is even, and hence P(x) = a_nc^n + a_{n - 1}c^{n - 1} + \dots + a_1c + a_0 will be odd by observation 1. So p(c) will always be odd for integers c, and therefore P(c)\neq 0, as desired.

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  • ...dude

    By batteredbutnotdefeated, on Mon Nov 30, 2009 4:02 pm

  • nice story, cliu Rotfl

    By gauss1181, on Sat Nov 21, 2009 12:40 pm

  • In response to the -1=1 post:

    OH NO! Math is pointless, and AoPS is going to go down in 2 hours! Send your last PMs, everyone. Sad

    JK! Mr. Green

    By biblioman, on Mon Nov 16, 2009 9:08 am

  • This is Bob Smile
    Bob can jump Jump
    Bob can rofl Rotfl
    Bob can clap...
    Oops, Bob was injured real bad and will now be taken to the hospital and go through all of these random procedures which include CPR and AED and probably surgery. Stretcher

    Unfortunately, Bob dies and then comes back to life as the cliu.

    the cliu can jump.
    TO BE CONTINUED...

    By the cliu, on Thu Nov 12, 2009 7:11 pm

  • in response to the blog title:

    YES, I AM! Ninja

    By biblioman, on Wed Nov 11, 2009 2:25 pm

  • ty. Very Happy

    Calvin and Hobbes rocks

    By Thunder365, on Tue Nov 03, 2009 9:11 am

  • I like your background. Wink

    By westiepaw, on Mon Nov 02, 2009 7:24 pm

  • sure

    By Thunder365, on Sat Oct 24, 2009 4:00 pm

  • contrib? (cause you contribute to mine)

    By PowerOfPi, on Sat Oct 24, 2009 2:07 pm

  • hi peoples
    how do you contribute?

    By chickenpotpi, on Fri Oct 09, 2009 9:51 am

58 Shouts

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  • Joined: 27 Feb 2009
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