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Answers ( 8 )
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    it's 23,54

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    banh

    In the beginning randomly pair the points and join the segments. Let SS be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting 2n2n points by nn segments, there are finitely many possible values of SS.) If two segments ABABand CDCDintersect at OO, then replace pairs ABABand CDCDby ACACand BDBD. Since

    AB+CD=AO+OB+CO+OD>AC+BDAB+CD=AO+OB+CO+OD>AC+BD

    by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease SS. Since there are only finitely many possible values of SS, so eventually there will not be any intersection.leuleueoeobucquagianroiundefined

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    My answer is no.

    At the begining there are 32 black pieces. We will prove that the number of black pieces left on the board is always even.

    In the single move, we change a row or column, assume that row or column has k black pieces and 8 - k while pieces, after change all black to white and while to black, that row or column will has 8-k black and k white. The difference between black pieces after a single move is (8 - k) - k = 8 - 2k. The difference is a even (8 - 2k). Therefore, after every move, the black pieces left is alaways even and cannot be one black piece at any time.

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    If we let x=1x=1 and o=–1o=–1, then note that consecutive symbols are replaced by their product. If we consider the product PP of the nine values before and after each operation, we will see that the new PP is the square of the old PP. Hence, PP will always equal 11 after an operation. So nine oo's yielding P=–1P=–1 can never happen.  

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    n the beginning, randomly divide the members into two groups. Let S be the sum of the number of the pairs of enemies in each group. If a member has at least two enemies in the same group, then the member has at most one enemy in the other group. Transferring the member to the other group, we will decrease S by at least one. Since S is a nonnegative integer, it cannot be decreased forever. So after finitely many transfers, each member can have at most one enemy in the same group. 

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    Let SS be the sum of all the mnmn numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most 2mn2mn tables. So SS can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then SS increases. Since SS has finitely many possible values, SS can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative. 

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    Tom 20 candy

    Tony 80 candy

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