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    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.

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    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}

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    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:

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

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

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

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

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    5 8 P O X Y H M

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

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

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

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

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

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    A B > < C

    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.

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    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.

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    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 \(x\in\) A0, 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.

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    Let Cn 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 An have differenr colors, then A1, ..., An can be colored in Cn ways, while An+1 can be colored in p - 2 ways. If A1 and An have the same color, then A1, ..., An can be colored in Cn-1 ways and An+1 can be colored in p - 1 ways. So Cn+1 = (p-2)Cn +(p-1)Cn-1 with n > 2 (*)

    (*) can be written as:  

         Cn+1 + Cn = (p - 1) (Cn + Cn-1)

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

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