[/quote]
,
be a cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
.
and 
meet at 
and the lines 
and 
meet at 
.
and 
be an acute triangle with circumcenter 
and orthocenter 
.
denote the circumcircle of triangle 
the midpoint of 
.
and 
,
.
and 
similarly.
such that for all positive integers 
and ![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
,
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of 
(which may depend on 
)
.
,
be the least prime that does not divide 
,
to be the product of all primes less than 
For 
,
.
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
.
.
,
,
such that, for each 
,
is an integer. Prove that there is a positive integer 
for all 
.
.
such that 
be the midpoint of 
and let 
and the circumcircle of 
.
passes through a fixed point.
and 
intersect each other at 
and 
.
and 
respectively. Let 
and 
be the reflection of 
intersect circles 
and 
(both distinct from 
is the second tangent to 
to 
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.
questions.
= 2∠AMP.
is the midpoint of the arc 
,
,
,
be the 
intersects 
.
is perpendicular to 
,
is the incenter of 
and the circle 
of radius 
Let 
The line 
intersects 
Let 
Let 
and 
to 
is a sequence of roads such that one can move from 
and 
denote the maximal number of paths in a diverse collection of paths from 
denote the maximal number of paths in a diverse collection of paths from 
holds if and only if the number of roads going out from 
Y by mmkkll, feranjos, CrazyFok, Littlelame, emmelin, KhaiMathAddict, cubres
Determine all composite integers 
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of 
with 
, then 
divides 
for every 
.
that satisfy the following property: if 
, 
, 
, 
are all the positive divisors of 
with 
, then 
divides 
for every 
.This post has been edited 7 times. Last edited by v_Enhance, Sep 18, 2023, 12:33 AM
Reason: minor latex pet peeve
Reason: minor latex pet peeve
be the smallest two prime divisors of 
.
for some 
and 
for some nonnegative integer 
.
,
and 
which is a contradiction.
.
,
and hence 
and hence 
and hence 
and hence 
and hence 
and hence 
etc.
and hence 
is a prime power which indeed works.
one can get that 
so 
,
in the same way, which leads to 
and 
for prime 
so 
and this is contradiction
for distinct prime divisors 
of 
for some prime divisor ![\[\frac{\frac{n}{p} + n}{\frac{n}{q}} = \frac{q(p+1)}{p},\]](http://latex.artofproblemsolving.com/0/5/c/05c967bf45ae5d33ffba891e1d9c652bc72acc03.png)
obviously not an integer.
and 
or 
;![\[ \frac{\frac{n}{p} + \frac{n}{p^2}}{\frac{n}{q}} = \frac{q(p+1)}{p^2}, \]](http://latex.artofproblemsolving.com/5/0/5/5059120c1c8d485fe7447ba05c92c0a5593df531.png)
obviously not an integer either.
.
Then, for some 
(specifically, the smallest 
)![\[\frac{n}{q}\large \mid\frac{n}{p^k}+\frac{n}{p^{k-1}}=\frac{n}{p^k}(k+1).\]](http://latex.artofproblemsolving.com/a/d/5/ad52ef76dad656f90976900413d51f04a0a18c4a.png)
This means that for some integer 
,![\[\frac{n\cdot l}{q}=\frac{n}{p^k}(p+1) \iff l\cdot p^k=pq+q\]](http://latex.artofproblemsolving.com/6/8/1/68157585856dc13142624a2d5fc58ee9edff7020.png)
so 
.
for all primes 
,
.![\[
\frac{n}{q} \mid \frac{n}{p^{k-i}} + \frac{n}{p^{k-i-1}},
\]](http://latex.artofproblemsolving.com/4/0/a/40aaa0ef28c0aca6ab6f5bb96699eb71b6d2046a.png)
but this is obviously false. We are done.
for distinct prime divisors 
for some prime divisor 
or 
;
.
is divisible by 
,
divides 
.
.
.
,
for some 
,
.
so 
are both divisible by 
is not.
for primes 
.
which is not an integer.
which are not integers. Thus only 
for some primes 
and some 
.
and hence
Which is impossible unless 
,
to be an integer, which is impossible.
,
and 
be the first two primes of 
be the largest number, such that 
![\[\frac{n}{p_2},\frac{n}{p_1^s},\frac{n}{p_1^{s-1}}\]](http://latex.artofproblemsolving.com/7/a/2/7a2e24917485ff73d99d64ee636ca0a168b0b005.png)
are consecutive divisors of 
Which is a contradiction, hence 
,
,
on both sides, the left side has 
,
,
This does not hold true
for some prime 
.
, 
.![\[d_i=p_{i-1}\mid p_i + p_{i+1}=d_{i+1}+d_{i+2}\]](http://latex.artofproblemsolving.com/d/7/e/d7ec55f74aa6c1f73d60b87f3a29dbf4713c30d3.png)
quite clearly.
for all 
for some 
and 
as the divisors are arranged in increasing order. Now, if 
which is a very clear contradiction since 
are both primes. Thus, there are no such 
where 
work evidently. We now show these are the only ones.
.
![$$\implies \boxed{d_{i}d_{i+1} \mid d_{i+2}(d_{i}+d_{i+1})} [\because i \mapsto m-i-1]$$](http://latex.artofproblemsolving.com/c/1/6/c16495b4b20ccabcb4b9e38e9198166de8adc88a.png)
Now 
gives, 
since 
.
is a prime number (say 
has any prime factor 
which is a blatant contradiction. Thus 
as it can not be 
otherwise 
again, a contradiction. Now by an easy strong induction and similar argument we can conclude that 
.
where 
)
.
-
:
,
.
,
and 
and 
,
.
and 
,
.
,
.
so we must have 
.
.
or 
.
,
,
for a prime 
,
,
is just 
,
always divides 
.
and 
.
.
,
.
to be an integer, but 
is clearly not an integer. There are no cases here.
for 
.
must be an integer, but this is clearly impossible.
be the smallest two prime divisors of 
Let 
be the two consecutive divisors. Hence, we get 
so 
or 
All perfect powers work.
so 
.
,
.
(easy to see why)
,
with 
are the only solutions (easy to check why these work).
be the largest integer such that 
and 
(assume 
as the case 
is trivial). One may notice that 
,
and 
.
From the above we say 
,
Hence applying this argument in general yields 
So 
So if some prime 
where 
So there is no other prime 
dividing 
It is easy to see that this works.
