AoPS Academy
+2 w
to job 
and has a better ability in job than candidate B, but A is assigned job and B is assigned job , then A will complain. Also, if it is possible to assign each job to a candidate with a higher ability, the director will complain. Show that the boss can assign the jobs so as to avoid any complaints.
be a fixed integer greater than 
. The sequence 
, 
, 
, 
is defined as follows:![\[x_i = \begin{cases}2^i&\text{if }0\leq i \leq m - 1;\\\sum_{j=1}^mx_{i-j}&\text{if }i\geq m.\end{cases}\]](http://latex.artofproblemsolving.com/3/c/6/3c6879be01c394c27712d3acf43e22f1be17db49.png)
Find the greatest 
for which the sequence contains consecutive terms divisible by .
be non-negative real numbers such that 
Prove that
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from 
to 
, and let 
be the reflection of across 
. Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
such that 
is an integer for every positive integer 
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment if and only if the following condition holds: distinct from 
and 
, there exists 
such[/center]
is directly similar to either 
or 
.[/center]
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to , 
to , and 
to 
.
such that for any two infinite sequences of positive integers 
and 
, such that 
for all 
, there always exists a real number 
such that 
for all 
.
such that for any two infinite sequences of positive integers and , such that for all , there always exists a real number such that for all .
and 
of any length are considered. The answer for 
will be slightly different.
satisfying the equation for all real numbers 
:![\[
f(x^2 + y + f(y)) = f(x)^2
\]](http://latex.artofproblemsolving.com/0/6/f/06f2c3b16aecf900a28c033c45993564afd80055.png)
,
.
,![\[\lfloor 2c\rfloor\equiv 1\pmod n\]](http://latex.artofproblemsolving.com/5/7/b/57bdbd10b6b9a77edf8c6a026f995b4acdc86073.png)
![\[\lfloor{4c}\rfloor\equiv 1\pmod n\]](http://latex.artofproblemsolving.com/5/3/3/533a5ca2b3609513078b264da52dd4d70643ee76.png)
contradiction as 
or 
.
,
.
contains a 
after the point, let ![\[\lfloor 2^k\{c\}\rfloor\equiv 0\pmod 2\]](http://latex.artofproblemsolving.com/2/9/0/290f9f1f284c79c37e19a505294b727b8fa75d90.png)
![\[\lfloor a_k c\rfloor\equiv 0\pmod 2\]](http://latex.artofproblemsolving.com/c/3/b/c3b71d0a1ca562d7101dc9fc604afcf845333d24.png)
contradiction. Otherwise, 
,
.
for all 
.![\[ \lfloor{ca_i}\rfloor\equiv b_i\pmod 2\]](http://latex.artofproblemsolving.com/c/9/f/c9f51bdb9e11d5647f6f0f691b896d6e83cc1721.png)
is equivalent to there exists 
such that![\[ \frac{2x_i+b_i}{a_i}\le c < \frac{2x_i+b_i+1}{a_i} \]](http://latex.artofproblemsolving.com/2/7/3/2734a7f8c2445d96bbf417e957e36656d4ca9a53.png)
Similarly, the ceiling condition is equivalent to![\[ \frac{2x_i+b_i}{a_i}< c \le \frac{2x_i+b_i+1}{a_i} \]](http://latex.artofproblemsolving.com/3/8/e/38e7c2db2972a69bc8e46a95689a907a31343ad8.png)
We will now show that there exists 
such that![\[ \frac{2x_1+b_1}{a_1}\le \frac{2x_2+b_2}{a_2}\le \cdots\]](http://latex.artofproblemsolving.com/5/b/d/5bdf9ebd97a6b68a612daf80d4d5dc1089613199.png)
and![\[ \frac{2x_1+b_1+1}{a_1}\ge \frac{2x_2+b_2+1}{a_2}\ge \cdots\]](http://latex.artofproblemsolving.com/4/e/d/4ed10f44de9ae3c5726bb82a967d65bdf90b5483.png)
Induct. If 
exists, then we find 
.![\[ (2x_i+b_i)d_{i+1} \le 2x_{i+1}+b_{i+1} \le (2x_i+b_i+1)d_{i+1}-1 \]](http://latex.artofproblemsolving.com/3/e/0/3e0d29acd659ff7c3eca93aed8492cae323fada7.png)
But 
so there exists at least two integers within the interval. One of them must be 
so 
is not eventually constant, take 
is not eventually constant, take 
strictly decreasing.