people be arranged in a circle, numbered clockwise from 
to 
denote the number of the last remaining person.
,
.
.
under certain conditions. Prove that the necessary and sufficient condition for this is that both 
and 
are prime numbers.
be the unit circle in the complex plane. Let 
be the map given by 
.
and 
for 
.
is called the period of 
.
of period 
.
)
and 
consecutive even numbers, such that the sum of their squares divides the square of their product.
and 
Then 
in terms of 
and 
for some real number 
and ![\[ a_n=na_{n-1}+(n-1)(n!(n-1)!-1) \]](http://latex.artofproblemsolving.com/4/1/f/41f0cc7b041a8bba056882b5c794503195c374fd.png)
for integers 
.
,
,
and 
are positive integers whose greatest common divisor is 
and 
are in the coordinate plane such that 
.
denote the locus of all points 
in the coordinate plane satisfying 
,
be the midpoint of 
.
and 
are on 
and 
.
can be expressed in the form 
,
and 
.
Prove that
Let 
Prove that
satisfying:
,
.
Y by sami1618
Ali is hosting a large party. Together with his 
friends, 
people are seated around a circular table in a fixed order. Ali places apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction).
Find all values of such that after some number of steps, the situation reaches a point where each person has exactly one apple.
friends, 
people are seated around a circular table in a fixed order. Ali places apples for serving directly in front of himself and wants to distribute them among everyone. Since Ali and his friends dislike eating alone and won't start unless everyone receives an apple at the same time, in each step, each person who has at least one apple passes one apple to the first person to their right who doesn't have an apple (in the clockwise direction).Find all values of
such that after some number of steps, the situation reaches a point where each person has exactly one apple.
work inductively as it goes from 
to two copies 
which are apart, this decays into all ones.
turn into a 
component with 
and 
zeros before them. The 
decays the same way so we can finish inductively to get that it never is all ones. Thus this ends up becoming 
inch worms in different states which never have the same all 
for all non-negative integers 
.
steps, everyone will have exactly one apple. We proceed by induction on 
is trivial. For the induction step, assume that the result holds for 
.
apples, the 
people will have exactly 
are even.
,
works, then 
also works.
for a non-negative integer 
.
