Why are there so many Graphs in China TST 2001?
by steven_zhang123, Mar 29, 2025, 11:44 PM
Let
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.
(1) Suppose the length of the longest path in graph
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.)
(2) For the case where
and graph
is connected, can we determine that the length of the longest path in graph
,
?
(3) For the case where
, is it necessary for graph
to have a path of length
(i.e., a Hamiltonian path)?








(1) Suppose the length of the longest path in graph





(2) For the case where




(3) For the case where



The Quest for Remainder
by steven_zhang123, Mar 29, 2025, 11:42 PM
Given sets
,
, 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.
(1) For what positive integer
, among all its positive divisors, are the numbers of
divisors and
divisors equal?
(2) For which positive integers
, 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?






(1) For what positive integer



(2) For which positive integers






Very chaotic
by steven_zhang123, Mar 29, 2025, 11:37 PM
Let
be the Euler's totient function.
1. For any given integer
, does there exist
such that for any
,
and
,
is a non-negative power of
?
2. For integer
, are there integers
and
satisfying:
And these two different
correspond to the same
and
as described in (1), yet
.
3. Define
as the number of elements in set
. For integer
, let
and
. Compare
with
.

1. For any given integer







2. For integer



![\[
\varphi(k_i) \in \left ( \frac{x}{a} ,x \right ], i = 1,2; \quad \varphi(k_1) \neq \varphi(k_2).
\]](http://latex.artofproblemsolving.com/6/c/b/6cbe59f15b51439d1202267f521b573a52f1fe67.png)




3. Define







This post has been edited 1 time. Last edited by steven_zhang123, 40 minutes ago
FE f(x)f(y)+1=f(x+y)+f(xy)+xy(x+y-2)
by steven_zhang123, Mar 29, 2025, 11:27 PM
Find all functions
such that for all
, we have
.



2025 TST 22
by EthanWYX2009, Mar 29, 2025, 2:50 PM
Let
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
Find the smallest real number
such that for any set
of 2025 positive real numbers, the following inequality holds:
where
denotes the largest element in
.






![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)


![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)


A and B play a game
by EthanWYX2009, Mar 29, 2025, 2:49 PM
Let
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.
Prove that there exists a positive number
, Alice has a strategy such that, no matter how Bob colors the edges, the game terminates with the number of colored edges not exceeding
![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](//latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)


