[/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 Burapha integer if it satisfy the following condition
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 cubres
Kritesh manages traffic on a 
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the grid.
Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.
What is the minimum value of that guarantees that Kritesh's job is possible?
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the grid.Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.What is the minimum value of
that guarantees that Kritesh's job is possible?
This post has been edited 2 times. Last edited by Tony_stark0094, Apr 13, 2025, 3:12 AM
Reason: typo
Reason: typo
is the answer:
is the answer. To prove that it is minimal, consider the situation where cars in left 22 rows face right and right 22 rows face left, and top 22 cars in the middle row face down.
is the answer. To prove that it is minimal, consider the situation where cars in left 22 rows face right and right 23 rows face left.
.
rows face downwards.
.
cars in the middle row face left.
cars must be manually removed in this configuration, as desired.
cars are removed.
with cars pointing down. Fill the lower 
cells of the remaining row with cars pointing left and the remaining 
.
for any 
grid for odd 
cars must be removed. In each row, no cars facing opposite directions can be remaining as they bump into each other. Thus, in each row either all the right-facing or left-facing cars must be removed, mandating a removal of at least![\[\frac{n(n-1)}{2} + \frac{n-1}{2} = \frac{n^2-1}{2}\]](http://latex.artofproblemsolving.com/f/1/d/f1d343d222521b6fc5fd518f231f37e5cd7d0c61.png)
cars as desired.
, 
, 
and ![\[(U+L)+(U+R)+(D+R)+(D+L) = 2n^2\]](http://latex.artofproblemsolving.com/3/b/3/3b3e882bed42cba0df71636a9527bc79de076f47.png)
we have that,![\[\min\{U+L, U+R , D+R , D+L\} \le \left \lfloor \frac{2n^2}{4} \right \rfloor = \frac{n^2-1}{2}\]](http://latex.artofproblemsolving.com/5/6/5/5654607babab37133b3f5b5b06a71fe71ee7124d.png)
which finishes the proof of the bound and we are done.
