There are $ n$ candidates on a table. Petrik and Mikola alternately take candies from the table according to the following rule. Petrik starts by taking one candy; then Mikola takes $ i$ candies, where $ i$ divides $ 2$, then Petrik takes $ j$ candies, where $ j$ divides $ 3$, and so on. The player who takes the last candy wins the game. Which player has a winning strategy?
Problem
Source: Ukraine 1997
Tags: combinatorics proposed, combinatorics
madness
22.07.2009 20:51
Case: n is even
Mikola can always ensure (by picking up one candy/an odd number of candies every time it is her turn) that when its Petrik's turn, he has to pick up an even number of candies to win, which is impossible and thus, she has the winning strategy.
Case: n is odd
Mikola can always ensure (by picking up an even number of candies once and odd number of candies every other time it is her turn) that when its Petrik's turn, he has to pick up an even number of candies to win, which is impossible and thus, she has the winning strategy.
Mahdi_Mashayekhi
18.12.2021 16:22
we have to cases : 1- n is odd : in this case player 2 wins because player 1 can only choose odd numbers and player 2 has to do this strategy " make the remaining candidates even" because he can choose 1 and 2. 2- n is even : in this case player 2 wins with same strategy.