MJC/2017 JC2 Prelim Marking Scheme/H3 Math (9820/01)/Math DeptPage 1 of 132017H3MATH(9820/01)JC2PRELIM–MARKINGSCHEMEQnSolutionMark Scheme1Mathematical Induction9(i)Let S(n) be the statement a player can always find a correct startingtoken to win the game when there arengreen tokens,nred tokens,n∈!+.S(1): 1 green token, 1 red token.Choose the green token first.Therefore, S(1) is true.Assume S(k) is true for somek∈!+. i.e. a player can always find acorrect starting token to win the game when there arekgreen tokensandkred tokens.S(1k+):1k+green tokens,1k+red tokens.Given any order of1k+green tokens and1k+red tokens, there mustexist an ordered pair of green and red tokens (in that clockwise orderwhere G first, then R). (In the worst case scenario, after a series of1k+consecutive green tokens, the next token must be red.)Remove the pair of green and red tokens. There arekgreen tokens andkred tokens remaining. From the inductive step, the player can find astarting point to win the game. The ordered pair of green and redtokens can be reinserted in the same spot where they were taken from.From the same starting spot, the player can still pick all22k+tokenswhile fulfilling the condition that at least as many green tokens as redtokens are picked up at all times.Therefore, S(k) is true implies S(1k+) is true.Since S(1) is true and S(k) is true implies S(1k+) is true, bymathematical induction, S(n) is true forn∈!+. A player can alwaysfind a correct starting token to win the game when there arengreentokens andnred tokens.B1 – statementB2 – correctconsideration ofS(1)B1 – S(k)assumptionB2 – someexplanation forinductive step(must mention Gfirst then R)B1 - conclusion(ii)No, the player might not be able to win the game for certainarrangements of tokens.A counter example for the case when2n=would be if the tokenswere arranged as follows.To win the game, the player must start with a green token and thenpicks up 2 red tokens consecutively, causing the player to not be ableto win the game.B1B1 -counterexampleRRGGRR