Prove that there exists a positive number

![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
This post has been edited 1 time. Last edited by EthanWYX2009, Yesterday at 2:56 PM
Monkeys have bananas
by nAalniaOMliO, Mar 28, 2025, 8:20 PM
Ten monkeys have 60 bananas. Each monkey has at least one banana and any two monkeys have different amounts of bananas.
Prove that any six monkeys can distribute their bananas between others such that all 4 remaining monkeys have the same amount of bananas.
Prove that any six monkeys can distribute their bananas between others such that all 4 remaining monkeys have the same amount of bananas.
The Principle of Inclusion-Exclusion (PIE)
by aoum, Mar 26, 2025, 12:18 AM
The Principle of Inclusion-Exclusion (PIE): Counting Overlapping Sets
The Principle of Inclusion-Exclusion (PIE) is a powerful combinatorial technique used to calculate the size of the union of overlapping sets. It corrects for overcounting by systematically adding and subtracting the sizes of various intersections.
1. The Basic Form of Inclusion-Exclusion
For two finite sets
and
, the size of their union is given by:
![\[
|A \cup B| = |A| + |B| - |A \cap B|.
\]](//latex.artofproblemsolving.com/6/f/3/6f3e15b9f2bdb4331acc41fa09951927b82d5014.png)
This formula accounts for the fact that elements in the intersection
are counted twice—once in
and once in
—so we subtract the overlap.
For three sets
,
, and
:
![\[
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|.
\]](//latex.artofproblemsolving.com/6/c/9/6c903bc32ca8e0072bc277d0f441b7ab7bfa3eca.png)
The pattern alternates between adding and subtracting the sizes of intersections of increasing complexity.
2. The General Formula of Inclusion-Exclusion
For
finite sets
, the size of their union is:
![\[
\left| \bigcup_{i=1}^n A_i \right| = \sum_{k=1}^n (-1)^{k+1} \sum_{1 \leq i_1 < \dots < i_k \leq n} |A_{i_1} \cap \dots \cap A_{i_k}|.
\]](//latex.artofproblemsolving.com/d/f/b/dfbfeb1be8ae686a72f94d98c0a9545baf7c6094.png)
In words:
3. Proof of the Inclusion-Exclusion Formula
Consider any element
that belongs to at least one of the sets. We need to ensure each element is counted exactly once.
Define:
![\[
m(x) = \text{Number of sets } A_i \text{ that contain } x.
\]](//latex.artofproblemsolving.com/7/5/2/752600a41711228882174ede71b5c691681dfeb1.png)
In the PIE formula:
By summing over all elements, the formula holds by correcting these overcounts systematically.
4. Example: Counting Multiples
Find how many integers between
and
are divisible by
,
, or
.
Let:
Step 1: Count each set:
![\[
|A| = \left\lfloor \frac{100}{2} \right\rfloor = 50, \quad |B| = \left\lfloor \frac{100}{3} \right\rfloor = 33, \quad |C| = \left\lfloor \frac{100}{5} \right\rfloor = 20
\]](//latex.artofproblemsolving.com/e/d/c/edcadd78cbc24e4a2cf63e321f0b3882a570c65e.png)
Step 2: Subtract pairwise intersections:
![\[
|A \cap B| = \left\lfloor \frac{100}{6} \right\rfloor = 16, \quad |A \cap C| = \left\lfloor \frac{100}{10} \right\rfloor = 10, \quad |B \cap C| = \left\lfloor \frac{100}{15} \right\rfloor = 6
\]](//latex.artofproblemsolving.com/7/4/5/745c9fa1f79f7c5225b905fc989c0e1f758e7f93.png)
Step 3: Add the triple intersection:
![\[
|A \cap B \cap C| = \left\lfloor \frac{100}{30} \right\rfloor = 3
\]](//latex.artofproblemsolving.com/e/e/b/eebf2c896136ef1c5193cdf26481e2fc2fbb0816.png)
Step 4: Apply PIE:
![\[
|A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74
\]](//latex.artofproblemsolving.com/5/1/1/51166af1a180dadbb4afedc762bcd952ae33b57b.png)
Thus, there are
numbers between
and
divisible by
,
, or
.
5. Applications of Inclusion-Exclusion
The principle of inclusion-exclusion has broad applications:
6. Advanced Generalization of Inclusion-Exclusion
For an arbitrary measure space
and measurable sets
, the inclusion-exclusion formula generalizes to:
![\[
\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{k=1}^n (-1)^{k+1} \sum_{1 \leq i_1 < \dots < i_k \leq n} \mu(A_{i_1} \cap \dots \cap A_{i_k}).
\]](//latex.artofproblemsolving.com/8/6/9/8693c12d0a177189733a29628374e7a4c51b5b7c.png)
This allows the principle to be applied to continuous spaces and probability measures.
7. Derangements and Inclusion-Exclusion
A classical application of PIE is counting derangements—permutations where no object appears in its original position.
Let
represent the number of derangements of
objects. Using PIE:
![\[
D_n = n! \sum_{k=0}^n \frac{(-1)^k}{k!},
\]](//latex.artofproblemsolving.com/b/2/f/b2f5e9eda7c3683161636aec79924d453e445c9e.png)
where the summation alternates signs based on how many fixed points to exclude.
8. Conclusion
The Principle of Inclusion-Exclusion is a fundamental tool in combinatorics, allowing precise counting of complex unions by carefully handling overlaps. It applies across multiple domains, including probability theory, graph theory, and set theory, providing a versatile method for solving intricate counting problems.
9. Principle of Inclusion-Exclusion Video by Sohil Rathi
References
The Principle of Inclusion-Exclusion (PIE) is a powerful combinatorial technique used to calculate the size of the union of overlapping sets. It corrects for overcounting by systematically adding and subtracting the sizes of various intersections.

Inclusion–exclusion illustrated by a Venn diagram for three sets
1. The Basic Form of Inclusion-Exclusion
For two finite sets


![\[
|A \cup B| = |A| + |B| - |A \cap B|.
\]](http://latex.artofproblemsolving.com/6/f/3/6f3e15b9f2bdb4331acc41fa09951927b82d5014.png)
This formula accounts for the fact that elements in the intersection



For three sets



![\[
|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|.
\]](http://latex.artofproblemsolving.com/6/c/9/6c903bc32ca8e0072bc277d0f441b7ab7bfa3eca.png)
The pattern alternates between adding and subtracting the sizes of intersections of increasing complexity.
2. The General Formula of Inclusion-Exclusion
For


![\[
\left| \bigcup_{i=1}^n A_i \right| = \sum_{k=1}^n (-1)^{k+1} \sum_{1 \leq i_1 < \dots < i_k \leq n} |A_{i_1} \cap \dots \cap A_{i_k}|.
\]](http://latex.artofproblemsolving.com/d/f/b/dfbfeb1be8ae686a72f94d98c0a9545baf7c6094.png)
In words:
- Add the sizes of all individual sets.
- Subtract the sizes of all pairwise intersections.
- Add the sizes of all three-way intersections.
- Continue this pattern, alternating signs, until you reach the intersection of all
sets.
3. Proof of the Inclusion-Exclusion Formula
Consider any element

