** STICKY ** League of Reason? PROVE YOURSELF!

[SchrodingersFinch]

A bit of a mathematical problem:
One fine day you visit your favourite restaurant and find out they're having a contest.
They're offering free dinner for a month to anyone who can give the right answer to a certain problem.
Of course you decide to participate, so here's the fun part:

The restaurant just created the first 4-dimensional cake and they ask you what the highest
possible number of pieces is you can get after cutting this 4d-cake ten times (straight cuts i.e. no figure skating with the dimensional blade ).

What answer wins you the contest (and how did you arrive at the solution)?

Hint 1:

Hint 2: The answer is neither 36, nor 42, nor 144.

I believe I got it, although my math is a bit rusty.

Let's say we have a D-dimensional cake and we make n cuts creating the maximum number of pieces possible, which is

.

The D-1 dimensional cuts must cut through each other in way that creates the maximum number of intersections between the cuts.
These intersections divide the cuts into the maximum number of regions. The maximum number of regions is the same as the maximum number of pieces in a D-1 dimensional cake with n-1 cuts, which is

.


When we make the nth cut it is also divided into the maximum number of regions,

.
Each region of the cut divides the piece it goes through into two, creating

new pieces.
Therefore,

From this it's easy to construct this table, where we can see that the answer is 386.

 

[Case]

Bingo. Finch, you have officially proven yourself. Congratulations.
:geek:

Now I don't own a restaurant, but if we ever meet in real life, I'll treat you to a good meal. Took 4 months for this to be solved, after all. :mrgreen:
For now, I'll just award you the first official reasoncake award:

 

[Case]

Since I was asked to offer another challenge right away, I decided to go with something comparatively easy.

One fine day, TruthfulChristian finds himself walking home with a collection of small pinecones in his bag, excited to guess their average size. Suddenly the earth opens up and he almost falls into the freshly created rift. He holds on tight to his delusions, empties the bag, fills it with lots of hot air by abusing the second law of thermodynamics and escapes his untimely demise. At home, his imaginary friend asks if his hunt for pinecones was successful. TC believes for a while and then replies: "I know I had the answer right there. I had so many pinecones to prove it, too." His imaginary friend asks "How many did you collect?" TC mumbles some incoherent words and finally proclaims: "God knows how many... but when I ordered them in groups of two, three, four, five or six, I would always be left with a single pinecone at the end. Only when I put them in groups of seven did I get equal groups."

How many pinecones did TC collect (lower bound)?

 

[Muto]

@ case: 961

Here is my callenge: (I hope it wasn't allready there):

We have two women sitting and talking:

Q: How old are your three children?

A:If you multiply their ages you get 36.

Q: I am not satisfyed. Can you give me more information?

A: Yes, if you add their ages you get the number on the sign across the street.

Q: I need more information to determine their age.

A: My oldest daughter has beautiful eyes.

After this discussion the questioner knows the age of the children.
Can anyone else determine it as well?

Note: The women are good at mathematics/logic. The age of each child is a positive integer.

 

[MuslimPonderings]

How many pinecones did TC collect (lower bound)?

721?

 

[MuslimPonderings]

After this discussion the questioner knows the age of the children.
Can anyone else determine it as well?

Note: The women are good at mathematics/logic. The age of each child is a positive integer.

9,2,2 ?

 

[MuslimPonderings]

721 @ Case

9, 2, 2@ Muto

Do I need to say why?

 

[MuslimPonderings]

How many pinecones did TC collect (lower bound)?

No, not 721...

301?

 

[Case]

@ case: 961

Wrong. Also, please add whatever way you took to solve this (e.g. equation).

 

[OGjimkenobi]

A bit of a mathematical problem:
One fine day you visit your favourite restaurant and find out they're having a contest.
They're offering free dinner for a month to anyone who can give the right answer to a certain problem.
Of course you decide to participate, so here's the fun part:

The restaurant just created the first 4-dimensional cake and they ask you what the highest
possible number of pieces is you can get after cutting this 4d-cake ten times (straight cuts i.e. no figure skating with the dimensional blade ).

