Mathematics!!

[Artinum]

OK, that's a bit more logic-based than just mathematics and it's fairly well-known but let me say it.

 

Let us assume that there is an island where of all the inhabitants there are exactly k of them with blue eyes. No one however knows the colour of their eyes, no one can tell them and they have no reflective surfaces to see the colour of their eyes (no, not even the water). They can only see the colour of the eyes of all other inhabitants. The population is not very big and thus they all meet together early every night. However, for some completely crazy and random law whoever has blue eyes has to leave the island at midnight. One day a new person visits and right after midnight makes the comment "At least one of you has blue eyes". The person is honest so they're telling the truth. The inhabitants are all reasonable and law-abiding so as soon as someone finds out that they have blue eyes they leave the island at midnight. 

 

What's going to happen after they learn that at least one person has blue eyes?

 

In any normal society, one could simply ask around, or look. But these weirdos can't talk to each other. So the logical conclusion is thus - everyone looks at everyone else's eyes. If there is exactly one person with blue eyes, they will not find any others - and therefore they will know they have blue eyes, and will leave the island.

 

The problem comes when two or more people have blue eyes...

[AugustaAdaByron]

In any normal society, one could simply ask around, or look. But these weirdos can't talk to each other. So the logical conclusion is thus - everyone looks at everyone else's eyes. If there is exactly one person with blue eyes, they will not find any others - and therefore they will know they have blue eyes, and will leave the island.

 

The problem comes when two or more people have blue eyes...

 

Yes... and what happens then?

 

If every person can form this thought then if a person sees someone with blue eyes that didn't leave the island on the first night then this means that the blue eyes person sees someone else on the island that has blue eyes...

[Artinum]

Yes... and what happens then?

 

If every person can form this thought then if a person sees someone with blue eyes that didn't leave the island on the first night then this means that the blue eyes person sees someone else on the island that has blue eyes...

 

Of course. So if no-one leaves the first night, there are at least two people. If you saw just one blue eyed person, you are the other - and both would leave that night. If there are three or more people, still no-one will leave, and they will know there are three and they will find each other the next day. And so on.

[Artinum]

Or, more likely, people start wearing false colour contacts...

[GoldLynx]

OK Guys help me out here.  This ought to be a cakewalk for you.  

 

Say I have a product and I'm going to retail it.  My cost is $100.  I was taught that a 30% markup would be (cost*1.3) = $130.  I just got off the phone with a lady who told me that's wrong, and it should be (cost/.7) = $142.86.  

 

The second version doesn't make any damn sense to me.  

 

Our not-too-bright ladyfriend approached the problem this way:

 

I have an item with cost C (in dollars). After a 30% markdown, the item was $100. How much did it cost originally?

She supposed, in other words, that marking an item up 30% to a higher cost is the same as marking a higher-cost item down by 30%. So she just pretended $100 was a sale price, 30% off the original. After a 30% markdown the item cost only 70%, or .7, of what it did originally. So $100, her markdown price, must be 70% of C.

100=.7C

She divided both sides by .7 to get $142.86.

She was wrong. You solved the problem correctly. The point is that if you mark an object up by a certain percent, and then mark it down by that same percent, you will not get back to your original price.

 

*Stop here if you are satisfied and don't want a math-induced headache*

 

As I was typing that up, I realized that a series of markdowns and markups could be thought of as a sequence. I start with $100, mark it up 30%, and get $130. I mark this down 30% and get $91, because 30% of $130 is of course more than 30% of $100. Mark up 30% again; you're at $118.30. Markdown: $82.81. It's slowly falling! It's calculus! But will it approach zero or start settling down? Let's think about what's really happening. I'm repeatedly multiply by 1.3, by .7, by 1.3, by .7. On and on forever. But the order of multiplication doesn't matter, so let's group the numbers: (1.3)^n and (.7)^n. We want to do this forever, so we ask ourselves the limit as n approaches infinity of (100)(1.3)^n*(.7)^n. Bring the 100 outside: 100 times the limit as n approaches infinity of (1.3)^n* (.7)^n. But (1.3)^n(.7)^n = (1.3*.7)^n. 1.3*.7 = .91 or 91/100. Now we're at 100 times the limit as n approaches infinity of (91/100)^n. This is a limit of the form x^n. Since x is less than one, the limit goes to zero, and of course 100 times our zero limit is also zero. So an infinite number of 30% markups and markdowns on any starting price will give you a final cost of zero (free stuff!).

