[/quote]
,
students and a teacher with 
marbles. The teacher then play a Marble distribution according to the following rules. At the start, each student receives at least 
marbles from the teacher. Then, the teacher chooses a student , who has never been chosen before, such that the number of marbles that he owns in a multiple of 
.
other students. The same goes on until the teacher is not able to choose anymore student.
satisfying the following condition: 
. Prove that
differents subjects, each consist of 
students. Given that for any 
different subjects, there exists a student compete in both subjects. Prove that there exists a student who compete in at least 
different subjects.
be positive real numbers such that 
.
,
the number of positive divisors of 
is an odd integer.
holds for every positive divisor 
of 
be real numbers such that 
for all 
.
.
be a quadrilateral with 
.
and 
meet at 
.
and 
denote the circumcircle and the circumcenter of triangle 
.
and 
denote the circumcircle and circumcenter of triangle 
.
meets 
and 
(other than 
and 
)
be the midpoints of minor arcs 
(not including 
(not including 
.
and 
be touch points of the incenter of 
at 
and 
,
and 
be the circumcenter of triangles 
and 
,
and 
concurrent.
of one variable that fullfill the following for all real numbers 
and 
:
.
if 
then 
.
,
such that 
.
be the unit circle in the complex plane. Let 
be the map given by 
.
and 
for 
.
is called the period of 
.
of period 
.
)
which satisfy the following equality for all 
![\[f(x)f(y)-f(x-1)-f(y+1)=f(xy)+2x-2y-4.\]](http://latex.artofproblemsolving.com/2/0/a/20ae7491750bac4c94deaed32d5b9aa66b26491f.png)
Proposed by Dániel Dobák, Budapest
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.
