AoPS Academy
be vertices of regular pentagon inscribed in a circle whose radius is 
and center is at 
. Find all possible values of 
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
![\[ a_n=2^n+3^n+6^n -1,\ n\geq 1. \]](http://latex.artofproblemsolving.com/6/4/9/6499d8f992f3e72765cab1f3afe15df99d01b12a.png)
and it is clear, since for any prime 
we have 
is divisible by 
by Fermat Small Theorem (
, while 
). For 
we have 
and so it is divisible by 6.
and it is clear, since for any prime we have is divisible by by Fermat Small Theorem (, while ). For we have and so it is divisible by 6.
because it doesn't seem ok (I mean 
would mean that 5 divides 15/2 from the definition of congruence).. So I wrote it like this 
and then we get 
which is obviously divisible by p (for p>=5), and beacuse 
and 
we see that there's no solution in prime numbers..
those residues coprime to are invertible, i.e. if 
and are coprime, then there exists 
such that 
modulo . So, if we write 
, we mean 
.
because it doesn't seem ok (I mean would mean that 5 divides 15/2 from the definition of congruence)..
! I think this is the easiest Number theory problem in IMO .
: 
is divisible by 
(It's proved easily by Fermat Little Theorem)
p>3
x=2^{p-2}, y=3^{p-2}, z=6^{p-2}
2x\equiv 3y\equiv 6z (\equiv 1) (\mod p)and
\gcd(p,2)=\gcd(p,3)=1
x\equiv 3z, y\equiv 2z (\mod p)
x+y+z\equiv 6z\equiv 1 (\mod p)or
p|a_{p-2}
.
, so the 4 solutions for the characteristic equality 
are 
by the explicit term, we get
and 
. is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution.
is unique in both directions modulo any prime, 0 will appear again. Therefore 1 is the only solution.
!!!!!!!!! 
,
for all 
,
,
is a complex number so we are done
![\[2^n + 3^n + 6^n - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/a/d/6ad29e27ec0af4da7cf507a4e6dd028d3b82747b.png)
Let 
.![\[\frac{1}{2} + \frac{1}{3} + \frac{1}{6} - 1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/6/b/a/6ba086076d7eeff9592a9116904281db53aa0def.png)
so 
![\[2^{p-2}+3^{p-2}+6^{p-2}\equiv2^{-1}+3^{-1}+6^{-1}\equiv1\pmod{p}.\]](http://latex.artofproblemsolving.com/d/e/a/deabeef3910647848a5e1104e94d4999f44bcc38.png)
for all 
,![\[a_{p-2} \equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0\pmod{p}\]](http://latex.artofproblemsolving.com/5/2/3/523533c71a0b86d911bd502c4e4f364aeb4cb041.png)
for all 
.
divides 
and 
respectively. Thus, the only such 
contains all primes.
then we have 
which is true for any 
then we have 
which is true for all even 
.
.
to get 
.
,
,
.
is a power of 
which is not relatively prime to 
that divides 
,
so 
,
,
,
,
,
,
,
,
Note that 
which completes the problem.
which obviously works. It clearly suffices to show that 
.
,
modulo 
don't work. Now, suppose that 
Therefore 
![\[2^{p-2}+3^{p-2}+6^{p-2}-1\equiv \frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1 \equiv 0 \pmod{p}\]](http://latex.artofproblemsolving.com/6/7/2/672f7522379f8e28a4f5f8d901177aa72acc5282.png)
Thus, 
for all primes 
satisfy this condition, but the cases where 
are messier. In particular any positive integer of the form 
does not satisfy this condition.
and 
By FLT, we can say 
where 
Because 
we can say 
so we are done.