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and 
be given as: 
, 
(where 
can be 0), and let 
be the count of numbers 
such that 
and 
(
can also be 0). (Here, 
is considered the same as 
).
is 
, then is called a perfect number. For example, the sum of the positive divisors of 6 is 
, hence 6 is a perfect number.
, where 
are positive integers, and 
are odd primes.
be positive integers, and let 
all be 
-th roots of unity, satisfying:![\[
\alpha_1^j + \alpha_2^j + \cdots + \alpha_n^j = 0 \quad \text{for any } j (0 < j < k).
\]](http://latex.artofproblemsolving.com/3/3/b/33be8eef29cd506bf1132a86def9992ec8d21cb4.png)
Prove that among , each -th root of unity appears the same number of times.
space stations, with a unidirectional pipeline connecting every pair of stations. A station directly reachable from station P without passing through any other station is called a directly reachable station of P. The number of stations jointly directly reachable by the station pair 
is to be examined. The plan requires that all station pairs have the same number of jointly directly reachable stations.
, then triangle ABC is called a unidirectional cyclic triangle.) stations meeting the above planning requirements be constructed for infinitely many integers 
?
of positive integers, with 
prime, such that 
is a perfect square.
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
: They take turns to perform operations, where each operation consists of coloring one or two edges that have not been colored yet. The game terminates if at any point there exists a triangle whose three edges are all colored.
, Alice has a strategy such that, no matter how Bob colors the edges, the game terminates with the number of colored edges not exceeding![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
, 
(where 
). For any given integer 
, consider a graph 
with 
vertices that contains a Hamiltonian cycle 
and has more than 
edges. For which lengths 
must the graph necessarily contain a cycle of length 
?
be a given integer, 
. Consider a graph 
with vertices satisfying the condition: for any two non-adjacent vertices 
and 
in graph , the sum of their degrees must satisfy 
. Please answer the following questions and prove your conclusions. is 
satisfying the inequality 
, does graph necessarily contain a cycle of length 
? (The length of a path or cycle refers to the number of edges that make up the path or cycle.)
and graph is connected, can we determine that the length of the longest path in graph , 
?
, is it necessary for graph to have a path of length 
(i.e., a Hamiltonian path)?
be a set of 2025 positive real numbers. For a subset 
, define 
as the median of 
when all elements of are arranged in increasing order, with the convention that 
. Define![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set of 2025 positive real numbers, the following inequality holds:![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in .
, 
, if a positive integer leaves a remainder (the smallest non-negative remainder) that belongs to when divided by 19, then that positive integer is called an 
number. If a positive integer leaves a remainder that belongs to when divided by 19, then that positive integer is called a 
number., among all its positive divisors, are the numbers of divisors and divisors equal?, are the numbers of divisors less than the numbers of divisors? For which positive integers , are the numbers of divisors greater than the numbers of divisors?
such that 
. Prove that 
May everyone not to upload solutions on this problem anymore until May 15/2025, this is an active problem on Mathematical Reflection! (Thank you Victoria_Discalceata1)
Hence, it's enough to prove 
then
Equality cases: 
such that . Prove that 
then 
We have![\[a^3b+b^3c+c^3a \leqslant a^3b+b^3c+c^3a + c(b+c)(a-b)(b-c) = b(c^3+abc+a^3).\]](http://latex.artofproblemsolving.com/a/3/b/a3b661208070cac04b6174c698638c1c5e427328.png)
Thus![\[\begin{aligned}
a^3b+b^3c+c^3a+9abc &\leqslant b(c^3+abc+a^3)+9abc \\
& = b[(c+a)^3-3ca(c+a)+(b+9)ca] \\
& = b[(3-b)^3-3ca(3-b)+(b+9)ca] \\
& = b[4bca-(b-3)^3] \\
& \leqslant b[b(c+a)^2-(b-3)^3] \\
& \leqslant 3b(b-3)^2.
\end{aligned}\]](http://latex.artofproblemsolving.com/9/8/8/988d5dd48718e002f93bdfa7ff8b211abd3c6055.png)
It remains to prove that![\[b(b-3)^2 \leqslant 4,\]](http://latex.artofproblemsolving.com/c/b/9/cb95af570bf9666e4c9d5213e55843b2c7f4dfc4.png)
or![\[(4-b)(b-1)^2 \geqslant 0.\]](http://latex.artofproblemsolving.com/f/c/7/fc77075fa28a0e371a000627785880fe6256de30.png)
Which is true.
, it suffices to show that![\[4(a+b+c)^3\ge27\left(a^3b+b^3c+c^3a\right)+81abc(a+b+c),\tag{1}\]](http://latex.artofproblemsolving.com/2/9/4/29424447811d740c1290746604007e21cbd3496a.png)
or![\[\sum_{\rm cyc}a^4+6\sum_{\rm cyc}a^2b^2-\frac{33}4\sum_{\rm cyc}a^2bc\ge\frac{11}4\sum_{\rm cyc}a^3b-4\sum_{\rm cyc}a^3c.\tag{2}\]](http://latex.artofproblemsolving.com/0/3/8/038fccd2b9294dab8674f08fa9f184b5eade1c7d.png)
![\[\sum_{\rm cyc}a^4+r\sum_{\rm cyc}a^2b^2+(p+q-r-1)\sum_{\rm cyc}a^2bc\ge p\sum_{\rm cyc}a^3b+q\sum_{\rm cyc}a^3c\]](http://latex.artofproblemsolving.com/4/5/3/453c0ef90919246a1d442fac357fa944cb1f281e.png)
holds for all real numbers 
, 
, 
if and only if 
. Since![\[3(1+6)=21\ge\left(\frac{11}4\right)^2+(-4)^2+\frac{11}4(-4)=\frac{201}{16},\]](http://latex.artofproblemsolving.com/6/4/a/64a8c14171fbab32f7d92c9f12f0bf9db93c3ad7.png)
then (2) indeed holds. We are done.![\[\frac18\sum_{\rm cyc}\left(4 a^2-2ab+9ac-7bc-4c^2\right)^2+\frac{45}8\sum_{\rm cyc}a^2(b-c)^2\ge0.\]](http://latex.artofproblemsolving.com/2/0/6/206d2dbd6999a8962e7610d31505505214234271.png)
holds for all 
with 
and 
Equality holds when 
or 
or 