 

Will this always happen? Let's find out. The reason our limit went to zero was because 1.3*.7 was less than one. 1.3 and .7 are 1+/- 30%, or 1+/-0.3. Let's generalize the markup/markdown decimal as n. We have (1+n)(1-n). If this ends up being less than one, our cost will go to zero. If it ends up greater than one, the cost will go to infinity. Also, n must be less than one--it wouldn't make sense to mark an item down by, say, 120%. But what is (1+n)(1-n)? 1-n^2. For real values of n, this will always give us a value less than 1, and therefore a cost approaching zero. Interestingly enough, our problem works for any percentage, and it doesn't matter whether you start by marking the item up or down since we grouped the terms anyway.

 

Thoughts, anyone?

[GoldLynx]

This means that three chickens (twice the number) will lay three eggs in those one and a half days. In three days, therefore, three chickens will lay six eggs. So in three days, one chicken will lay a third that number, or two eggs. Two eggs in three days =2/3 egg/day*chicken.

 

 

Also, eggs per day-chicken is literally the best unit I've ever used.

 

[GoldLynx]

OK, that's a bit more logic-based than just mathematics and it's fairly well-known but let me say it.

 

Let us assume that there is an island where of all the inhabitants there are exactly k of them with blue eyes. No one however knows the colour of their eyes, no one can tell them and they have no reflective surfaces to see the colour of their eyes (no, not even the water). They can only see the colour of the eyes of all other inhabitants. The population is not very big and thus they all meet together early every night. However, for some completely crazy and random law whoever has blue eyes has to leave the island at midnight. One day a new person visits and right after midnight makes the comment "At least one of you has blue eyes". The person is honest so they're telling the truth. The inhabitants are all reasonable and law-abiding so as soon as someone finds out that they have blue eyes they leave the island at midnight. 

 

What's going to happen after they learn that at least one person has blue eyes?

 

Twist answer: there was only one blue-eyed person on the island. The new person saw that almost everyone on the island had horrible, ugly, muddy-poop-colored eyes. He said, "Well, at least one of you has blue eyes." They figured out what he meant and all looked at each other. The person with blue eyes saw that no one else had blue eyes. She left the island at midnight.

With the stranger.

They fell in love.

[Penguinmom]

OK, that's a bit more logic-based than just mathematics and it's fairly well-known but let me say it.

 

Let us assume that there is an island where of all the inhabitants there are exactly k of them with blue eyes. No one however knows the colour of their eyes, no one can tell them and they have no reflective surfaces to see the colour of their eyes (no, not even the water). They can only see the colour of the eyes of all other inhabitants. The population is not very big and thus they all meet together early every night. However, for some completely crazy and random law whoever has blue eyes has to leave the island at midnight. One day a new person visits and right after midnight makes the comment "At least one of you has blue eyes". The person is honest so they're telling the truth. The inhabitants are all reasonable and law-abiding so as soon as someone finds out that they have blue eyes they leave the island at midnight. 

 

What's going to happen after they learn that at least one person has blue eyes?

 

Case 1: 1 person has blue eyes.

Evening 1:

As Artinum said, "If there is exactly one person with blue eyes, they will not find any others - and therefore they will know they have blue eyes, and will leave the island."

 

Case 2: 2 people have blue eyes:

Evening 1: People with Blue eyes know that there is at least 1 person with blue eyes.  People without blue eyes know there are at least 2 people with blue eyes.

Evening 2: A person with blue eyes sees that the other person with blue eyes didn't leave on Night 1.  Therefore there must be at least 2 people with blue eyes. Since the blue-eyed person only sees 1 other blue-eyed person, they conclude that they, them-self, must have blue eyes.  The other person with blue eyes realizes the same, and they both leave at midnight.

 

Case 3: 3 people have blue eyes

Evening 1: People with Blue eyes know there are at least 2 people with blue eyes.  People without blue eyes know there are at least 3 people with blue eyes.  People with Blue eyes think either they're someone with non-blue eyes from Case 2, else there are 3 people with blue eyes, the 3rd person being themselves. 

Proof by contradiction:

Assume that all of the blue-eyed people think they have non-blue eyes (and therefore are in Case 2).  They will expect the other 2 people with blue eyes to leave at midnight after Evening 2.  However, when they meet on evening 3, they will see that the other 2 didn't.  Therefore, they realize they have blue eyes, and will leave at midnight after Evening 3.

 

... you know, I never understood it until I worked it out case-by-case with a proof by contradiction just now!  Cool!

 

By the way, I'm new here (just started this week), and I got my Bachelor's in Mathematics.

[AugustaAdaByron]

Welcome!

 

Yup, it's a proof of contradiction combined with induction, in order to prove it for any k.

You prove it for 1 person, assume it's true for k-1 and then prove it for k.

[GoldLynx]

Case 1: 1 person has blue eyes.

Evening 1:

As Artinum said, "If there is exactly one person with blue eyes, they will not find any others - and therefore they will know they have blue eyes, and will leave the island."

 

Case 2: 2 people have blue eyes:

Evening 1: People with Blue eyes know that there is at least 1 person with blue eyes.  People without blue eyes know there are at least 2 people with blue eyes.

