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.
be a function satisfying the following conditions:
, if 
and 
, then 
;
is surjective.
.
be a triangle with 
and 
. Let 
and 
be the midpoints of 
and 
respectively. The point 
is inside of 
such that 
is equilateral. Let 
and 
. Prove that 
.
. This has to be covered by three types of tiles - red, blue and black. The red tiles are of size 
, the blue tiles are of size and the black tiles are of size 
. Let 
denote the number of ways this can be done. For example, clearly 
because we can have either a red or a blue tile. Also 
since we could have tiled the plot as: two red tiles, two blue tiles, a red tile on the left and a blue tile on the right, a blue tile on the left and a red tile on the right, or a single black tile.
for all 
.
for all 
.![\[ \binom{m}{r} = \begin{cases}
\dfrac{m!}{r!(m-r)!}, &\text{ if $0 \le r \le m$,} \\
0, &\text{ otherwise}
\end{cases}\]](http://latex.artofproblemsolving.com/b/8/d/b8d653661e421db1fce7d93837e7bce31337bb30.png)
for integers 
.
can be written in the form
where 
for some non-negative 
such that
for every 
.
be real numbers such that ![\[ab^2+bc^2+ca^2=6\sqrt{3}+ac^2+cb^2+ba^2.\]](http://latex.artofproblemsolving.com/3/d/0/3d015292c249a69a44d5f923092201d7a271b896.png)
Find the smallest possible value of 
. such that the following statement holds: Suppose real numbers 
, 
, 
, 
, 
, 
, , 
satisfy 
for all 
. Then there exists 
, 
, , 
, each of which is either 
or 
, such that![\[ \left| \sum_{i=1}^n \varepsilon_i a_i \right| + \left| \sum_{i=1}^n \varepsilon_i b_i \right| \le 1. \]](http://latex.artofproblemsolving.com/6/5/8/658ddc5e8626e7d5ef29809c8c3ac641e3a87757.png)
vertices is given. To each vertex, a natural number from the set 
is assigned. Prove that there are 
vertices 
such that if the numbers 
are assigned to them respectively, then 
and 
is a parallelogram.
such that 
. Show that 
. When will equality hold?
.
has 
obtuse. Let , , be the feet of the altitudes from 
, 
, 
respectively. Let 
, 
, 
be arbitrary points on 
, , 
respectively. The circles with diameter 
, 
, 
are drawn. Show that the lengths of the tangents from the orthocentre of to these circles are equal.
![\[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]](http://latex.artofproblemsolving.com/6/4/9/6499d8f992f3e72765cab1f3afe15df99d01b12a.png)
and it is clear, since for any prime 
we have 
is divisible by 
by Fermat Small Theorem (
, while 
). For 
we have 
and so it is divisible by 6.
and it is clear, since for any prime we have is divisible by by Fermat Small Theorem (, while ). For we have and so it is divisible by 6.
because it doesn't seem ok (I mean 
would mean that 5 divides 15/2 from the definition of congruence).. So I wrote it like this 
and then we get 
which is obviously divisible by p (for p>=5), and beacuse 
and 
we see that there's no solution in prime numbers..
those residues coprime to are invertible, i.e. if 
and are coprime, then there exists 
such that 
modulo . So, if we write 
, we mean 
.
because it doesn't seem ok (I mean would mean that 5 divides 15/2 from the definition of congruence)..
! I think this is the easiest Number theory problem in IMO .
: 
is divisible by 
(It's proved easily by Fermat Little Theorem)
p>3
x=2^{p-2}, y=3^{p-2}, z=6^{p-2}
2x\equiv 3y\equiv 6z (\equiv 1) (\mod p)and
\gcd(p,2)=\gcd(p,3)=1
x\equiv 3z, y\equiv 2z (\mod p)
x+y+z\equiv 6z\equiv 1 (\mod p)or
p|a_{p-2}
.
, so the 4 solutions for the characteristic equality 
are 
by the explicit term, we get
and 
. is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution.
is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution.
!!!!!!!!! 
such that 
.
.
,
is a complex number so we are done
![\[2^n + 3^n + 6^n - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/a/d/6ad29e27ec0af4da7cf507a4e6dd028d3b82747b.png)
Let 
.![\[\frac{1}{2} + \frac{1}{3} + \frac{1}{6} - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/b/a/6ba086076d7eeff9592a9116904281db53aa0def.png)
so 
![\[2^{p-2}+3^{p-2}+6^{p-2}\equiv2^{-1}+3^{-1}+6^{-1}\equiv1\pmod{p}.\]](http://latex.artofproblemsolving.com/d/e/a/deabeef3910647848a5e1104e94d4999f44bcc38.png)
for all 
,![\[a_{p-2} \equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/5/2/3/523533c71a0b86d911bd502c4e4f364aeb4cb041.png)
for all 
.
and 
divides 
contains all primes.
then we have 
which is true for any 
then we have 
which is true for all even 
.
.
to get 
.
,
,
.
is a power of 
which is not relatively prime to 
that divides 
,
so 
,
,
,
,
,
,
,
,
Note that 
which completes the problem.
which obviously works. It clearly suffices to show that 
.
,
modulo 
don't work. Now, suppose that 
Therefore 
![\[2^{p-2}+3^{p-2}+6^{p-2}-1\equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0 \pmod{p}\]](http://latex.artofproblemsolving.com/6/7/2/672f7522379f8e28a4f5f8d901177aa72acc5282.png)
Thus, 
for all primes 
satisfy this condition, but the cases where 
are messier. In particular any positive integer of the form 
does not satisfy this condition.
and 
By FLT, we can say 
where 
Because 
we can say 
so we are done.