be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
,
such that 
.
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
,
are all greater than 
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 
,
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
,
to 
,
to 
.
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
is inscribed in circle 
with 
and 
.
.
meets 
(other than 
lies on arc 
of 
is perpendicular to 
.
denote the midpoint of chord 
.
be a 
grid of unit squares.
or 
rectangle. A subset 
edges of the grid squares.
Y by pbj2006, justJen, HamstPan38825, megarnie, Adventure10, Mango247, Sedro
Each cell of an 
board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:
(i) The difference between any two adjacent numbers is either
or 
.
(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to.
Determine the number of distinct gardens in terms of
and 
.
board is filled with some nonnegative integer. Two numbers in the filling are said to be adjacent if their cells share a common side. (Note that two numbers in cells that share only a corner are not adjacent). The filling is called a garden if it satisfies the following two conditions:(i) The difference between any two adjacent numbers is either
or 
.(ii) If a number is less than or equal to all of its adjacent numbers, then it is equal to
.Determine the number of distinct gardens in terms of
and 
.
,
grid with 
squares and operating as necessary.
(since we excluded all 1s).
.
until the entire grid is filled.
so at the end, all of the squares will be filled in. Furthermore, each 
so each number satisfies condition (ii). To prove it satisfies condition (i), notice that the smallest number that a square containing number 
(
can also be adjacent to it, but nothing above that(as if there was a square with number 
.
where we choose one of the numbers in the adjacent square that a square is greater than. However, notice that this inequality chain has to end at some square, and this square has a number less than all the squares in the grid, so it must be a 0, a contradiction.
.
. Thanks in advance!
.
,
'
'
have no adjacent squares so 
.
for 
have been placed and they have been placed such that, after all 
'
'
adjacent to a 
by (ii). But that contradicts our induction hypothesis, therefore, 
because there are 
be the distance to the nearest 0, and let 
be the value assigned to A.
because there exists a path with 
.
.
,
.
which is absurd.
all other squares by starting from some square and going to a neighbor that is smaller, thus there is always at least one '0'.
and we will never find an adjacent cell numbered less than 
which means 
are adjacent to a square with distance 
from 
,
.
.
number of 
,
and so on. This effectively forces all other numbers on the board.
where 
denotes the number of cells to the left of 
denotes the number of cells below 
as 
.
,
,
with a 
denote the set of all cells labeled with 
as the set of all cells not in any 
for valid 
and adjacent to a cell in 
for all positive integers 
must be labeled with a 
;
is labeled with 
denote the set of cells not in 
.
labeled with 
or 
or 
It follows that the answer is indeed 
