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0

Let's draw a circumference of a circle with center O. If we draw two diameters perpendicular to each other then we can construct the square ABCD in the circle. Lets draw the bisectors of the sides AB and CD which cut the circle at points F and G. Then we draw the straight lines AF,FB and CG,GD. So A,C,G,D,B,F is the required hexagon.


2

Let's see if black has a good way to prevent Black can try So, it seems that plan doesn't quite work. However, white has another way to deliver mate:


2

The first thing that came to mind for me was... Unfortunately, I'm not sure of the missing color because... Maybe someone else will be able to get the correct color from here.


2

Answer (if "hexagon with different side lengths" means "at least 1 side has a different length" and not "no side has the same length as another one"): Reason:


4

If n = 2k + 1 is odd then If n = 2k is even then


4

For odd values of $n$: For $n=2$: For $n=4$: This generalises to all even $n$. Proof:


-2

Let's assume $n, k>1$. Let's call the last turn $M_x$ and $M_x\leq n^k$. If the number of turns is even then the first player loses; if the number of turns is odd the first player wins.


2

As with the other question, we first find out how many valid ways there are to place the numbers. Now to get the probability:


0

I got the same answer as aschepler I found


19

The solution reads: Solved crossword image: Words found: Finally:


7

Let's first see in how many ways the numbers can be placed with all neighbours coprime. Now for the probability:


5

Using the top left corner as (1,1) and the bottom right corner as (16,16) — (row, column) It looks like


7

The words are all And Heres the completed grid: And listing them (see El-Guests answer for coordinates!):


2

The missing color could be Because


4

Using: The left coloumn is: If the right column is also such a set, the missing colour is: So what's this all about?


14

Here is a straightforward escape route.


7



2

A quick and dirty approach, slightly different to the ones given. Again, [no-computers].


11

Due to the 6, we can place the following few pieces: After a little bit of trial and error from here, I found the solution: 4 is attacked by: 6 is attacked by all but


7

A general [no-computers] solution: First of all: To summarize: Let's now consider iterating this. The following diagram may help to illustrate what happens: Let's now apply this to the present case.


10

For each cycle, the number of prisoners executed is exactly A more mathematical approach:


8

For this one, you don't even need to rotate the pieces: And the final ones, which are closely related:


7

Here's one solution: Other solutions may be possible.


4

I thought of approaching this mainly in a number-theoretic way, consider divisibility, to heavily constrain the heuristic search so it can be done by hand: $(a+1)$ is a member of >=1 PT $a$ is a member of >=2 distinct PTs moreover, considering the left-hand side of the square: call the LHS side $b$, $b$ is in one of those PTs involving $a$, WLOG call ...


6

The general formulas for creating right angle triangles with integer sides are A=m^2+n^2, B=2mn, c=m^2-n^2 if m=n+1 then two sides of the triangle are consecutive. In your problem the smallest values for m,n are m=4 and n=3. So a+1=25 and a=24. First triangle has sides 7,24,25 Second triangle has sides 18,24,30 Third triangle has sides 25,25,30.


10

The minimum value for $a$ is: Visualization: Proof:


11

There are two ways to do this: the algebraic way and the 'clever' way. The algebraic way The clever way


29

Let $$f(a,b) := \frac{\sqrt{a^2+3ab+b^2-2a-2b+4}}{ab+4}.$$ John's procedure is now to repeatedly replace the leftmost two values $a,b$ on the blackboard by the single value $f(a,b)$. I claim that John never writes a negative value onto the blackboard. Proof: the blackboard always begins with positive values. Suppose $a$ and $b$ are positive values John ...


3

The number is First step: Now a little bit of brute force, which takes a few minutes with a pocket calculator:


3

Take the string as "abcbc" and k=5. Now, according to the algorithm we have "aabbccbc"(three extra characters). But, due to the repitition of "bc", we can take the string as "abcbcbc"(which has just two characters extra). This will give us exactly 5 subsequences of "abcbc".


1

Here's one possibility: Our strategy dictates the final string should be But we can do better:


10

The words (which are indeed connected & relevant) are: Here's the final image: And sorry, forgot to mention that...


5

This is a linear recursion with characteristic polynomial $x^4 - x^3 - x^2 - x - 1,$ which has roots $ z_1 = 1.928, z_2 = -0.775, z_{3,4} = -0.076 \pm 0.815i.$ The general solution is $x_n = \sum\limits c_i z_i^n,$ and the number of digits is $\lfloor \log_{10}(x_n) \rfloor + 1.$ Since $|z_2|, |z_3|, |z_4|<1,$ we have $|x_{2014} - c_1 z_1^{2014}| \le |...


2

Partial answer and progress: So I put this into Wolfram Alpha, with the terms of f(n)=f(n-1)+f(n-2)+f(n-3)+f(n-4), f(1)=1, f(2)=13,f(3)=169,f(4)=2014, which can be found here. I noticed that in the graph provided (below), the general rule was 10^(n+2) = f(n*3.5), for n larger than 2. This would make f(2014) = 10^577.4285714, since 2014 = 575.4285714 * 3.5 ...


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