[/quote]
,
be a prime number and 
.![\[\left| p^m - (p - 2)! \right| > p^2.\]](http://latex.artofproblemsolving.com/2/1/c/21ca8bb6e5727d48b18b7ec3b127029c4c97694f.png)
Proposed by Miloš Milićev
Determine 
such that 
Determine 
children, each connected to each other with the same type of rope, to be cured. Two experiments are said to be of the same type, if each of the ropes connecting the children has the same medicine imbued in it. She then unties them and lets them go back home.
be the minimum m such that Myla can perform at least 
experiments of the same type. Prove that:
For every 
there exists a 
and 
such that for all 
,![\[f(n, k) = a_kn + b_k.\]](http://latex.artofproblemsolving.com/6/6/0/660699c6cec811eee25f727076665a6df3257f57.png)
Find the value of 
for every 
be a prime number. Prove that if 
,
,
are integers (not necessarily positive) satisfying the equations ![\[ a^n + pb = b^n + pc = c^n + pa\]](http://latex.artofproblemsolving.com/9/2/2/922cd81d65164e887b05cc9bc6026998d6a80ce6.png)
then 
.
is an even integer. At the start, a line joining two vertices is chosen arbitrarily and one of its endpoints is chosen as its pivot. Now, Aloo rotates the line around the pivot either clockwise or anti-clockwise until it passes through another vertex of the 
be such that 
and 
.
be a triangle with orthocentre 
.
be the foot of altitudes of 
onto the opposite sides, respectively. Consider 
,
about line 
.
cut 
,
be the orthocenter of 
.
are concyclic.
of natural numbers such that for every pair of natural numbers 
and 
,![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
=2$$](http://latex.artofproblemsolving.com/0/c/d/0cd423e22e36e5b16ad19841c5702dd1fefbc354.png)
be the circumcenter and orthocenter of triangle 
be intersections of 
with 
.
be orthocenter and circumcenter of triangle 
.
![$f:\mathbb (0;1] \to\mathbb R$](http://latex.artofproblemsolving.com/7/9/9/799657d25f079cf2f55cd420ee41aa57bebc68a9.png)
such that![$$f(x)+f(1-x)=0\; \forall x \in (0;1] $$](http://latex.artofproblemsolving.com/7/d/e/7de74760f30216f280a168cd1e8c5737cb6bff0c.png)
be a positive integer. Consider necklaces of length n whose beads are colored in one of three colors, say red, green, or blue, with exactly two beads of each color (so 
)
of such distinct necklaces (up to rotation and reflection).
and 
meet at two distinct points 
and 
.
through 
(other than 
.
be the midpoint of the minor arc 
(i.e., the point on 
subtends the arc 
be the midpoint of the minor arc 
.![\[
MN \parallel \text{(radical axis of } \omega_1, \omega_2).
\]](http://latex.artofproblemsolving.com/6/6/4/6647f0f7182bdf5096ec5f2956a272803082f4f7.png)
Y by NO_SQUARES
A 
-sequence is a sequence of 
numbers 
each equal to either 
or 
. Determine the largest 
so that, for any -sequence, there exists an integer 
and indices 
so that 
for all 
, and 
-sequence is a sequence of 
numbers 
each equal to either 
or 
. Determine the largest 
so that, for any -sequence, there exists an integer 
and indices 
so that 
for all 
, and 
This post has been edited 7 times. Last edited by ike.chen, Aug 6, 2023, 10:00 PM
Reason: Typo
Reason: Typo
.
.
,
of the 
blocks with entries 
entries 
.
,
entries 
.
positive numbers in the initial sequence. We then choose all entries 
.
then one of the two adjacent 
must be landed on, therefore at least half a kind must be landed on, so with 
buckets in a string of length 
red buckets and 
,
,
.
ones. We will choose the 
as follows: divide the sequence into blocks of 
'
of the 
,
doesn't work, we will take the sequence 
,
and 
.
.
.
.
(since we have an unit block).
,
.
,
in a gap(so say if we have 
we go through indices 
)
be the number of 
![\[C\le \text{max}\left(x-\frac{2022-x}{2},2022-x-\frac{x}{2}\right) = \text{max}\left(\frac{3x-2022}{2},\frac{4044-3x}{2}\right).\]](http://latex.artofproblemsolving.com/6/a/5/6a5f591b76bc00b336a54b13be6c0f53529a63fd.png)
This is the maximum of two numbers that sum up to 
,
,
is repeated 
is added to it. Notice the maximal sum we can get in each block of 
is 
is 
.
is 
,
be the number of ones and 
.
numbers 
,
numbers 
In a block of 
numbers. Therefore, sum of 
So for every 
.
