Mathcounts marathon!!!!

**Posted:** Thu Oct 15, 2009 12:54 pm

_________________
linux + macintosh $\implies$ li + mac = limac.
Q.E.D.

the cliu

An ancient philosopher was pondering one day what to call his newly discovered subject. He then realized that he wanted all of the letters of this new concept to be different, and four letters long. There just happened to be 26 letters in his alphabet at the time. What is the probability that he called his newly discovered subject "math"?

**Posted:** Sun Oct 18, 2009 7:17 pm

vliu

well since there are 26 letters in an alphabet you can get a total of 26*25*24*23 4 letter words. next since you can only spell math 1 way the answer in 1/(26*25*24*23) which is 1/ 358800

**Posted:** Sun Oct 18, 2009 7:23 pm

Thunder365

NP:
(my old one)
Six students are going to sit in a row for Countdown Round in Mathcounts. Thunder wants to sit next to cliu. Athunder does not want to sit next to Alex. The other two people can sit anywhere. How many ways can these six students be seated?

**Posted:** Mon Oct 19, 2009 1:14 pm

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I'm not dumb. I just have a command of thoroughly useless information. Mr. Green

the cliu

gahhh thunder why is ur problem so difficult?

New problem: sry thunder, ur problem was too hard Neutral

There are 2 students left in the Countdown Round. The first person to get to 3 points wins (lol ik this is wrong but didn't wanna make it too hard). cliu has a probability of $\frac {1}{4}$ to get a point against thunder, who has a $\frac {3}{4}$ probability of getting a point against cliu. Currently the score is tied 1-1. What is the probability that cliu wins?

**Posted:** Wed Oct 21, 2009 6:36 pm

Thunder365

The probability is 1...

jk, but in reality it would be 1.

So the outcome either goes CTC,CC,TCC.

1st-3/64
2nd-4/64
3rd-3/54
total 10/64=5/32

NP:
How many ways are there to roll 5 dice to get a sum of 10?

**Posted:** Thu Oct 22, 2009 3:21 am

_________________
I'm not dumb. I just have a command of thoroughly useless information. Mr. Green

the cliu

solution

I do not think that my solution is correct. Please identify any errors and correct them for me, thanks Smile

New problem

AndrewTom

Solution

New problem

**Posted:** Sat Oct 24, 2009 9:49 am

Thunder365

Solution:
We have:
$x^{4} - 4x^{2} + x + 2 = 0$
$x^2(x^2 - 4) + 1(x + 2) = 0$
$x^2(x - 2)(x + 2) + 1(x + 2) = 0$
$(x^2[x - 2])(x + 2) = 0$
so one solution is x=-2. We can also have:
$(x^2[x - 2]) = 0$
$x^3 - 2x^2 + 1 = 0$
Seeing that one solution is x=1, we can synthetic divide by 1 to get the following:
$x^2 - x - 1 = 0$ Which has solutions $\frac {1 + \sqrt {5}}{2}$ and $\frac {1 - \sqrt {5}}{2}$ , both of which are greater than -2, so our solution is $\boxed{x = - 2}$

NP:
Let $f(x)$ be a linear function for which $f(7) - f(3) = 16$ . What is $f(13) - f(3)$ ?

@ cliu: The answer you gave is slightly off. You can set up an equation: $a + b + c + d + e = 5$ (we have the sum as 5 not 10 because each dice MUST be at minimum a 1, so we get 10-5=5)
Now how many non-negative solutions does that equation have by balls and urns? And how did this not overcount the total solutions?

**Posted:** Sat Oct 24, 2009 10:17 am

_________________
I'm not dumb. I just have a command of thoroughly useless information. Mr. Green

AndrewTom

Solution

New problem

**Posted:** Sat Oct 24, 2009 10:59 am

Humbleman · New Member Offline Joined: 06 Jan 2008 Posts: 9

**Posted:** Mon Oct 26, 2009 7:58 pm

fishythefish

Andrea is 14, so Kelly is 36, because if Kelly were anything else, Andrea would not be 14.

Because Kelly is 36, Mark is 29.