What answer wins you the contest (and how did you arrive at the solution)?

Hint 1:

Hint 2: The answer is neither 36, nor 42, nor 144.

I believe I got it, although my math is a bit rusty.

Let's say we have a D-dimensional cake and we make n cuts creating the maximum number of pieces possible, which is

.

The D-1 dimensional cuts must cut through each other in way that creates the maximum number of intersections between the cuts.
These intersections divide the cuts into the maximum number of regions. The maximum number of regions is the same as the maximum number of pieces in a D-1 dimensional cake with n-1 cuts, which is

.


When we make the nth cut it is also divided into the maximum number of regions,

.
Each region of the cut divides the piece it goes through into two, creating

new pieces.
Therefore,

From this it's easy to construct this table, where we can see that the answer is 386.

Oh damn, you totally beat me to it, I just finished coming up with this exact same answer and was about to post it when I saw your reply, it's all good though, it only took me 1 minute and 17 seconds to come up with the answer if you include the time I spent typing the whole thing up on Microsoft Word. I was actually able to instantly visualize the entire process in less than a fraction of a second. I then formulated a detailed reply and explanation in 7 different languages over the following 4 and a half seconds which took me near the 5 second mark. The remaining 1 minute and 12 seconds were spent drilling away at the old keyboard as my typing speed is a bit rusty.

It kinda sucks cause I recently upgraded my 4 dimensional calculator up to 5 dimensions and it would of been kind of neat to be able to check my math with it but oh well, what can ya do, ya know?

Congrats on beating me to this one, I'll handle the next one for sure.

 

[Nelson]

One fine day, TruthfulChristian finds himself walking home with a collection of small pinecones in his bag, excited to guess their average size. Suddenly the earth opens up and he almost falls into the freshly created rift. He holds on tight to his delusions, empties the bag, fills it with lots of hot air by abusing the second law of thermodynamics and escapes his untimely demise. At home, his imaginary friend asks if his hunt for pinecones was successful. TC believes for a while and then replies: "I know I had the answer right there. I had so many pinecones to prove it, too." His imaginary friend asks "How many did you collect?" TC mumbles some incoherent words and finally proclaims: "God knows how many... but when I ordered them in groups of two, three, four, five or six, I would always be left with a single pinecone at the end. Only when I put them in groups of seven did I get equal groups."

I believe the answer is 721. Here is my explanation:

We have some total amount of pinecones T. When we take 1 away from this we have a number that is divisible by 2,3,4,5, and 6. The number itself is divisible by 7. So we know that:

where X is some unknown factor. We also know that:

where Y is another unknown factor. I moved the 1 to the right side in the first equation set both equations equal and solved for Y to get:

I was planning to then write a quick program to find the lowest value of Y that was an integer over an array of X values that are integers, but it was easy enough to spot that X equal to 1 gives a number divisible by 7. So, there you go. Final answer: 721.

 

[Case]

No. ^^

 

[Nelson]

Ok, I think I found my mistake. The smallest integer that is a multiple of 2,3,4,5, and 6 should be used, not 720. This is 60, so from this I get:

301 pinecones

 

[Case]

I just saw MuslimPonderings beat you to it. It's indeed 301.

The mathematical solution is really long... and you guys already got the answer so I'm not gonna bother.

Next: TruthfulChristian has been locked in virtual reality again and this time, he roams the grid of a cube (you know... 8 vertices, 12 lines). He's at one end of the cubic grid while a tasty pinecone lounges at the other side. TC moves at random from one vertex to one of the connected vertices, this process takes 5 minutes. How long will it take him, on average, to get to the tasty pinecone?

 

[Muto]

@MuslimPonderings: correct

 

[SchrodingersFinch]

Next: TruthfulChristian has been locked in virtual reality again and this time, he roams the grid of a cube (you know... 8 vertices, 12 lines). He's at one end of the cubic grid while a tasty pinecone lounges at the other side. TC moves at random from one vertex to one of the connected vertices, this process takes 5 minutes. How long will it take him, on average, to get to the tasty pinecone?

