Each time, a person takes 1, 2, 3, or 4 chips. The strategy for a player to win the game is taking the number of chips so that the number of chips left is the form of 5k + 1 .

If It is my turn, and there are 2014 chips in the pile, I shoud take 3 chips and the left is 2011 (=5.k +1). And if the opponent takes x chips (x = 1, 2, 3, 4) , i win take 5 - x chips to guarantee the left is always in the form 5k + 1. And finally, the opponent will take the last chip.

5, 10, 15, 20 should be in different groups, since otherwise, if two of them belong to the same group, the sum of the two elements is divisble by 5.

Therefore, k must \(\ge4\). We confirm that k = 4 is the least value, since we can group the integers from 1 to 20 into 4 subsets that satifies the problem as follows:

{1, 6, 11, 16, 5}, {2, 7, 12, 17, 10}, {3, 8, 13, 18, 15} , {4, 9, 14, 19, 20}

We have:

 \(a_9x^9+a_8x^8+...+a_1x+a_0=\left(x+2\right)\left(x+3\right)...\left(x+9\right)\left(x+10\right)\)

With x = 1, we have:

  ​ ​​ ​\(a_9+a_8+a_7+...+a_1+a_0=3.4...10.11\)       (1)

With a = -1, we have:

  ​ ​\(-a_9+a_8-a_7+...+a_2-a_1+a_0=1.2.3...8.9\)   (2)

subtracting (1) and (2), we have:

  \(2\left(a_9+a_7+..+a_3+a_1\right)=3.4...10.11-1.2...8.9\)

 \(2\left(a_1+a_3+a_5+a_7+a_9\right)=3.4...8.9\left(10.11-1.2\right)\)

  \(2\left(a_1+a_3+a_5+a_7+a_9\right)=3.4..8.9.108\)

 \(a_1+a_3+a_5+a_7+a_9=3.4...8.9.54=1.2.3...8.9.27=9!27\)

\(PO=PM+MO=8+5=13\)

\(PX=PY=\sqrt{PO^2-OX^2}=\sqrt{13^2-5^2}=12\)

\(PX.XO=PO.XH\left(=2.area\left(XPO\right)\right)\)

=> \(XH=\dfrac{PX.XO}{PO}=\dfrac{12.5}{13}=\dfrac{60}{13}\)

\(XY=2.XH=2.\dfrac{60}{13}=\dfrac{120}{13}\)

Suppose that it takes Alice x hours to go from A to B. Then, it takes Bill x + 2.5 hours to go from B to A. We get that Alice’s speed is d/x and Bill’s speed is d/(x + 2.5), where d is the distance from A to B. Since Alice and Bill met 3 hours after they started walking, we have:

  \(3\left(\dfrac{d}{x}+\dfrac{d}{x+2,5}\right)=d\)

\(\Rightarrow x=5\).

So Bill take 5 + 2,5 = 7,5 hours to go from B to A.

Assume x be the correct seuqence in {a, b, c, d, e}, we infer the distance between x and the other sequences would be in a set of {1, 2, 3, 4}. (the distance here is the number of different positions in the two sequences)

We have:

+ With sequence a: distance(a,b) = 2, distance(a,c) = 3, distance(a,d) = 3, distance(a,e) = 2.

  => a is not correct sequence because {2, 3, 3, 2} differs {1, 2, 3, 4}

+ With sequence b: distance(b,a) = 2, distance(b,c) = 1, distance(b,d) = 3, distance(b,e) = 4.

  => b should be the correct sequence because {2, 1, 3, 4} = {1, 2, 3, 4}.

+ With sequence c: distance(c,a) = 3, distance(c,b) = 1, distance(c,d) = 4, distance(c,e) = 5.

  => c is not correct sequence because {2, 1, 4, 5} differs {1, 2, 3, 4}

+ With sequence d: distance(d,a) = 3, distance(d,b) = 3, distance(d,c) = 4, distance(d,e) = 1.

  => d is not correct sequence because {3, 3, 4, 1} differs {1, 2, 3, 4}

+ With sequence e: distance(e,a) = 2, distance(e,b) = 4, distance(e,c) = 5, distance(e,d) = 1.

  => d is not correct sequence because {2, 4, 5, 1} differs {1, 2, 3, 4}.

So the correct sequence is b.

Let A₀ be the subset of S = {1, 2, ...3000} containing all numbers of the form 4ⁿk, where n is a nonnegative integer and k is an odd positive integer. Then no two elements of A₀ have ratio 2. A simple count shows A₀ has 1999 elements. Now for each \(x\in\) A_0, form a set \(S_x=\left\{x,3x\right\}\cap S\). Note the union of all \(S_x\)'s contains \(S,S_0\), by the Dirichlet principle, any subset of S having more than 1999 elements must contain a pair in some \(S_x\), hence of ratio 2. So no subset of 2000 numbers in S has the property.

Let C_n be the answer for n points. We have C₁ = p, C₂ = p(p-1) and C₃=p(p-1)(p-2).

For n + 1 points, if A₁ and A_n have differenr colors, then A₁, ..., A_n can be colored in C_n ways, while A_n+1 can be colored in p - 2 ways. If A₁ and A_n have the same color, then A₁, ..., A_n can be colored in C_n-1 ways and A_n+1 can be colored in p - 1 ways. So C_n+1 = (p-2)C_n +(p-1)C_n-1 with n > 2 (*)

(*) can be written as:  

     C_n+1 + C_n = (p - 1) (C_n + C_n-1)

=> C_n+1 + C_n = (p - 1)^n-2 (C₃ + C₂) = p(p - 1)ⁿ.

By induction we infer that C_n = (p - 1)ⁿ + (-1)ⁿ (p - 1)