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Author Topic: Omnimath Challenge  (Read 18888 times)

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Offline apcalc

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Re: Omnimath Challenge
« Reply #15 on: December 26, 2010, 09:14:11 pm »
p(d)=0.20*(34-d)

Solve for x:
∫p(d)dd on the range from 0 to x = 50

p(d)=6.8-.2d
∫ p(d) dd = 6.8d-.1x^2

(6.8x-.1x^2) - 0 = 50
-.1x^2+6.8x-50=0

x=(-6.8±√(6.8^2-4*.1*50))/(2*-0.1)
x=8.388 or x=59.612

It will take a bit more than 8 days, so 9 days of uploading, assuming he started today, that would be January 3rd.
« Last Edit: December 26, 2010, 09:15:34 pm by apcalc »
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Offline Munchor

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Re: Omnimath Challenge
« Reply #16 on: December 26, 2010, 09:18:33 pm »
The Java answer was perfect for me, simple and good!
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Offline nemo

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Re: Omnimath Challenge
« Reply #17 on: December 26, 2010, 09:20:49 pm »
Quote from: ScoutDavid on December 26, 2010, 09:18:33 pm
The Java answer was perfect for me, simple and good!

i didn't know how to explain what i did in my code so i just copied it.
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Offline jnesselr

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Re: Omnimath Challenge
« Reply #18 on: December 26, 2010, 09:21:24 pm »
Interesting solution.
So, starting today, day N. N=34.
.2(N) is the amount of money made per day.
So, for day one is .2N
For day two, it is .2N+.2(N-1)=.4N-.2=.2(2N-1)
day three: .4N-.2+.2(N-2)=.6N-.6=.2(3N-3)
day four: .6N-.6+.2(N-3)=.8N-.12=.2(4N-4)
It's strange that from day two to day three, I get a jump from 2N-1 to 3N-3. This part doesn't make sense.
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Offline yunhua98

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Re: Omnimath Challenge
« Reply #19 on: December 27, 2010, 10:30:58 pm »
Problem:
How many games were lost today?

Spoiler For answer:
>9000!
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Offline Munchor

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Re: Omnimath Challenge
« Reply #20 on: December 31, 2010, 11:58:57 am »
Quote from: yunhua98 on December 27, 2010, 10:30:58 pm
Problem:
How many games were lost today?

Spoiler For answer:
>9000!
I wish my teacher doesn't put stuff like that in my next maths test.

Anybody has another problem?
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Offline apcalc

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Re: Omnimath Challenge
« Reply #21 on: December 31, 2010, 12:16:19 pm »
I think this one requires calculus (and a calculator, but we all have that ;)) to solve.  It was the only fair question I could think of. :(

The number of new members who join Omnimaga on any given day since January 1, 9001 is defined by the function m(t)=√(√(x)), where t is the time in days since 1/1/9001.  How many members will Omnimaga have at midnight on 2/1/9001, assuming that there were already 666 members on 1/1/9001?  Show your work, but you can use a calculator to do some of the math that is almost impossible to do by hand.  Have fun! :)
« Last Edit: December 31, 2010, 12:17:23 pm by apcalc »
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Offline squidgetx

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Re: Omnimath Challenge
« Reply #22 on: December 31, 2010, 12:31:52 pm »
A variation on the first problem

DJ's posting rate can be modeled by the equation posts per day= 30sin(days since Dec 31 2010)+30. He has 21362 posts on the 0th day after 12/31/10 (today).
Scouts's posting rate can be modeled by the equ posts per day= 70/{1+e^[(days since Dec 31 2010)+40]}. Scout has 2232 posts on the 0th day after 12/31/10

When will Scouts's number of posts exceed DJ's? You can give your answer in terms of days past Dec 31 2010 if you like

edit; apcalc ninja'd me with a different problem D:<

edit2:
Spoiler For apcalc:
I got 725 members after rounding to the nearest whole number while saying that 2/1 is 31 days after 1/1 :P At first I was like....how the *** do you take the antiderivative of that thing...but then I realized it was just t^(1/4) :P
« Last Edit: December 31, 2010, 01:27:33 pm by squidgetx »
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