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Hello!
I require help with a certain mathematical problem. I am playing a game with my friend, and we are inevitably stuck.
The game is simple, we have all the numbers from 1 to 500, my friend makes the first move and removes one number (He does not tell me the number, so I suppose it is "x"), and the second move is mine, I remove one number to. From then I must play by myself, I must come to a solution where the last 2 numbers, when you (+) them between themselves can be divided by 3. If they can't be divided by 3 then I have lost the game. How can I make an equation to solve this?
Best regards,
Alexander
I require help with a certain mathematical problem. I am playing a game with my friend, and we are inevitably stuck.
The game is simple, we have all the numbers from 1 to 500, my friend makes the first move and removes one number (He does not tell me the number, so I suppose it is "x"), and the second move is mine, I remove one number to. From then I must play by myself, I must come to a solution where the last 2 numbers, when you (+) them between themselves can be divided by 3. If they can't be divided by 3 then I have lost the game. How can I make an equation to solve this?
Best regards,
Alexander
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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17 Mar 2011, 13:40
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See from 1 to 500, you have 166 numbers which are divided by 3, and 167 numbers which are divided by 3 with the remainder of 1 and 2 correspondingly. Then you strategy is:
If you partner removes a number which is divided by 3 without a remainder, then you do this too (for example, he removes 99, then you removes 111)
If you partner removes a number which is divided by 3 with the remainder of 1 then you remove one with the remainder of 2 and vice versa (for example, he chooses 1 - you chose 13, he chooses 302 - you choose 61)
This strategy guarantees that after each round the sum of all numbers is divisible by 3. Moreover, since there is even number of numbers divided by 3 and equal number of numbers divided by 3 with remainder of 1 and 2, this will work at each step.
You have the dominant strategy! Congratulations)
If you partner removes a number which is divided by 3 without a remainder, then you do this too (for example, he removes 99, then you removes 111)
If you partner removes a number which is divided by 3 with the remainder of 1 then you remove one with the remainder of 2 and vice versa (for example, he chooses 1 - you chose 13, he chooses 302 - you choose 61)
This strategy guarantees that after each round the sum of all numbers is divisible by 3. Moreover, since there is even number of numbers divided by 3 and equal number of numbers divided by 3 with remainder of 1 and 2, this will work at each step.
You have the dominant strategy! Congratulations)
Thank you! That is perfection itself!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