Evening 2: A person with blue eyes sees that the other person with blue eyes didn't leave on Night 1.  Therefore there must be at least 2 people with blue eyes. Since the blue-eyed person only sees 1 other blue-eyed person, they conclude that they, them-self, must have blue eyes.  The other person with blue eyes realizes the same, and they both leave at midnight.

 

Case 3: 3 people have blue eyes

Evening 1: People with Blue eyes know there are at least 2 people with blue eyes.  People without blue eyes know there are at least 3 people with blue eyes.  People with Blue eyes think either they're someone with non-blue eyes from Case 2, else there are 3 people with blue eyes, the 3rd person being themselves. 

Proof by contradiction:

Assume that all of the blue-eyed people think they have non-blue eyes (and therefore are in Case 2).  They will expect the other 2 people with blue eyes to leave at midnight after Evening 2.  However, when they meet on evening 3, they will see that the other 2 didn't.  Therefore, they realize they have blue eyes, and will leave at midnight after Evening 3.

 

... you know, I never understood it until I worked it out case-by-case with a proof by contradiction just now!  Cool!

 

By the way, I'm new here (just started this week), and I got my Bachelor's in Mathematics.

 

That was complicated, but I followed you! Impressive! I wasn't even going to begin to try figuring it out.

[Sam Ashen]

Inspired by the huge prize in the Power Ball, how about....

 

Some lottery has a one in 1.0 gazillion probability of winning the grand prize.  Coincidentally, 1.0 gazillion tickets are sold, exactly the inverse of the odds of winning.  Assuming each ticket sold is a random combination, approximately what is the probability that no sold ticket wins the grand prize?

 

Note 1:  One Gazillion is not defined.  Rest assured, it is a very large number.

Note 2:  The odds of winning the Power Ball = 26 x 69 x 68 x 67 x 66 x 65 / 120 = 292,201,338.

 

Pretty simple, but fun!

[Gainsdalf the Whey]

[(1.0 gazillion-1)/(1.0 gazillion)]^1.0 gazillion

Sent from my iPhone using Tapatalk

[Sam Ashen]

That should be pretty close to exp(-1) or 36.79%.  

[Artinum]

The odds of a winning ticket in that lot are compounded by there being any number of winning tickets. It's therefore easier to consider the case where no tickets win, and invert the odds.

 

Odds of winning - 1/x (as x is easier to type than "gazillion").

Odds of not winning (including runner up prizes) - 1-1/x, or (x-1)/x.

Number of tickets - x

 

Odds of no winning tickets - ( (x-1)/x) ^ x, or (x-1)^x / x^x.

 

In a draw with odds of one in four, the chance of four tickets all losing would thus be (4-1)^4 / 4^4, or 81/256 (about 31.6%). In other words, at least one of those four tickets will be a winner a little over two thirds of the time.

 

Odds of one in eight? 7^8 / 8^8 = 5764801/16777216 = about 34% likelihood of no winning tickets (or, again, you'll get at least one winner two thirds of the time).

 

My calculator is not capable of working out the Powerball odds accurately, but it does seem to come close to 36.79% likelihood of no winners. Unfortunately my logarithm days were well over a decade ago!

[GoldLynx]

Did you know the Russians multiply differently than we do?

Write the numbers side by side. For the number on the left, continually divide by two, writing the answer below and ignoring remainders, until you get to 1. For the one on the right, multiply by two until your columns are of equal length. Say it's 19x44.

 

19        44

9          88

4        176

2        352

1        704

 

Now, paying attention to the column on the right, cross out any numbers opposite an even number on the left.

 

19       44

9         88

4       176

2       352

1       704

 

Add up what's left on the right: 44+88+704=836. And voila, 19x44=836.

It's awesome (though tedious), but does anyone know why this works?

[Penguinmom]

Did you know the Russians multiply differently than we do?

Write the numbers side by side. For the number on the left, continually divide by two, writing the answer below and ignoring remainders, until you get to 1. For the one on the right, multiply by two until your columns are of equal length. Say it's 19x44.

 

19        44

9          88

4        176

2        352

1        704

 

Now, paying attention to the column on the right, cross out any numbers opposite an even number on the left.

 

19       44

9         88

4       176

2       352

1       704

 

Add up what's left on the right: 44+88+704=836. And voila, 19x44=836.

It's awesome (though tedious), but does anyone know why this works?

 

Hmm... working backwards, the numbers that are added in the rightmost column can be expressed as 44(2⁰+2¹+2⁴) = 44(1+2+16) = 44(19).

 

Or, another way of putting it would be 19 = 1(2⁰) + 1(2¹) + 0(2²) + 0(2³) + 1(2⁴).

