-gon. In each turn, Mursalin and Agnishom has to draw a diagonal that doesn't intersect any of the diagonals that have been already formed. Mursalin goes first. The person who is unable to make a move loses. For how many possible values of , does Agnishom have a winning strategy if ?
Mursalin and Agnishom decide to play a game where they alternately draw diagonals between the vertices of a regular
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For any n-sided polygon, the number of diagonals that can be formed without intersection = (n - 3). So, a four sided polygon has 1 such diagonal, a pentagon has 2 such diagonals, a hexagon has 3 such diagonals and so on. Now, if Mursalin starts first, then he can win if the number of diagonals formed are odd. Agnishom wins if number of diagonals formed are even. From 457 to 1003 sided polygons, number of diagonals which do not intersect for each polygon will range from 454 diagonals to 1000 diagonals respectively. The even numbers from 454 to 1000(both included) is 274. As Agnishom wins only when even number of diagonals are formed, the number of winning chances that he has is 274.