Difference between revisions of "2000 AIME II Problems/Problem 15"

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== Problem ==
 
== Problem ==
A stack of <math>2000</math> cards is labelled with the integers from <math>1</math> to <math>2000,</math> with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: <math>1,2,3,\ldots,1999,2000.</math> In the original stack of cards, how many cards were above the card labeled 1999?
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Find the least positive integer <math>n</math> such that <center><math>\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.</math></center>
  
 
== Solution ==
 
== Solution ==

Revision as of 18:44, 11 November 2007

Problem

Find the least positive integer $n$ such that

$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.$

Solution

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See also

2000 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
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