Define:
![\[
m(x) = \text{Number of sets } A_i \text{ that contain } x.
\]](http://latex.artofproblemsolving.com/7/5/2/752600a41711228882174ede71b5c691681dfeb1.png)
In the PIE formula:
- Each element is counted once for every set it belongs to.
- Each pairwise intersection removes elements counted twice.
- Each triple-wise intersection adds back elements removed too often.
- This alternation continues, ensuring each element is counted exactly once.
By summing over all elements, the formula holds by correcting these overcounts systematically.
4. Example: Counting Multiples
Find how many integers between





Let:
Step 1: Count each set:
![\[
|A| = \left\lfloor \frac{100}{2} \right\rfloor = 50, \quad |B| = \left\lfloor \frac{100}{3} \right\rfloor = 33, \quad |C| = \left\lfloor \frac{100}{5} \right\rfloor = 20
\]](http://latex.artofproblemsolving.com/e/d/c/edcadd78cbc24e4a2cf63e321f0b3882a570c65e.png)
Step 2: Subtract pairwise intersections:
![\[
|A \cap B| = \left\lfloor \frac{100}{6} \right\rfloor = 16, \quad |A \cap C| = \left\lfloor \frac{100}{10} \right\rfloor = 10, \quad |B \cap C| = \left\lfloor \frac{100}{15} \right\rfloor = 6
\]](http://latex.artofproblemsolving.com/7/4/5/745c9fa1f79f7c5225b905fc989c0e1f758e7f93.png)
Step 3: Add the triple intersection:
![\[
|A \cap B \cap C| = \left\lfloor \frac{100}{30} \right\rfloor = 3
\]](http://latex.artofproblemsolving.com/e/e/b/eebf2c896136ef1c5193cdf26481e2fc2fbb0816.png)
Step 4: Apply PIE:
![\[
|A \cup B \cup C| = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 74
\]](http://latex.artofproblemsolving.com/5/1/1/51166af1a180dadbb4afedc762bcd952ae33b57b.png)
Thus, there are






5. Applications of Inclusion-Exclusion
The principle of inclusion-exclusion has broad applications:
- Counting Problems: Solving problems involving overlapping categories.
- Probability Theory: Finding the probability of the union of events.
- Combinatorics: Counting permutations with restrictions.
- Set Theory: Analyzing intersections and unions of large collections.
- Graph Theory: Counting subgraphs and other combinatorial objects.
6. Advanced Generalization of Inclusion-Exclusion
For an arbitrary measure space


![\[
\mu \left( \bigcup_{i=1}^n A_i \right) = \sum_{k=1}^n (-1)^{k+1} \sum_{1 \leq i_1 < \dots < i_k \leq n} \mu(A_{i_1} \cap \dots \cap A_{i_k}).
\]](http://latex.artofproblemsolving.com/8/6/9/8693c12d0a177189733a29628374e7a4c51b5b7c.png)
This allows the principle to be applied to continuous spaces and probability measures.
7. Derangements and Inclusion-Exclusion
A classical application of PIE is counting derangements—permutations where no object appears in its original position.
Let


![\[
D_n = n! \sum_{k=0}^n \frac{(-1)^k}{k!},
\]](http://latex.artofproblemsolving.com/b/2/f/b2f5e9eda7c3683161636aec79924d453e445c9e.png)
where the summation alternates signs based on how many fixed points to exclude.
8. Conclusion
The Principle of Inclusion-Exclusion is a fundamental tool in combinatorics, allowing precise counting of complex unions by carefully handling overlaps. It applies across multiple domains, including probability theory, graph theory, and set theory, providing a versatile method for solving intricate counting problems.
9. Principle of Inclusion-Exclusion Video by Sohil Rathi
References
- R. Graham, D. Knuth, and O. Patashnik, Concrete Mathematics, 2nd Edition (Addison-Wesley, 1994).
- I. Anderson, A First Course in Discrete Mathematics (Springer, 2001).
- Wikipedia: Inclusion-Exclusion Principle.
- AoPS: Principle of Inclusion-Exclusion.
How many cases did you check?
by avisioner, Jul 17, 2024, 12:01 PM
Determine all ordered pairs
of positive integers, with
prime, such that
is a perfect square.
Proposed by Tahjib Hossain Khan, Bangladesh



Proposed by Tahjib Hossain Khan, Bangladesh
This post has been edited 1 time. Last edited by avisioner, Jul 20, 2024, 4:57 PM
Reason: Proposer name added
Reason: Proposer name added
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