Consider an integer $n \ge 4 $ and a sequence of real numbers $x_1, x_2, x_3,..., x_n$. An operation consists in eliminating all numbers not having the rank of the form $4k + 3$, thus leaving only the numbers $x_3. x_7. x_{11}, ...$(for example, the sequence $4,5,9,3,6, 6,1, 8$ produces the sequence $9,1$). Upon the sequence $1, 2, 3, ..., 1024 $ the operation is performed successively for $5$ times. Show that at the end only one number remains and find this number.
Problem
Source: JBMO 2008 Shortlist A9
Tags: JBMO, algebra
pco
15.10.2017 13:43
parmenides51 wrote: Consider an integer $n \ge 4 $ and a sequence of real numbers $x_1, x_2, x_3,..., x_n$. An operation consists in eliminating all numbers not having the rank of the form $4k + 3$, thus leaving only the numbers $x_3. x_7. x_{11}, ...$(for example, the sequence $4,5,9,3,6, 6,1, 8$ produces the sequence $9,1$). Upon the sequence $1, 2, 3, ..., 1024 $ the operation is performed successively for $5$ times. Show that at the end only one number remains and find this number. Easy induction shows that starting from infinite sequence of natural numbers, the element of rank $k$ after $n$ operations is $4^nk-\frac{4^n-1}3$ Setting there $n=5$, the two first elements are $683,1707$ So, in the specific case of a sequence initially limited to $1024$ : $\boxed{\text{The fifth operation gives a unique number }683}$
Marinchoo
15.08.2020 19:38
Simpler solution: we know that the number must be of the form $22...2223$ in base $4$ so as $n < 1025 = 100001$(base$4$) the only such number is $22223$(base $4$) which is $683$.