Rothschild the benefactor has a certain number of coins. A man comes, and Rothschild wants to share his coins with him. If he has an even number of coins, he gives half of them to the man and goes away. If he has an odd number of coins, he donates one coin to charity so he can have an even number of coins, but meanwhile another man comes. So now he has to share his coins with two other people. If it is possible to do so evenly, he does so and goes away. Otherwise, he again donates a few coins to charity (no more than 3). Meanwhile, yet another man comes. This goes on until Rothschild is able to divide his coins evenly or until he runs out of money. Does there exist a natural number $N$ such that if Rothschild has at least $N$ coins in the beginning, he will end with at least one coin?
Problem
Source: Israel Autumn 2016 TST1/2
Tags: number theory, number theory solved
28.07.2019 07:33
I think this problem needs further explanation.
28.07.2019 07:45
junioragd wrote: I think this problem needs further explanation. What's not clear?
28.07.2019 08:21
The first and last statement is bit clumsy. You could just choose an $N \equiv 0 [2^e]$??
28.07.2019 08:27
Mr.Chagol wrote: The first and last statement is bit clumsy. You could just choose an $N \equiv 0 [2^e]$?? The question is whether this will be true for all large enough N
28.07.2019 10:30
Well,if he needs to donate some number of coins in the second phase(no more than 3),he can pick an appropriate number mod4 and secure division with 4 people?
28.07.2019 10:31
Maybe he always donates $1$ coin,I dont know?
28.07.2019 10:35
Or if the number is of the form $3k+1$ he donates $1$ coin,if it is of the form $3k+2$ he donates $2$ coins.
30.07.2019 09:48
Any answers?