So I don't know where that 25 came from, but Kelly is 36 and Mark is 29.

**Posted:** Tue Oct 27, 2009 12:42 pm

_________________
I am NOBODY. NOBODY is PERFECT. Therefore, I am PERFECT. Ninja

There are 3 types of people in the world. Those who can count and those who can't. maybe

ATTENTION ALL CALCOHOLICS!!! Know your limits. Don't drink and derive. Rotfl

AwesomeToad

Post a new problem.

NP

**Posted:** Wed Oct 28, 2009 8:18 am

gauss1181

$x+\frac{1}{x}=5$

$x^2+2+\frac{1}{x^2}=25$

$x^2+\frac{1}{x^2}=23$

Squaring that:

$x^4+2+\frac{1}{x^4}=529$

$x^4+\frac{1}{x^4}=527$

$(x^4+\frac{1}{x^4})(x+\frac{1}{x})=x^5+x^3+\frac{1}{x^3}+\frac{1}{x^5}=2635$

Now to find $x^3+\frac{1}{x^3}$ :

$x^3+3x+\frac{3}{x}+\frac{1}{x^3}=125$

$x^3+15+\frac{1}{x^3}=125$

$x^3+\frac{1}{x^3}=110$

$x^5+\frac{1}{x^5}=\boxed{2525}$

**Posted:** Wed Oct 28, 2009 10:40 am

_________________
My Olympiad Compilation! (plus some bits of randomness too)
Latest Problems Typed: 1979 ISL number theory problem

$LaTeX$

Post a new problem.

New Problem

**Posted:** Wed Oct 28, 2009 1:43 pm

_________________
$I,^{{\LaTeX}^{{Shall}^{{Pwn}^{ALL!}}}}$
Wallbash Red

DOPE!!! I FAIL AT EVERYTHING!!!
I hope this isn't Spam

!!!!! Please rate this post at least a $4$ . If not, please pm me and tell me why.

Thunder365

Solution:
If we have:
B_________E___S where bs=10 and eb=8, then es=2

NP:
7 red pegs, 6 blue pegs, 5 white pegs, 4 orange pegs, 3 yellow pegs, 2 black pegs, and 1 brown peg are place in the following spaces such that no two pegs of the same color are in the same row or column. How many ways are there to arrange the pegs?
DIAGRAM:
_
_ _
_ _ _
_ _ _ _
_ _ _ _ _
_ _ _ _ _ _
_ _ _ _ _ _ _

**Posted:** Wed Oct 28, 2009 2:39 pm

_________________
I'm not dumb. I just have a command of thoroughly useless information. Mr. Green

Scamper

Solution

NP

**Posted:** Wed Oct 28, 2009 3:43 pm

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If it's unsatisfactory, PM me on how to improve my posts!
$\mathfrak{\sim Scamper}$

$LaTeX$

How is this possible?

There are 65 people who have a candy.

There are 71 people who have chocolate.

Isn't chocolate a candy?

New Problem

**Posted:** Wed Oct 28, 2009 3:46 pm

_________________
$I,^{{\LaTeX}^{{Shall}^{{Pwn}^{ALL!}}}}$
Wallbash Red

DOPE!!! I FAIL AT EVERYTHING!!!
I hope this isn't Spam

!!!!! Please rate this post at least a $4$ . If not, please pm me and tell me why.

fishythefish

Assume that chocolate is not a candy.

Would it be 14? 7 for all common years (1 per day that Jan. 1st can fall on) and 7 more for the same thing with leap years?

Unless there is a day that cannot begin a year...

**Posted:** Thu Oct 29, 2009 12:39 pm

_________________
I am NOBODY. NOBODY is PERFECT. Therefore, I am PERFECT. Ninja

There are 3 types of people in the world. Those who can count and those who can't. maybe

ATTENTION ALL CALCOHOLICS!!! Know your limits. Don't drink and derive. Rotfl

Thunder365

**Posted:** Thu Oct 29, 2009 1:42 pm

_________________
I'm not dumb. I just have a command of thoroughly useless information. Mr. Green