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.
is composite. 
- all divisors of . It is known, that 
are all divisors for some 
(except 
). Find all such .
be a positive integer, with divisors 
. Prove that 
is always less than 
, and determine when it is a divisor of .
we consider the set of ten circles, each of which contains three of the vertices of the pentagon on its circumference. Is it possible that none of these circles contains the pentagon? Prove your answer.
is tangent to the circle 
at 
. 
and 
cut the circle at 
and 
respectively. 
is the in-centre of 
, and is on the line 
. 
and 
intersect at 
, while 
and 
intersect at 
. Prove that the points 
lie on a circle.
be an acute triangle and 
its orthocenter. Let be a point on the parallel through 
to such that 
. Define 
and 
as points on the parallels through 
to and through 
to similarly. If 
are positioned around the sides of as in the given configuration, prove that are collinear.
be an acute triangle, its orthocenter and the center of its nine point circle. Let be a point on the parallel through to such that 
and and are on different sides of and a point on the parallel through to such that 
and and are on different sides of . If 
and 
are the reflections of over and respectively, 
and 
are the intersections of 
and 
respectively with the circumcircle of , prove that the intersection of lines 
and 
lies on 
.
, 
is the angle bisector of 
. The circumcircle of 
intersects 
in 
respectively and the line through parallel to cuts 
in 
respectively. Prove that point 
lie on a circle.
kids in a class. Is it true that for all positive integers 
it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
be an integer. In a configuration of an 
board, each of the cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of , the maximum number of good cells over all possible starting configurations.
lamps 
in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp 
and its neighbours (only one neighbour for 
or 
, two neighbours for other 
) are in the same state, then is switched off; – otherwise, is switched on.
Prove that there are infinitely many integers 
for which all the lamps will eventually be off.
Prove that there are infinitely many integers for which the lamps will never be all off.
integers 
, each less than or equal to 
, prove that at least one of them is divisible by some other member of the set.
be positive integers such that no one of them is divisible by any other member of the sequence. Then![\[
a_1 \geq 2^k,
\]](http://latex.artofproblemsolving.com/1/f/e/1fecb6d6811bfc4aff7ccefa20412f0a5b860066.png)
where 
is defined by the inequalities![\[
3^k < 2n < 3^{k+1}.
\]](http://latex.artofproblemsolving.com/2/8/f/28f12a76029727a828167f1cb8e118b84d710d62.png)
This estimate for 
is the best possible.
be a sequence of positive integers. Then![\[
\max \left( (a_i, a_j) \right) > \frac{38n}{147} - c,
\]](http://latex.artofproblemsolving.com/6/5/d/65df2fa2c65d90d0548cb3c11dd36ea42c4060e6.png)
where 
is a constant independent of 
, and 
denotes the greatest common divisor of 
and 
. This is the best possible estimate.
, there exists only degree polynomial 
satisfying 
and ![$ (x + 1)[f(x)]^2 - 1$](http://latex.artofproblemsolving.com/9/c/1/9c1e80d035c923dd13711bff624d8a1a3e6a3b15.png)
is an odd function.
be the sequence 
so that 
for each 
and so that the decimal number 
is divisible by 
for each 
. Show that the decimal 
is irrational.
, then![\[2^{m+kn}\mid\overline{\underbrace{b_n\cdots b_1\ldots b_n\cdots b_1}_{k\text{ groups}}a_m\cdots a_1}.\]](http://latex.artofproblemsolving.com/2/4/d/24da518ef9cbdd1a586ddcbf96ed11c624b787a4.png)
Let 
, 
, then for all natural numbers , we have![\[2^{m+kn}\mid 2^mt+\sum_{i=0}^{k-1}10^{m+in}s,\]](http://latex.artofproblemsolving.com/6/5/b/65b8cf593896270e409398548b18b3a970da9e5e.png)
so![\[2^{kn}\mid t+5^m\sum_{i=0}^{k-1}10^{in}s=t+5^ms\cdot\frac{10^{kn}-1}{10^n-1}.\]](http://latex.artofproblemsolving.com/7/5/6/756e3b847083b40e993eb16623ba45459cceba86.png)
Thus, 
, so![\[2^{kn}\mid\left(10^n-1\right)t-5^{ms}.\]](http://latex.artofproblemsolving.com/9/2/a/92afc9a5d20487e2daf14014560ab9e56db1f56d.png)
This is not true for sufficiently large , since the right side is a fixed non-zero integer. Contradiction.
