Omnimaga

General Discussion => Other Discussions => Math and Science => Topic started by: Munchor on December 26, 2010, 02:14:34 am

Title: Omnimath Challenge
Post by: Munchor on December 26, 2010, 02:14:34 am
 :evillaugh:

DJ Omnimaga:

21114 (24.753 per day)

2071 (29.600 per day)

Today is 26th of December

If both DJ and I keep at this rythm and there is no other influence in post numbers, what precise day will I have more posts than him??


I haven't solved it yet, good luck!
Title: Re: Omnimath Challenge
Post by: DJ Omnimaga on December 26, 2010, 02:19:15 am
We should maybe do it based on 2010 daily post rate but that would require database operations and stuff. I'm sure my average for 2010 is in the 50s :P
Title: Re: Omnimath Challenge
Post by: Munchor on December 26, 2010, 08:27:43 am
Quote from: DJ Omnimaga on December 26, 2010, 02:19:15 am
We should maybe do it based on 2010 daily post rate but that would require database operations and stuff. I'm sure my average for 2010 is in the 50s :P

What matters is the exercise, not the content. I have started solving it and think I have an answer!!
Title: Re: Omnimath Challenge
Post by: apcalc on December 26, 2010, 09:04:59 am
Exactly 3928.82 days. :P

On September 27, 2021!

Let sp=Scout's Total Posts
Let dp=DJ's Total Posts
Let d=total number of days (starting at day 0, right now)

sp=24.753*d +21114
dp=29.600*d + 2071

Find value for d where sp=dp, therefore:

24.753*d +21114 = 29.600*d + 2071
19043 = 4.847*d
d=3928.82
Title: Re: Omnimath Challenge
Post by: Munchor on December 26, 2010, 09:20:07 am
New Problem:

If graphmastur has 10 000 peanuts and each one costs 22 cents, how long would it take for him to make 1000 dollars if he sells 666 peanuts per day, and only works 2 days per week? Start a 1 January (it is a Sunday, and Sunday is the 1st day of the week).

Concerning the first challenge... I got to the number of days, but how'd you get the date?
Title: Re: Omnimath Challenge
Post by: apcalc on December 26, 2010, 02:53:41 pm
Quote from: ScoutDavid on December 26, 2010, 09:20:07 am
New Problem:

If graphmastur has 10 000 peanuts and each one costs 22 cents, how long would it take for him to make 1000 dollars if he sells 666 peanuts per day, and only works 2 days per week? Start a 1 January (it is a Sunday, and Sunday is the 1st day of the week).

Concerning the first challenge... I got to the number of days, but how'd you get the date?

I got the date by using an online date calculator. :P

Second Challenge:

First, I am going to assume that the two days that graphmastur works each week are Sunday and Monday.

(666 peanuts/day)*(2day/week)*(0.22 dollars/peanut)=293.04 dollars/week

(1000 dollars)/(293.04dollars/week)=3.41250... weeks

Three weeks + a bit less than half a week=6 days+1day=7 work days

Starting from January 1st, we go through three weeks entirely.  Then, the first day of the fourth week he works, he will cross 1000.  Assuming this day is a Sunday, it will be January 22.
Title: Re: Omnimath Challenge
Post by: Munchor on December 26, 2010, 03:19:25 pm
100 000 cents is 1000 dollars

666*2*x*22 = 100 000
29304x = 100 000
x = 3.41

x = weeks
weeks = 3.41

3 weeks and almost 1 day (0.82 of a day) to make 1000 dollars.
So, it would be January the 22nd!

I got there too apcalc, this one was easier than the other ??? I'm not very good at creating problems. Anyone want to post a harder challenge?
Title: Re: Omnimath Challenge
Post by: DJ Omnimaga on December 26, 2010, 03:23:05 pm
Quote from: apcalc on December 26, 2010, 09:04:59 am
Exactly 3928.82 days. :P

On September 27, 2021!

Let sp=Scout's Total Posts
Let dp=DJ's Total Posts
Let d=total number of days (starting at day 0, right now)

sp=24.753*d +21114
dp=29.600*d + 2071

Find value for d where sp=dp, therefore:

24.753*d +21114 = 29.600*d + 2071
19043 = 4.847*d
d=3928.82
Woah I expected something like 4-5 years O.O
Title: Re: Omnimath Challenge
Post by: Ashbad on December 26, 2010, 03:27:39 pm
I expected like 5 days... :)
Title: Re: Omnimath Challenge
Post by: jnesselr on December 26, 2010, 05:00:19 pm
Yeah, I make $146.52 a day (In the problem, that is). I love being included in these.

They are all simple algebra problems. I'll have to come up with something a lot harder later.
Title: Re: Omnimath Challenge
Post by: Munchor on December 26, 2010, 08:54:52 pm
Ashbad got a new link cable which costed him 50$, since Texas is evil.

He got it today. Today he uploads 34 games. Tomorow 33, the day after 32 and minus 1 every day until the day he uploads 0.

If he makes 0.20$ per each game he makes, when will he have made profit (>50$)?
Title: Re: Omnimath Challenge
Post by: nemo on December 26, 2010, 08:59:25 pm
jan 3rd
Title: Re: Omnimath Challenge
Post by: Ashbad on December 26, 2010, 09:02:35 pm
Lol never

Eh if only people liked my games then I could get money... :(
Title: Re: Omnimath Challenge
Post by: Munchor on December 26, 2010, 09:03:44 pm
Quote from: nemo on December 26, 2010, 08:59:25 pm
jan 3rd

Why?
Title: Re: Omnimath Challenge
Post by: nemo on December 26, 2010, 09:07:46 pm
Quote from: ScoutDavid on December 26, 2010, 09:03:44 pm
Quote from: nemo on December 26, 2010, 08:59:25 pm
jan 3rd

Why?

Code: [Select]
public class Solve{
public static void main(String[] args){
double sum = 0;
int i = 34;
for(; sum < 50; i--)
sum += i * .2;
System.out.println(Math.abs(i-34) + " days, including today");
}
}
Calendar (http://www.timeanddate.com/calendar/)
Title: Re: Omnimath Challenge
Post by: apcalc 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.
Title: Re: Omnimath Challenge
Post by: Munchor on December 26, 2010, 09:18:33 pm
The Java answer was perfect for me, simple and good!
Title: Re: Omnimath Challenge
Post by: nemo 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.
Title: Re: Omnimath Challenge
Post by: jnesselr 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.
Title: Re: Omnimath Challenge
Post by: yunhua98 on December 27, 2010, 10:30:58 pm
Problem:
How many games were lost today?

Spoiler For answer:
>9000!
Title: Re: Omnimath Challenge
Post by: Munchor 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?
Title: Re: Omnimath Challenge
Post by: apcalc 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! :)
Title: Re: Omnimath Challenge
Post by: squidgetx 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