Try Winning This Game!
In a game, Bob goes first, and he has to say a positive integer less than or equal to 16.
Then, Allison must add a positive integer less than or equal to 16 to Bob’s number, at which point Bob must add a positive integer less than or equal to 16, and so on. The winner is whomever says the number 2015 .
What number must Bob say first to ensure that he will win the game if both players play optimally?
The answer is 9.
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4 solutions
Well explained.
Another good answer is 14. Because 2015 - 14 = 2001. 2001 = (17) 1 + (16) 124. So the strategy is that Bob say 14, Annison say 1<=x<=16, Bob say 16-x (or 16 if Annison choice 16) and finally Bob is sure to say 2015
If Bob has to win and say 2 0 1 5 then he must say 1 9 9 8 (Why?). Because then maximum Allison can say is 2 0 1 4 and bob will say 2 0 1 5 and minimum that allison can say is 1 9 9 9 and then also bob can say 2 0 1 5 ( 1 9 9 9 + 1 6 = 2 0 1 5 ).
But to say 1 9 9 8 bob has to say 1 9 8 1 with same reasoning.Observe that the numbers which bob must say are of the type 2 0 1 5 − ( 1 7 n ) .So smallest positive integer of the form 2 0 1 5 − ( 1 7 n ) is 9 for n = 1 1 8 .
Moderator note:
Be careful with the explanation. While you have the correct ideas, this can be cleaned up slightly. Using the language of invariance principle or winning positions make it easier to express what you are thinking.
Exactly Same Method.
Just a simple division but great logic behind it!
Can you explain your reasoning?
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Sir, is there any other way to get the answer??? Is it possible to get it using probability in some way??
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Nope. Whatever Allison says, Bob can say a number that is 17 more than the last one; he has a 100% chance of winning this way.
I don't believe there is a solution involving probability. You should check out the answers by @Gautam Sharma and @Chris Callahan .
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@Eli Ross – Lets say that Jack also joins the game and Bob goes first, and he has to say a positive integer less than or equal to 16 to Jack .Then, Jack must add a positive integer less than or equal to 16 to Bob’s number, and has pass it on to Allison at the same time he must add a positive integer less than or equal to 16, and so on. The winner is whomever says the number 2015.
What number must Bob say first to ensure that he will win the game if all the players play optimally?
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@Rohan K – My teacher played this game with us on new session in class 10.The game was one who reaches 50 fifty first loses and we had to add 6 in every turn.In real life it can be used to troll a kid.
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@Gautam Sharma – Nice example bro!!
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@Rohan K – Yes you can play the game with your parents or friends. The winner would be the one who reaches 100 first and we can add positive integers from 1 to 10. In that case if you start with 1 no one can ever beat you provided that you play smartly.
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@Kushagra Sahni – Lets say that Jack also joins the game and Bob goes first, and he has to say a positive integer less than or equal to 16 to Jack .Then, Jack must add a positive integer less than or equal to 16 to Bob’s number, and has pass it on to Allison at the same time he must add a positive integer less than or equal to 16, and so on. The winner is whomever says the number 2015.
What number must Bob say first to ensure that he will win the game if all the players play optimally?
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@Rohan K – I dont think it is possible with Jack, he can disrupt the entire game. Since I did it using just like Gautam Sharma's method, instead of subtracting 17, you will think about 34 but since we do not know what Jack will say we cannot say who will win. This is what I think, I might even be wrong and please correct me if I am wrong.
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@Satyajit Ghosh – We can't think of 34 can we?. What if both the other players say 1 and then the chance comes to Bob again? Then game won't be finished there. We need to think of it in some other way.
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@Kushagra Sahni – That's what I'm saying thinking of 34 would be wrong (it might not have come into your mind but it was in the back of my mind while thinking about this situation)
@Satyajit Ghosh – If there's three playing, and one doesn't care about winning, but only about sabotaging one of the others, then that other one doesn't stand a chance. The question is, though, what if they ALL play optimally?
2 0 1 5 ≡ 9 m o d 1 7
Bob can take advantage of the fact that Allison can only add natural numbers less than or equal to 16. He can do this by constantly making sure that the sum of the number he adds and the number Allison adds is 17. Therefore if Bob starts with some number x, he can make sure that next turn he says x+17, and then x+34, and then x+51, and so on, regardless of what Allison says. So all he has to do is start with some natural number less than or equal to 16 that when added with some multiple of 17 become 2015. So we look for the number x that satisfies 1≤x≤16, and x+17n=2015, with n∈N. The only such number is 9 (in this case n=118). Therefore, if Bob chooses 9, he can ensure victory by controlling the sum of the number he adds and the number Allison adds.