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02 Aug 2006, 11:50
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Louis and Fatima are on a long road trip, so Fatima proposes a game to pass the time. Louis will start by saying an integer from 1 to 15, and Fatima and Louis will take turns calling out an integer that is from 1 to 15 greater than the number they have just heard. For example, if Fatima hears Louis say 49, she can say 50 or 55 or 64, but not 48 or 65. The winner of the game is the person who can say 2006. If Louis begins by saying 4, what number should Fatima reply with as part of a fail-safe strategy to win the game?
(A) 6 (B) 8 (C) 11 (D) 14 (E) 19
(A) 6 (B) 8 (C) 11 (D) 14 (E) 19
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02 Aug 2006, 21:28
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I am not sure of my solution. But this one looks more like a puzzle to me!
Firstly if Louis if denoted by L and Fatima by F, then the order of events is
LFLFLF...............LF.
Secondly Fatima should win under any circumstances after she names her first number. Fatima can name any number between 5 and 19 (4+1 to 4+15).
I backtracked from 2006. If Fatima is to be safe she needs to let Louis finish in the range (1991,1992,...2005) so that she is able to win by saying any number that is more than Louis's number from this set by 1 to 15.
So that means the number that Fatima ought to say before Louis choses a number from (1991,1992,...2005) is 1990. Again she needs to let Louis hover in the set (1975,1976,...1989) for her to stop at 1974.
So the pattern for Fatima could be 2006,1990,1974...(diminishing by 16). You should be able to reach 6 using this principle (16*125 =2000 and 125 is an odd number --> L and F together had an even number of chances as F was the person to finish the competition).
So is the answer '6'?
Firstly if Louis if denoted by L and Fatima by F, then the order of events is
LFLFLF...............LF.
Secondly Fatima should win under any circumstances after she names her first number. Fatima can name any number between 5 and 19 (4+1 to 4+15).
I backtracked from 2006. If Fatima is to be safe she needs to let Louis finish in the range (1991,1992,...2005) so that she is able to win by saying any number that is more than Louis's number from this set by 1 to 15.
So that means the number that Fatima ought to say before Louis choses a number from (1991,1992,...2005) is 1990. Again she needs to let Louis hover in the set (1975,1976,...1989) for her to stop at 1974.
So the pattern for Fatima could be 2006,1990,1974...(diminishing by 16). You should be able to reach 6 using this principle (16*125 =2000 and 125 is an odd number --> L and F together had an even number of chances as F was the person to finish the competition).
So is the answer '6'?
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 Problem Solving (PS) 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!
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