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16 votes (50%)
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5 votes (15.62%)
10 votes (31.25%)
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32 members have voted

chevy
chevy
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December 29th, 2020 at 8:52:17 AM permalink
Quote: rsactuary

If I were Adam or Chris in this situation, I would vote against it, knowing that Bob and Dave for sure will vote against it. Then Ernie walks the plank, and they have a better chance of more coins. Seems to satisfy the requirements, yes?



If you are referencing my second answer, you can go back and look at my first attempt. I explain my reasoning there (albeit with a flawed final answer, hence the revision).

I think Adam or Chris would only get less in future rounds, so priority rule #1 would have them vote yes and accept Ernie's offer.
Wizard
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Wizard
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December 29th, 2020 at 1:04:59 PM permalink
Quote: rsactuary

If I were Adam or Chris in this situation, I would vote against it, knowing that Bob and Dave for sure will vote against it. Then Ernie walks the plank, and they have a better chance of more coins. Seems to satisfy the requirements, yes?



As a reminder, the correct answer for Ernie's suggestion, starting with Adam, is 2-0-1-0-997. It will get a vote of Y-N-Y-N-Y.

If Chris votes against it, then Dave would suggest 1-2-0-997, which would get a vote of Y-Y-N-Y. In that scenario, Chris gets nothing. He is better off voting yes to Ernie's suggestion and getting one gold coin.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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December 29th, 2020 at 1:14:01 PM permalink
In the following figure there are five rectangles, all of equal area, inscribed in a square. The short side of the yellow rectangle is 1. What is the area of each rectangle?

It's not whether you win or lose; it's whether or not you had a good bet.
chevy
chevy
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December 29th, 2020 at 2:24:32 PM permalink


Area of each rectangle = 16/5

a=width of pink
b=width of green
c=height of pink
d=height of green

Then

A(yellow)=A(pink)=A(green)=A(red)=A(blue)

1*(a+b) = ac = bd = (c-d)*b = (1+c)*(1+c-a-b)

Four equations (one for each equal sign). solve

a=16/15, b=32/15, c=3, d=3/2

Common area = ac = 48/15 = 16/5 (area of pink for example)

ThatDonGuy
ThatDonGuy
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December 29th, 2020 at 2:36:21 PM permalink

Let x^2 be the area of each rectangle; the area of the square is 5 x^2, so the area of each rectangle = sqrt(5) x.

The area of the yellow rectangle = 1 * the long side length = x^2, so the yellow triangle's long side has length x^2.
This means the short side of the blue rectangle has length sqrt(5) x - x^2.
The long side of the blue triangle has length sqrt(5) x, so its area = sqrt (5) x * (sqrt(5) x - x^2) = x^2.
sqrt(5) x^2 * (sqrt(5) - x) = x^2
sqrt(5) * (sqrt(5) - x) = 1 = 5 - sqrt(5) x
sqrt(5) x = 4
x = 4 / sqrt(5)
The area of each rectangle = x^2 = 16/5

Wizard
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Wizard
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December 29th, 2020 at 4:05:01 PM permalink
Both chevy and Don are correct! Well done gentlemen.

I got this one from one of my favorite YouTubers, Presh Talwalkar.


Direct: https://www.youtube.com/watch?v=lrND8AeL23s


Let a = Short side of the blue rectangle.

The long side of the yellow rectangle must be 4a, because the area of the sum of the yellow, pink, green, and red rectangles is four times the blue rectangle and both that sum of those four rectangles and the blue rectangle have the same height.

The area of the yellow rectangle is thus 1*4a = 4a.

The length of a side of the square is a + 4a = 5a.

The area of the whole square is (5a)^2 = 25a^2.

Five times the yellow rectangle equals the whole square:

5*4a = 25a^2

Divide both sides by 5a:

4 = 5a

a = 4/5.

The area of the whole square is thus 25*(4/5)^2 = 16.

The area of each rectangle is thus 16/5 = 3.2
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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December 29th, 2020 at 5:59:17 PM permalink
Here's a Slightly Tougher Math Problem: it does require some calculus.

You are asked, as a homework problem, "What is the largest number with Googol (i.e. 10^100) decimal digits?", and you decide that, rather than just say, "Googolplex - 1", you want to "write out" the answer to your hard drive - all 10^100 digits (all of which are 9s) of it.
If you start now, your hard drive can write 10^10 decimal digits per second, so it would take 10^90 seconds, or a little more than 3 x 10^82 years, to write the whole thing. However, as technology improves, if you wait, you can then use a faster hard drive. The speed of hard drives is increasing exponentially, and doubles every 10^8 seconds, which is a little more than 3 years. In other words, if you wait t seconds before starting, the hard drive speed will be 10^10 x 2^(t / 10^8) digits per second. Eventually, you will reach a point where any additional waiting time would be longer than the time saved by waiting.
What is the shortest amount of total time (waiting + writing) in which the number can be written in full?

Bonus question (some math if you want there to be, but mainly physics): what's the primary flaw with this idea?
gordonm888
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gordonm888
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December 29th, 2020 at 6:18:54 PM permalink


In general, the flaw is that the time to store information cannot decrease endlessly. A specific limit is the plank time (approximately 5.4x 10-44 seconds) at which the uncertainty in the information becomes larger than the information.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
ThatDonGuy
ThatDonGuy
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December 29th, 2020 at 8:32:08 PM permalink
Quote: gordonm888



In general, the flaw is that the time to store information cannot decrease endlessly. A specific limit is the plank time (approximately 5.4x 10-44 seconds) at which the uncertainty in the information becomes larger than the information.



10^100 decimal digits take 3.322 x 10^100 bits to store.
Since the radius of a sphere varies as the cube root of its volume, multiplying the radius of a sphere by 10^(100/3) multiplies the volume by 10^100, so the minimum possible radius of a spherical hard drive needed to store the entire number = 2.1544 x 10^33 times the radius of whatever is needed to store a bit.
Assuming the hard drive consists entirely of neutrons tightly packed together (assuming this is even possible), a neutron has a radius of 8 x 10^(-16) m, so the radius of the hard drive is 1.7235 x 10^15 km, or about 182 light years. Even if you could build one that large, it would take 182 light years for the last few digits to be written to the far edges of the drive.

chevy
chevy
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December 29th, 2020 at 8:33:12 PM permalink


T = total time = t + 10^100 / (10^10 * 2^(t/10^8)) = t + 10^90 * 2^(-t/10^8)

Set dT/dt=0
1+10^90 * (-1/10^8) ln(2) 2^(-t/10^8) = 0
1=10^82 ln(2) 2^(-t/10^8)

2^(t/10^8) = 10^82 ln(2)

(t/10^8) ln(2) = ln[10^82 ln(2)]

t = 10^8 ln[10^82 ln(2)] / ln(2)

t=271.869 * 10^8 sec.............this looks suspiciously close to e * 10^10........maybe rounding error in my calc?

So total time
T = 271.869 * 10^8 + 1/ln(2) * 10^8
T= 273.312 * 10^8 sec

T = 2.73312*10^10 sec ~ 866.667 years


Comments:
1) it is local minimum since ThatDonGuy said it was....and because this is less than with wait time = 0 and as wait time --> infinity
2) If my algebra is correct and write time reduces to 1/ln(2) * 10^8.....it suggest my algebra was not the most efficient and perhaps an easier simplification in solving for t exists.



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