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American
09/03/2017 at 09:39

hghfghfgh 26/03/2017 at 20:16
At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a  b, we see that:
+ if a odd and b even, or a even and b odd then a + b or a  b is still odd
+ if a and b are both even then a + b or a b is still even
+ If a and are both odd then a + b or a  b is even
So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).

Faded 19/01/2018 at 14:52
At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a  b, we see that:
+ if a odd and b even, or a even and b odd then a + b or a  b is still odd
+ if a and b are both even then a + b or a b is still even
+ If a and are both odd then a + b or a  b is even
So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even). [haha]

An Duong 10/03/2017 at 14:14
At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a  b, we see that:
+ if a odd and b even, or a even and b odd then a + b or a  b is still odd
+ if a and b are both even then a + b or a b is still even
+ If a and are both odd then a + b or a  b is even
So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).
falcon handsome moderators
09/03/2017 at 12:55

demo acc 10/03/2017 at 14:42
very diffcult

Ace Legona 09/03/2017 at 19:14
In the beginning randomly pair the points and join the segments. Let \(S \) be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting \(2n \) points by \(n \) segments, there are finitely many possible values of \(S\).) If two segments \(AB\)and \(CD\)intersect at \(O\), then replace pairs \(AB\)and \(CD\)by \(AC\)and \(BD\). Since
\(AB + CD = AO + OB + CO + OD > AC + BD\)
by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease \(S\). Since there are only finitely many possible values of \(S\), so eventually there will not be any intersection.

Faded 19/01/2018 at 14:52
In the beginning randomly pair the points and join the segments. Let S be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting 2n points by n segments, there are finitely many possible values of S.) If two segments ABand CDintersect at O, then replace pairs ABand CDby ACand BD
. Since
AB+CD=AO+OB+CO+OD>AC+BD
by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease S
. Since there are only finitely many possible values of S, so eventually there will not be any intersection.

Vũ Việt Vương 01/04/2017 at 16:59
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.