such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
is divisible by 
for all positive integers 
.
and 
.
be a polynomial in one variable with integer coefficients. Prove that the number of pairs 
of positive integers such that 
,
,
.
such that![\[\mathcal{E}(B \cup C) = B \cup C.\]](http://latex.artofproblemsolving.com/e/c/3/ec34115c9e6e169c19c739220065f647934a1a5f.png)
.
and ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
be a strictly monotonous function such that 
.
.
for which there exists a scrictly monotonous function 
such that 
,
, ∀x∈[0,1].
a distance 
from the start. What is the maximum time it could have taken to travel the distance 
Y by Mango247
In a language
an alphabet with 
letters is used
words are exactly all sequences of 
not necessarily different 
letters of length 
Two ends of a paper strip are glued so that the strip forms a ring the strip bears a sequence of 
letters
Say that a word is singular if one can cut a piece bearing exactly that word from the strip but one cannot cut out two such non-overlapping pieces It is known that one can cut out 
non-overlapping pieces each containing the same word Determine the largest possible number of singular words
(Bogdanov I.)
an alphabet with 
letters is used
words are exactly all sequences of 
not necessarily different 
letters of length 
Two ends of a paper strip are glued so that the strip forms a ring the strip bears a sequence of 
letters
Say that a word is singular if one can cut a piece bearing exactly that word from the strip but one cannot cut out two such non-overlapping pieces It is known that one can cut out 
non-overlapping pieces each containing the same word Determine the largest possible number of singular words(Bogdanov I.)
This post has been edited 1 time. Last edited by Hopeooooo, May 23, 2022, 2:19 PM
Reason: 1
Reason: 1
By a piece we always mean a piece of the strip containing exactly 
consecutive letters; Different piece may contain the same word. Say that a piece is singular if the word it contain is such. We start with constructing an example containing 
singular words. Define a word 
This will be the word having 
non-overlapping copies on the strip. There exist exactly 
possible 
l
Put them onto the strip in an arbitrary order, separating each two sequences by an instance of 
Each segment of the strip containing one 
letters. Clearly, the obtained strip contains 
non-overlapping copies of 
different positions (starting from the 
s
n
t
singular words. Consider an arbitrary piece 
containing a a word 
Either this piece contains a unique non empty prefix which coincides with some suffix of 
or there is no such prefix only in this case we will say that such prefix is empty. Let 
be the length of the defined prefix. Define similarly a suffix of 
which coincides with a prefix of 
Notice that the defined prefix and suffix do not overlap whenever 
(if 
we have 
)
If the piece contains a part then 
and 
Thus, piece 
and in this case the position of the part at 
so this piece is singular. We have proven that the constructed example works. It remains to prove that the number of singular words cannot exceed 
Enumerate the positions in the strip successively by 
(the numeration is cyclic module 
denote the piece starting at positions 
and let 
be the word on that piece. Let 
be a positive such that the pieces 
are pairwise disjoint and contain the same word 
(from the problem statement). Clearly, those pieces are not singular. For 
and 
we say that a piece 
is a rank 
follower, while 
is a rank 
followers of all ranks, at most 
non-singular pieces, which implies the desired bound. Thus, we are left to prove the claim quoted above. For any rank 
define its tail as its suffix of length 
We regard a tail as a sequence of letters). We show by induction on 
that for every sequence 
consisting of 
letters, there are no more than 
followers whose tails contain 
The base case 
is obvious; If a follower with tail 
then every singular follower's tail starting with 
For every 
there are at most 
such followers, by the inductive hypothesis. So the total number of such followers does not exceed 
as desired. Finally, if there is a singular follower 
whose tail is 
of 
itself) is at most 
as desired again.