I've heard that one before so I won't ruin it for the rest. But I do have a math puzzle of my own:

There is a closed curve C around a point O. The curve is completely smooth, so on every point on the curve you can draw a tangent line.
How can you easily* show that on the curve there exists at least one point P whose tangent line is perpendicular to the line OP?
Extra: what is the minimum number of such points? Draw a picture.

*I'm sure there are many ways to do it, but what is the simplest you can come up with?

 

[MuslimPonderings]

Next: TruthfulChristian has been locked in virtual reality again and this time, he roams the grid of a cube (you know... 8 vertices, 12 lines). He's at one end of the cubic grid while a tasty pinecone lounges at the other side. TC moves at random from one vertex to one of the connected vertices, this process takes 5 minutes. How long will it take him, on average, to get to the tasty pinecone?

Let T be where TruthfulChristian begins, P the position of the tasty pinecone, and S the connected vertex TruthChristian has moved to after 5 minutes.
If the average time from T to P is sought:
TP = (TQ^2 + PQ^2)^0.5
= ((((ST^2 + SQ^2)^0.5)^2) + PQ^2)^0.5
= (5^2 + 5^2) + 5^2)^0.5
= 75^0.5
= 9mins (1 sig. fig.)

If the average time from his latter position, S, is sought:
SP = (SQ^2 + PQ^2)^0.5
= (25 + 25) ^0.5
= 7mins (1 sig. fig.)

If the average time from his latter position, S, is sought but taking into account his initial move:
5 + SP = 12mins (1 sig. fig.)

 

[Pulsar]

There is a closed curve C around a point O. The curve is completely smooth, so on every point on the curve you can draw a tangent line.
How can you easily* show that on the curve there exists at least one point P whose tangent line is perpendicular to the line OP?
Extra: what is the minimum number of such points? Draw a picture.

Here's my take: follow the curve in an anti-clockwise manner (more technically, the curve can be parametrized by a single variable), and plot the distance d between each point on the curve and O. The result will be a function as shown at the bottom. This function will have a number of local minima and maxima. I've highlighted four of them, with corresponding distances d1, d2, d3 and d4. So, what are these local extrema? Draw a circle with centre O through such a point (see the figure on the left). If the point is a local minimum, this means that every surrounding point will lie outside this circle. If the point is a local maximum then every surrounding point lies inside the circle. This is nothing else than an alternative definition of a tangent point, and the perpendicular line goes through O (its the radius of the circle). In other words, a smooth curve will have as much of points with the desired property as the number of local minima and maxima of its "distance function" with O.

There is one special curve for which the distance function will be constant: a circle with centre O. Of course, every point of that curve will have the property (every point is an extremum).

What's the minimum number of possible points with the desired property? Two: a distance function with one minimum and one maximum. An example would be a circle, with O NOT at the centre, as shown in the figure on the right. The points with distances d1 and d2 have the desired property, the others don't.

I don't know if this is the simplest solution, but I like it

 

[Case]

@Muslim: Nope. ^^ Moving from one vertex to another already takes 5 minutes, if, by chance, he moves the direct, fastest path (well, one of the... six), it will take him 15 minutes already. So it can't be less than 15 on average by definition, so to speak.

 

[SchrodingersFinch]

Very good Pulsar. My approach is a little different, but I don't know if it's any simpler.

First you draw a circle inside the curve with O as its centerpoint. Then you expand the circle until it touches the curve on at least one point P. Because the curve is smooth it will have the same tangent line as the circle at the point P. And on a circle the tangent line is always perpendicular to the radius.

For the extra question we keep on expanding the circle. Now the circle will intersect the curve at two points (A and B). One arc will be outside the curve and the other inside. As we keep on expanding, the intersection points A and B eventually get closer, until they finally meet at the point D. By now the rest of the circle is outside the curve, so D can't be an intersection point. And obviously, the tangent line at the point D must be perpendicular to OD, the radius of the circle.