 

 

Regarding the left column...

I can see that when n mod₂ = 0 on the left, the coefficient is 0 for the right, and likewise if n mod₂ = 1 on the left, the coefficient is 1 for the right.

 

It seems like an interesting trick to convert a number into binary, but I'm not quite sure why it works.

[Artinum]

x * y = x/2 * 2y = x/4 * 4y and so on. If x is a power of two, you'd end up crossing off all the ys but the last one...

No change. Substitute x' = x/2 and y' = 2y.

 

If x=2n+1, however, the next x' = n = (x-1)/2.

x * y = (x-1)/2 * 2y + y.

If we substitute x'=(x-1)/2 and y'=2y, we're going to "lose" that y value. This is why we cross off the even ones and keep the "remainder" for the odd x values.

 

Iterate to the next step with x and y replaced by x' and y' values. Continue until x=1.

[Sam Ashen]

Did you know the Russians multiply differently than we do?

Write the numbers side by side. For the number on the left, continually divide by two, writing the answer below and ignoring remainders, until you get to 1. For the one on the right, multiply by two until your columns are of equal length. Say it's 19x44.

 

19        44

9          88

4        176

2        352

1        704

 

Now, paying attention to the column on the right, cross out any numbers opposite an even number on the left.

 

19       44

9         88

4       176

2       352

1       704

 

Add up what's left on the right: 44+88+704=836. And voila, 19x44=836.

It's awesome (though tedious), but does anyone know why this works?

 

19 x 44 = 9 x 88 + 44

9 x 88 = 4 x 176 + 88

4 x 176 = 2 x 352

2 x 352 = 704

 

So 704 + 88 + 44 = 836 = 19 x 44

 

Does that help?

[GoldLynx]

Hmm... working backwards, the numbers that are added in the rightmost column can be expressed as 44(2⁰+2¹+2⁴) = 44(1+2+16) = 44(19).

 

Or, another way of putting it would be 19 = 1(2⁰) + 1(2¹) + 0(2²) + 0(2³) + 1(2⁴).

 

 

Regarding the left column...

I can see that when n mod₂ = 0 on the left, the coefficient is 0 for the right, and likewise if n mod₂ = 1 on the left, the coefficient is 1 for the right.

 

It seems like an interesting trick to convert a number into binary, but I'm not quite sure why it works.

 

I need to brush up on my binary. But I trust you've done something very smart.

[GoldLynx]

x * y = x/2 * 2y = x/4 * 4y and so on. If x is a power of two, you'd end up crossing off all the ys but the last one...

No change. Substitute x' = x/2 and y' = 2y.

 

If x=2n+1, however, the next x' = n = (x-1)/2.

x * y = (x-1)/2 * 2y + y.

If we substitute x'=(x-1)/2 and y'=2y, we're going to "lose" that y value. This is why we cross off the even ones and keep the "remainder" for the odd x values.

 

Iterate to the next step with x and y replaced by x' and y' values. Continue until x=1.

 

You got the closest of anyone to explaining this in a way that made sense to me. Why does x*y=(x-1)/2 * 2y+y? Bear with me...I tutor algebra and I've completed calculus but I'm still not sure what you're saying. Almost got it... *brain sweats*

 

[AugustaAdaByron]

You got the closest of anyone to explaining this in a way that made sense to me. Why does x*y=(x-1)/2 * 2y+y? Bear with me...I tutor algebra and I've completed calculus but I'm still not sure what you're saying. Almost got it... *brain sweats*

 

 

x*y= (x-1)*y +y = (x-1)*y*(1/2)*2 +y= ((x-1)/2) * (2*y) + y

[Sam Ashen]

Multiplying integer numbers A and B:

 

If A is even, then A x B = A/2 x 2B and nothing happens.  That's why the Ruskies have you cross off B and just go to the next line.

 

If A is odd, then A x B = (A/2-0.5) x 2B + B - and that is why it gets added to the sum.

[GoldLynx]

Multiplying integer numbers A and B:

 

If A is even, then A x B = A/2 x 2B and nothing happens.  That's why the Ruskies have you cross off B and just go to the next line.

 

If A is odd, then A x B = (A/2-0.5) x 2B + B - and that is why it gets added to the sum.

 

Making more sense. Thank you again. I'll write it out and sort it out.

 

[Sam Ashen]

How about......

 

Given r² = x² + y², how could I mentally solve for r, given x and y?  (especially if x and y are quite a bit different)

 

For example, x = 15 and y = 5.25?

 

Just some fun.  

[GoldLynx]

How about......

 

Given r² = x² + y², how could I mentally solve for r, given x and y?  (especially if x and y are quite a bit different)

 

For example, x = 15 and y = 5.25?

 

Just some fun.  

 

I'm assuming there's a trick to this you're asking us to figure out?