Problem: { a k } k = 1 2011 { a k } k = 1 2 0 1 1
a 1 = 0.201 , a 2 = ( 0.2011 ) a 1 , a 3 = ( 0.20101 ) a 2 , a 4 = ( 0.201011 ) a 3 a 1 = 0 . 2 0 1 , a 2 = ( 0 . 2 0 1 1 ) a 1 , a 3 = ( 0 . 2 0 1 0 1 ) a 2 , a 4 = ( 0 . 2 0 1 0 1 1 ) a 3
and more generally
a k = { ( 0. 20101 … 0101 ⏟ k + 2 digits ) a k − 1 , if k is odd, ( 0. 20101 … 01011 ⏟ k + 2 digits ) a k − 1 , if k is even. a k = ⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ ( 0 . k + 2 digits 2 0 1 0 1 … 0 1 0 1 ) a k − 1 , if k is odd, ( 0 . k + 2 digits 2 0 1 0 1 … 0 1 0 1 1 ) a k − 1 , if k is even.
Rearranging the numbers in the sequence { a k } k = 1 2011 { a k } k = 1 2 0 1 1 { b k } k = 1 2011 { b k } k = 1 2 0 1 1 k , 1 ≤ k ≤ 2011 k , 1 ≤ k ≤ 2 0 1 1 a k = b k a k = b k
Answer Choices:
A. 671 6 7 1 1006 1 0 0 6 1341 1 3 4 1 2011 2 0 1 1 2012 2 0 1 2 Solution:
Because y = a x y = a x 0 < a < 1 0 < a < 1 y = x b y = x b [ 0 , ∞ ) [ 0 , ∞ ) b > 0 b > 0
1 > a 2 = ( 0.2011 ) a 1 > ( 0.201 ) a 1 > ( 0.201 ) 1 = a 1 , a 3 = ( 0.20101 ) a 2 < ( 0.2011 ) a 2 < ( 0.2011 ) a 1 = a 2 , 1 > a 2 = ( 0 . 2 0 1 1 ) a 1 > ( 0 . 2 0 1 ) a 1 > ( 0 . 2 0 1 ) 1 = a 1 , a 3 = ( 0 . 2 0 1 0 1 ) a 2 < ( 0 . 2 0 1 1 ) a 2 < ( 0 . 2 0 1 1 ) a 1 = a 2 ,
and
a 3 = ( 0.20101 ) a 2 > ( 0.201 ) a 2 > ( 0.201 ) 1 = a 1 . a 3 = ( 0 . 2 0 1 0 1 ) a 2 > ( 0 . 2 0 1 ) a 2 > ( 0 . 2 0 1 ) 1 = a 1 .
Therefore 1 > a 2 > a 3 > a 1 > 0 1 > a 2 > a 3 > a 1 > 0
1 > b 1 = a 2 > b 2 = a 4 > ⋯ > b 1005 = a 2010 1 > b 1 = a 2 > b 2 = a 4 > ⋯ > b 1 0 0 5 = a 2 0 1 0
> b 1006 = a 2011 > b 1007 = a 2009 > ⋯ > b 2011 = a 1 > 0 > b 1 0 0 6 = a 2 0 1 1 > b 1 0 0 7 = a 2 0 0 9 > ⋯ > b 2 0 1 1 = a 1 > 0
Hence a k = b k a k = b k 2 ( k − 1006 ) = 2011 − k 2 ( k − 1 0 0 6 ) = 2 0 1 1 − k k = 1341 k = 1 3 4 1
The problems on this page are the property of the MAA's American Mathematics Competitions