.
denote the number of 1s and -1s respectively. For each block of 
of them. Then the problem follows since![\[
c(1) - \sum_{\text{block of } x} \left\lfloor \frac x2 \right\rfloor \geq 1011 - \left\lfloor \frac{1011}{2} \right\rfloor = 506,
\]](http://latex.artofproblemsolving.com/c/f/e/cfea26dd28791bf66b1bd3133071a94e083c9bff.png)
as desired.
.
,
We will first prove a crucial lemma.
that satisfy the condition in the problems. Then, for any positive integer 
,
is in 
,
.
,
.
,
,
.
a contradiction. 
is maximized. Clearly 
,
gives the absolute value of the sum to be 
,
.
and 
as 
,
in the sequence of indices (i.e. making the new sequence of indices 
)
increase by 
over all 
,
which will make the absolute value of the sum bigger, contradicting the maximality of it. So we must have 
.
,
for which 
,
since 
since 
is 
and 
by the minimality of 
must be in 
.
is positive, so its absolute value is at least that of 
,
is maximal as well, so we may assume that 
,
and get a maximal 
,
.
for all 
.
,
,
are in 
'
has 
if 
if 
for some 
.
is at most 
,
,
of them. By the lemma, we have that for each odd 
,
,
,
.
,
of 
as desired.
.
for all 
,
so 
Now, note that 
is a valid sequence since 
,
for all 
,
Thus, we have that 
as desired.
so for any 
.
.
which would have at least as many 
,
for some 
for the same 
.
.
in order, where 
is the set of all 
.
,
,
'
odd numbers in 
,
so since all the 
'
Thus, we have that 
as desired.
works, so we're done. 
denote the sum in question; we seek the minimum of 
over all sequences 
.
,
terms; by symmetry, we can maximize the raw value of 
are among the chosen 
,
adjacent to but not among the chosen terms, we can include the terms 
whilst preserving conditions and incrementing 
.
,![\[S\le1-1+1+1-1+1+1-\dots-1+1+1=506,\]](http://latex.artofproblemsolving.com/a/4/e/a4edb6655d8673f0313b3ee5cb49f401abc79c1e.png)
as claimed.
denote the lengths of contiguous blocks of consecutive 
similarly. To maximize the magnitude of 
.
.![\[|S|_1=\sum_{i=1}^mr_i-\sum_{i=1}^n\left\lfloor\frac{b_i}{2}\right\rfloor\quad\text{and}\quad|S|_2=\sum_{i=1}^nb_i-\sum_{i=1}^m\left\lfloor\frac{r_i}{2}\right\rfloor\]](http://latex.artofproblemsolving.com/9/e/e/9eee5587035a47838e0dd5531df0dc0070f75850.png)
are both attainable. Summing, we have![\[|S|_1+|S|_2=\left(\sum_{i=1}^mr_i+\sum_{i=1}^nb_i\right)-\left(\sum_{i=1}^n\left\lfloor\frac{b_i}{2}\right\rfloor+\sum_{i=1}^m\left\lfloor\frac{r_i}{2}\right\rfloor\right)\ge2022-1011=1011,\]](http://latex.artofproblemsolving.com/5/1/2/51203d84ec9ce9aa552193e894e9702c98107368.png)
whence one of 
are 
,
By a straightforward check the exact bound is obtained for the sequence 
numbers respectively 
probably 
or 
equal to 
)
As a strategy lets pick all 
and in each block of 
instead of 
in the sigma notation?![\[\left| \sum_{i=1}^{k} a_{t_i} \right| \geqslant C.\]](http://latex.artofproblemsolving.com/7/7/b/77b6ddf8f9cbcab9c87ada489f8de339cf90f594.png)
in addition. In each 
we can get at most 
.
are the number of 
total then we have on a fixed sequence:
,
.
.
.
and the minimum is 
.![\[M+(-m)=\left \lceil \frac{a}{2}\right \rceil + \left \lceil \frac{b}{2}\right \rceil\]](http://latex.artofproblemsolving.com/b/0/c/b0c4a1ab52e28530605566f241d3e6381d38ed29.png)
which is equal to 
.
