Math Thread

In the premise, we are given some basic knowledge:
E = a + b
Y = ab
a > 1, b > 1

We can also establish at this point that Yumemi's product cannot be prime.

Statement 1:
Yumemi doesn't know - therefore her product cannot have just two prime factors.
This therefore means that at least one of the integers cannot be prime.

Statement 2:
Eirin says she already knew that Yumemi didn't know. The only way she can be sure of this is if her sum cannot be the summation of two primes. If any of the potential pairs of integers that add to her sum are two primes, then Eirin could not be certain that Yumemi doesn't know.

Statement 3:
Now that Eirin has said this, Yumemi is suddenly able to figure it out. The only information that Yumemi has gained from statement 2 is that the sum Eirin has cannot be the summation of two primes. This heavily reduces the potential amount of sums that Eirin could have. However, multiplying the potential pairs that make up the sum (e.g. 2*9, 3*8, 4*7, 5*6 for the sum 11) will not necessarily give a unique product, when compared with other sum's pair's products (e.g. 5*6 = 30, but also, 2*15 = 30). For Yumemi to have realised the two integers, she must have a product that is unique amongst the possible products. The method I used to find this was pretty much just brute force. (In order to find the sums to work with in the first place, a simple formula was '2 + x = sum' where x is a non-prime odd integer).
The only sums that you needed to work with in order to find an answer were 11, 17, 23, 27, 29, 35 and 37.
My working to find a potential answer is shown below (warning, large image):



Now that I've found 4 and 13 (as has Yumemi), I should be able to say this is the answer, right? Not so. I've only tested the cases up to a certain point, for all I know there could be another pair that is equally as likely. This is where the fourth statement comes in.

Statement 4
Before the third statement, Eirin did not know the integers. Yumemi discovered using the second statement's information, in combination with the product that she knows, the two integers Keine gave. With statement 4, Eirin boldly declares that she knows the integer pair. It's important to remember at this point that the two girls are perfect mathematicians. If there was even just one other possibility than (4,13) that existed, Eirin would be uncertain of the answer and so could not make this statement. Therefore, while earlier 4 and 13 was a possibility, with the fourth statement it is a certainty.

Answer: 4 and 13 are the integers Keine gave.

First two statements, combined with Goldbach's conjecture, indicate that the sum must be 2 + composite number. Which I would consider to be a "valid sum". All I had to do after that was go through each valid sum one by one, and check how many of its possible pairs will form a product that can be factored into a different pair that forms another valid sum. Although this is technically still brute force, I ended up finding the solution after checking only the second valid sum (17), so there was no need to iterate on Excel or try enforcing more assumptions to narrow down the possible pairs. I didn't actually know about Goldbach's conjecture at that time though, so it took me a while to realize that none of the even numbers I tested appeared to be valid sums before I decided to assume that all even numbers are invalid sums.

In any case, using Goldbach's conjecture to help narrow down solutions is fine. It could easily be shown that (4,13) works as a solution without having to resort to Goldbach's conjecture, so (4,13) is still a formally proven solution.

Basically, Prisoner A converts the Magic Tile location into a 6-bit binary number. Prisoner A then reads the state of the coins to generate a 6-bit number and compares it with the Magic Tile number. Prisoner A flips exactly one coin to ensure that the 6-bit number from the state of the coins matches with the Magic Tile number. Prisoner B reads the states of the coins to generate the 6-bit number and deduces the Magic Tile from just that. Details:

Converting Magic Tile Location into a 6-bit number

Each tile can be assigned a unique number from 0 to 63. This can be achieved by simply counting up the numbers as you read through all the tiles in any manner. The number is then converted into 6-bit binary (2^6 = 64, so 0 to 63 can be accommodated with only 6 bits). For example, A1 = 000000, A2 = 000001, A3 = 000010, ... H7 = 111110, and H8 = 111111.

Converting the state of the coins into a 6-bit number

This one is a bit more complicated. Here's how I'll explain it. A 6-bit binary has 6 bit positions, right? Let's number them from position 0 to position 5 (with position 0 being the LSB i.e. last bit). Now, each tile is supposed to represent a unique combination of numbers from 0 to 5. The number of such possible combinations is

Combinations of 0 numbers + Comb. of 1 number + Comb. of 2 numbers + Comb. of 3 numbers + Comb. of 4 numbers + Comb. of 5 numbers + Comb. of 6 numbers
= C(6,0) + C(6,1) + C(6,2) + C(6,3) + C(6,4) + C(6,5) + C(6,6)
= 1 + 6 + 15 + 20 + 15 + 6 + 1
= 64

which is exactly the number of tiles there are. Therefore, each tile can have a unique combination. One possible example of a setup is:



So how does this help convert the state of 64 coins into a single 6-bit string? Well, the idea is to calculate the value of each bit, one at a time. Each coin state represents a 1 (heads) or a 0 (tails). Each of the six bits are calculated by adding the coin states of every tile that represents that bit position. For example, Bit #4 is found by adding the coin states of every tile that has the number 4 in its unique combination. When I say add coin states, I mean to add them in mod 2 aka taking the odd parity aka applying XOR on the coin states.

