AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
such that 
for all 
be a positive integer. A board with a format 
is divided in equal squares.Determine all integers ≥3 such that the board can be covered in 
(or 
) pieces so that there is exactly one empty square in each row and each column.
be a positive integer. Given a sequence 
, 
, 
with 
or 
for each 
, , 
, the sequences 
, , 
and 
, , 
are constructed by the following rules: ![\[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\]](http://latex.artofproblemsolving.com/d/f/d/dfd49833078b6b676cd261cc358e8114b2c4b7f4.png)
![\[\begin{array}{lll}
a_{i+1} =
\begin{cases}
2a_{i-1} + 3a_i, \\
3a_{i-1} + a_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_i = 0, \\
\text{if } \varepsilon_i = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1, \\[15pt]
b_{i+1}=
\begin{cases}
2b_{i-1} + 3b_i, \\
3b_{i-1} + b_i,
\end{cases} &
\begin{array}{l}
\text{if } \varepsilon_{n-i} = 0, \\
\text{if } \varepsilon_{n-i} = 1, \end{array}
& \text{for each } i = 1, \dots, n - 1.
\end{array}\]](http://latex.artofproblemsolving.com/8/9/0/8908021c60992e39acdb4fcf6b587f96b88539e0.png)
Prove that 
.
and be natural numbers. Every one of the 
squares of the board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this: and can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
be a real or complex inner product space. Suppose that 
and that 
Then
be an integer, and let 
be positive integers. Show that there exist positive integers 
satisfying the following three conditions:
for 
the remainders of on division by are pairwise different; and
denotes the integer part of real number 
, that is, the largest integer that does not exceed .)
, the angle bisector at 
meet the opposite sides at 
, respectively. Prove that if the quadrilateral 
is cyclic, then
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number , divisible by , such that there exists an divisor table, or prove that such does not exist. be a right-angled triangle where ∠
=90°.Let 
be an altitude of that triangle and points 
and 
be the midpoints of and 
, respectively.If 
is the circumcenter of the triangle 
, prove that 
and are paralel.
, let 
be a point (
). Let 
be the anticomplement of , and let 
be the 
-anticomplementary conjugate of . Prove that the line 
is parallel to the real asymptote of the circular pivotal isocubic with pivot .
and 
. Prove that there always exists a positive integer 
such that 
. is a power of a prime number, but for arbitrary then well.....
be a positive integer. Define a sequence by setting 
and, for each 
, letting 
be the unique integer in the range 
for which 
is divisible by 
. For instance, when 
the obtained sequence is 
. Prove that for any the sequence 
eventually becomes constant.
be complex numbers satisfying 
. Show that there exist some among the complex numbers such that the modulus of the sum of these complex numbers is not less than 
.
hence your claim is done.
as vectors, and put them all in the plane starting at the origin. By pidgeonhole, one of our three sectors has at least 1/3 of the total length of the vectors. Wlog this sector is the one with rays in the 0 radian and 2pi/3 radian directions as its boundaries.
