AoPS Academy
be an odd prime number. Let 
be the sequence defined as follows: 
, and for 
, 
is the smallest integer greater than 
such that in 
there are no arithmetic progressions of length . We say that a positive integer is a ghost if it doesn’t appear in 
.?
with 
, prove that
be an integer. In a configuration of an 
board, each of the 
cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate 
counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of 
, the maximum number of good cells over all possible starting configurations.
such that the equation in the unknown 
:
admits 
roots: 
s.t. 
.
be a function, two times derivable on 
for which there exist 
such that![\[ \frac { f(b)-f(a) }{b-a} \neq f'(c) ,\]](http://latex.artofproblemsolving.com/a/3/1/a31861ef909a6e17ed8b495b9b9b16d887ba573f.png)
for all 
.
.
, 
and 
.
be complex numbers with 
. Find the greatest possible value of the expression 
be a positive root of the equation 
. Prove that the following sequence converges: 
(0<x<2π)
with domain ![$[0,2025]$](http://latex.artofproblemsolving.com/d/1/7/d17ac41d54c4b1888fb8bb43cf11b372691694a0.png)
satisfying 
and 
is strictly increasing on Compute smallest real M such that all functions in this set ,
.
have local extrema at 
respectively. find 
. Here 
are constants .
be the region in the complex plane enclosed by curve 
for 
. Compute perimeter of .
and 
such that![\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]](http://latex.artofproblemsolving.com/6/4/b/64b5f5919fd3710cc15e96724918eae216c8e4f0.png)
be a prime divisor of 
. Notice that 
for any positive integer 
that is not a multiple of 
, we get that either is a multiple of , in which case 
, or 
.
, and thus 
. In particular, it would imply 
, giving the solution 
.. Hence, all 
give 
. Either 
is the only such prime, or is even. In the latter case, 
. So, by a similar logic to above, unless , the only possibility is that is a power of two. In this case, if 
, then the left hand side is 
, which is a contradiction. Hence, again the only possibility is .
. Suppose is even, and let 
Note that for 
, we have 
. Therefore 
. Now consider the LHS mod 
. As 
, for all 
we have![\[a ^ n \equiv \begin{cases}
0 \pmod {2^{k+1}} & \text{if } 2 \mid a \\
1 \pmod {2^{k+1}} & \text{otherwise.}
\end{cases}\]](http://latex.artofproblemsolving.com/9/3/a/93acfff5bc3e7dfd4c5f59eecae5d329e20a7612.png)
Therefore,![\[1^n + 2^n + 3^n + \cdots + n^n \equiv \frac{n}{2} \not\equiv 0 \pmod {2^{k+1}},\]](http://latex.artofproblemsolving.com/9/7/0/9706fe8e82d8b582b4e63e358393b6bead716741.png)
is odd. Consider LHS mod 
. We notice that![\[\left(\frac{n-1}{2}\right)^n + 1 \equiv x! \equiv 0 \pmod {n-1}.\]](http://latex.artofproblemsolving.com/7/4/8/748703dd1e76d0b3fab6b09d01aaedfa91766c13.png)
, we know![\[\frac{n-1}{2} \Biggm| \left(\frac{n-1}{2}\right)^n + 1 \implies \frac{n-1}{2} \mid 1.\]](http://latex.artofproblemsolving.com/1/6/3/1630826d15c225d4ae03781ab4c278cd392b6e2f.png)
, which does not work. Therefore, we have 
. Checking each case by hand we see that 
gives 
and hence we are done. 
n is even
n is odd
is the only odd satisfying given condition
Then mod 
yields, by Euler's Theorem, 
So 
contradiction. Hence is odd. But if 
so by pairing 
contradiction. So 
and testing yields the only solution 
though?
but for me it was motivated by sum of powers and pairing to cancel things out, as is odd.
though? but for me it was motivated by sum of powers and pairing to cancel things out, as is odd.
