Who wins the game?
Heather, Sam and Theo are playing a game involving 2018 cards, each with a different integer between 1 and 2018 on it. They take turns to take a card. Heather goes first, then Sam, then Theo, and so on. After each of Heather's turns, the sum of the digits on all previously chosen cards must be odd. Sam and Theo win if the last card has the number 2 on it(and no other digits), Heather wins if she cannot move, or if the last card is anything else. Who wins the game?
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1 solution
First notice that between 1 and 2000 there are 1000 numbers with an odd digit sum and 1000 with an even digit sum, since for every number 1 ≤ x ≤ 1 0 0 0 , the number x + 1 0 0 0 has a digit sum of the opposite parity. Now you can count that there are 10 numbers with odd digit sums and 8 numbers with even digit sums between 2001 and 2018. This means that on the cards there are 1010 cards with an odd digit sum and 1008 with an even digit sum. When Heather has her first go, she must choose a card with an odd digit sum, and then the next 3 moves must have digit sums that sum to an even number, and so do the 3 after that, and then the 3 after that... So every 3 consecutive turns that start with Sam must either take away 3 even digit sum cards or 1 even digit sum card and 2 odd digit sum cards. This means that the parity of the number of cards with odd digit sums is always the same after Heather's go, and in fact it is always odd. This means that if there is a last card, it must have an odd digit sum and so cannot have the number 2 on it, so Sam and Theo cannot win. Therefore Heather wins.