Here 
which on summing would give 
where 
are the sequence of 
are the sequences of 
,![\[ \sum \left\lfloor x_i/2 \right\rfloor \le \left\lfloor (x_1+x_2+\dots)/2 \right\rfloor \le 505, \]](http://latex.artofproblemsolving.com/b/a/d/badff2a6857ed7f674c8e55113ee4bccc04a9f71.png)
so adding in all of the positive terms yields that 
![\[\underbrace{+1, \, +1, \, \ldots, \, +1}_{1011}, \, \underbrace{-1, \, -1, \, \ldots, \, -1}_{1011}.\]](http://latex.artofproblemsolving.com/1/6/5/165b995f3f164decf633b4aeced6b2bc9c2edf1f.png)
Then our subsequence 
;
.
,
,
Clearly it is optimal for Sally to begin on a ``block" of red buckets as if she begins on blue, she may just shift one ``block" back. Thus say she attempts to maximize 
.
,
,
and blocks of blue buckets 
,
,
.
to begin at, and a block 
to end at.
.
Then we guarantee at least,
and hence we are done. 
for subsequence 
to be the sum of the lengths of consecutive subsequences of 
in subsequence 
is the number of 
,
.
,
.
satisfies the conditions. Observe that both 
and 
are equal to 
followed by 
negative ones. Then, in each consecutive group of negative ones we can eliminate at least half of them so there are at most 
negative ones left. Hence, the sum is at least 
,
of them. So there exists a sequence with at least 
.
Now for the lower bound, suppose the number of 
through 
and similarly the number of 
.
and thus,
.
.
.
and 
pick 
and if 
pick 
.
.
.
so by repeating the same algorithm from the start (without forgetting about the last number) we can choose every positive number and skip at least 
negative numbers so the sum will be at least 
.
with 
.
of these 
.
of these 
of 
of them and hence 
,
.
,
we have we can guarantee a sum at least 
é 
o 
e 
o 
-
e 
alternadamente entre os termos de 
começando com 
.
e, para cada 
Como queríamos mostrar.
?
,
of the 
is at least 
.
,
In particular, consider the sequence
Where the first term is 
somehow (if we can find 
,
)
,
.
WLOG.
.
.
.![$[a_1, a_{t_k}]$](http://latex.artofproblemsolving.com/2/d/8/2d83cb5e1158d65c0ff5d94068275011262f7565.png)
;
,
.
But note that 
,
.
.
(the leftmost group of plusses aren't countable in the range), so 
yet again as desired.
Consider the following algorithm 
with 
.
is positive, let 
,
.
.
groups of length 
respectively.
while the other one gives us 
. Therefore, the average value of 
Since the average of 
,
and 
respectively. WLOG assume 
.
.
Here there are 
Similar work holds if we let the number of 
ones in our sequence, so that there are 
negative ones. As before, we choose our sequence 
where 
In a run of 
of them, requiring us to choose 
of them. Therefore, when we choose the sequence which satisfies this, we get our sum is 
Since the sum is an integer, it is at least 
showing sufficiency.
as claimed.
Consider all the 
)
It is easy to check that the maximum possible sum in this case is 506.
There are 
chunks with an anomaly at the end. Say we want to maximise the number of 
more 
are, we can always achieve
for some 
We do so according to the following algorithm,
![\[ a_i = \begin{cases}1 & i = 1 \\ -1 & i = 2022 \\ -1 & i \equiv 2, 3 \pmod 4 \text{ and } i \neq 2022 \\ 1&i \equiv 0, 1 \pmod 4 \text{ and } i \neq 1.\end{cases}\]](http://latex.artofproblemsolving.com/a/6/5/a65bbd9b4d8aecea0e9ca6c7ae820a1a09ee6994.png)
This configuration is symmetric because 
for all 
so without loss of generality, it is sufficient to prove that 
for all possible configurations of the 
for each 
,
so we must take a 
,
If we choose to take 
so that we get 
so our maximum would be 
Otherwise, our maximum would be 
Assume without loss of generality that the number of 
is at least 
(If instead the number of 
apply a similar strategy as described below but with signs flipped.)
Then "fill in the gaps" where needed. If, in 
there is a run of 
instances of 
instances of 
of them. Now, 
by our without-loss-of-generality, so we can take ![\[ \sum\limits_{i=1}^{k} a_{t_i} \ge 1011 \cdot 1 + 505 \cdot -1 = 506.\]](http://latex.artofproblemsolving.com/3/8/3/3831e1e3163fe53396d37cf72637d66daeacc76a.png)