Prisoner A flipping exactly one coin to make the coin state 6-bit string equal to the Magic Tile 6-bit string

This is pretty simple. After computing both of these bit strings, Prisoner A will compare them with each other and note down which bit positions are different. Prisoner A will then flip the tile that represents exactly the combination of unmatched bit positions. For example, if the Magic Tile is at 011010 and the Coin State bit string is calculated as 110011, then only bit positions 0, 3, and 5 are different. Prisoner A will flip the coin on the tile that represents the combination 035, which is D7 in the spoilered setup example.

If the two bit strings already match, then Prisoner A will flip the coin on the tile that represents the empty combination, which is A1 in the spoilered example. This coin's state is not involved in any of the calculations so it will not alter any results.

Prisoner B finding the magic tile

Prisoner B will compute the bit string corresponding to the coin states. Thanks to A's flipping, this should now become equal to the bit string for Magic Tile. Prisoner B can now convert the Magic Tile bit string into its coordinates and identify the Magic Tile.

Hmmm, yeah, there are apparently other solutions. I edited the original question to add a constraint that the sum of the numbers must be less or equal to 100. I haven't checked whether there are any other solutions myself, but I've eliminated all the other possible solutions that I heard of from elsewhere.

Did you instead change the problem into 3 wolves and 5 sheep, which results in the same problem (since just as sheep can't be in a wolf's range, it also means that wolves can't be in a sheep's range). It's easier to place three queens for five safe spaces rather than place five queens for three safe spaces, despite them being the same problem.

Hmm, this took a bit of trial and error, but I finally got it. White queens represent the wolves, white pawns represent illegal spaces, black pawns represent the sheep.

1. Begin. (8, 0, 0)
2. Fill the 5 liter bucket. (3, 5, 0)
3. Fill the 3 liter bucket with water from the 5 liter bucket. (3, 2, 3)
4. Empty the 3 liter bucket into the 8 liter bucket. (6, 2, 0)
5. Transfer water from the 5 liter bucket to the 3 liter bucket. (6, 0, 2)
6. Fill the 5 liter bucket with water from the 8 liter bucket. (1, 5, 2)
7. Fill the 3 liter bucket with water from the 5 liter bucket. (1, 4, 3)
8. Done!

Note that the intersection area is formed by two equilateral triangles of side length 1 and four segments.

Area of equilateral triangle: sqrt(3)/4
Area of 60 deg sector: pi/6
Segment area: pi/6 - sqrt(3)/4
Area of intersection: 2*triangle + 4*segment = sqrt(3)/2 + 2pi/3 - sqrt(3) = 2pi/3 - sqrt(3)/2

The two sides of the triangle with 2x are equal, as the configuration is basically the "angle at the centre of the circle is twice angle at the circumference of the circle" theorem.
Thus, red angle = blue angle = 90-x.
Then, 180 = x + 4(90-x)
180 = 360 - 3x
X = 60

For the sake of simplicity, say the large quadrant has a radius of 2.

Area of small quadrant: pi/4
Red area: pi/4 - (1-pi/4) = pi/2 - 1
Non-blue area: pi - (pi/2 - 1) = pi/2 + 1
Blue area: pi - (pi/2 + 1) = pi/2 - 1

Hence, ratio is 1:1.

No, I actually didn't notice that. I was just fooling around with initial placements of one or two wolves then attempting to find possible places for the rest.

Alright. In that case:
red + blue = k
2x + k = x + 2k = 180
x = k
3x = 180
x = 60

Imagine the small quadrant (of radius 1) is bounded by a square of side 1. Then let's call the area in the square but not in the quadrant A (A = 1 - pi/4).
Thus, the red area is simply 1 - 2A, or pi/4 - A, which is pi/2 - 1.

Consider half the area of intersection - it is just a segment of the unit circle. To find the angle T at the intersection, just consider when cosT = 1/2. T is pi/3, and so the full angle of the segment is 2pi/3.
Finally, calculate the area of the segment and multiply by two => A = 2*(pi/3 - sqrt3/4) = 2pi/3 - sqrt3/2

Nice and simple simultaneous equations, here. Label the red angle 'y' and the blue angle 'z'. (I also used degrees, just to simplify typing it out.)
180 - y - z = 2x (1)
180 - 2y - 2z = x (2)
2*(1): 360 - 2y - 2z = 4x (3)
(3) - (2): 180 = 3x
x = 60

Label the blue area A, and the red area B. Give the semi-circles radius x, and therefore the quarter circle has radius 2x.
A = pi(2x)^2/4 - pi(x)^2/2 - pi(x)^2/2 + B
Simplifies to A = B, therefore the ratio is 1.

x = 15 - 6sqrt5

Let A be the area and P the perimeter.
A = 2(pi(r)^2/2) + (1/2)(4r)(4r)(sqrt3/2) = pi(r^2) + 4sqrt3(r^2)
P = 2((2pi(r))/2) + 2(4r) = 2pi(r) + 8r

A and P are equal:
2pi(r) + 8r = pi(r^2) + 4sqrt3(r^2)
(pi + 4sqrt3)(r^2) - (2pi + 8)r = 0

Solve for r:
r = (8 + 2pi)/(4sqrt3 + pi)
or r = 0

Did this on paper, but can't be bothered to type it up: r = 30 - 20sqrt2