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.
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.
such that there exists a sequence of positive real numbers 
, 
, with the property that for every it holds that:![\[
a_1 + \cdots + a_{n+1} < \alpha \cdot a_n.
\]](http://latex.artofproblemsolving.com/9/b/7/9b73deb46032844b324288b02470ff0623ac0ff1.png)
Proposed by Pavle Martinović
of the set 
, define its colorfulness as the greatest natural number 
such that:
, , if 
, then 
.
? Determine how many such permutations have maximal colorfulness.
such that:
is strictly increasing,
such that 
for all 
,
, there exists 
such that![\[
f(y) = \frac{f(x) + f(x + 2024)}{2}.
\]](http://latex.artofproblemsolving.com/b/e/2/be26213154bb74bd5a35b8d160011351871bfa9b.png)
Proposed by Pavle Martinović
be an acute triangle. Let 
be the reflection of point over the line 
. Let 
and 
be the circumcenter and the orthocenter of triangle , respectively, and let 
be the midpoint of segment 
. Let 
and 
be the points where the reflection of line 
with respect to line 
intersects the circumcircle of triangle , where point lies on the arc not containing . If 
is the foot of the perpendicular from to 
, prove that 
.
+1 w
be a prime number and 
. Prove that![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
, 
, 
, 
of real numbers satisfies![\[
a_{2n-1} + a_{2n} > a_{2n+1} + a_{2n+2} \qquad \mbox{and} \qquad a_{2n} + a_{2n+1} < a_{2n+2} + a_{2n+3}
\]](http://latex.artofproblemsolving.com/4/d/a/4da3b88a20c42c1798141b8db086de341cfb9a67.png)
for every positive integer . Prove that there exists a real number 
such that 
for every positive integer .
such that 
is divisible by 
and 
. Prove that is a composite number.
satisfying
Prove
+2 w![$P(x) \in \mathbb{R}[x]$](http://latex.artofproblemsolving.com/4/5/3/453a624c3b002c1b0e78e0023b24dd22ddd03557.png)
such that:
with 
is an th prime and 
is an arbitrary positive integer.
such that for all real numbers 
and 
,
be a triangle and let 
its circumscribed circle. Let be the midpoint of the side and let be the point of intersection of the line 
with . By a straight line is drawn parallel to , which intersects at a point . Let 
be the midpoint of the segment 
and let 
be the point of intersection of 
with . Show that 
.
be triangle, inscribed in parabola. Tangents in points 
forms triangle . Prove that 
. (
is area of triangle 
).
be triangle, inscribed in parabola. Tangents in points forms triangle . Prove that . ( is area of triangle ).
be any point on the Steiner circumellipse of 
and let 
be the cevian triangle of WRT 
Then ![$[DEF]=2 \cdot [ABC].$](http://latex.artofproblemsolving.com/3/3/e/33ec9a123e2b3b63c128a4fe7fe4aa12551261a1.png)
be the antimedial triangle of (
against 
). The Steiner circumellipse 
of is obviously the Steiner inellipse of tangent to its sides at 
are collinear on the polar of WRT and similarly 
and 
are respectively collinear. Thus by Pappus theorem for the hexagon 
it follows that 
is at infinity, i.e. 
and similarly we get 
![$[DEF]=[DYZ]=[CYZ]=2\cdot [ABC]. \ \blacksquare$](http://latex.artofproblemsolving.com/b/4/0/b4028b012dc90ae29c0bc707cbf447409f6ab058.png)
be the perspector of the parabola. Since the center of any inconic is the isotomcomplement of its perspector (in the parabola case at infinity), then it follows then that the isotomic conjugate of WRT must be on the line at infinity is on the Steiner circumellipse of and the result follows from the lemma.
Prove that
Where 
