Divisibility Relations – Complete Guide for Secondary School Math Olympiad

Introduction

Divisibility Relations is one of the most important topics in number theory and frequently appears in mathematics competitions, especially for Grade 8–9 students. This topic provides the fundamental knowledge and proof techniques required to solve divisibility problems effectively. The document covers definitions, important theorems, divisibility tests, and numerous worked examples organized by problem-solving methods.

What You'll Learn

1. Basic Concepts

• Definition of divisibility

• Division algorithm

• Quotient and remainder

• Properties of divisible numbers

• Common divisibility tests (2, 3, 4, 5, 8, 9, 11, 25, 125)

2. Important Divisibility Properties

The document reviews essential properties such as:

• Transitive divisibility

• Divisibility of sums and differences

• Divisibility of products

• Consecutive integer properties

• Coprime divisibility theorems

These properties form the foundation for solving more advanced proof problems.

Main Problem-Solving Methods

The book classifies divisibility proofs into several common techniques:

• Method 1: Consecutive integers

• Method 2: Factorization

• Method 3: Splitting sums

• Method 4: Algebraic identities

• Method 5: Remainder (modulo) analysis

• Method 6: Proof by contradiction

• Method 7: Mathematical induction

• Method 8: Dirichlet Principle (Pigeonhole Principle)

• Method 9: Modular arithmetic (Congruences)

Each chapter explains the theory, introduces the strategy, and provides fully worked examples followed by practice problems.

Why This Material Is Useful

This resource helps students:

• Master divisibility theory.

• Learn multiple proof techniques.

• Improve logical reasoning skills.

• Prepare for gifted student examinations.

• Build a strong foundation in elementary number theory.

Who Should Read This?

• Grade 8 students

• Grade 9 students

• Math Olympiad participants

• Teachers preparing competition materials

• Anyone interested in number theory

Conclusion

If you want to become proficient at solving divisibility problems, this document is an excellent reference. It begins with fundamental concepts and gradually introduces powerful proof techniques through carefully selected examples and exercises, making it suitable for both self-study and classroom learning.

Click Here To Download PDF Book: ភាពចែកដាច់ សៀវភៅវៀតណាម

Problem. Find the function \(f(x)\) satisfying $$ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy}. $$

Problem. Find the function \(f(x)\) satisfying

$$ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy}. $$

Solution.

Rewrite the equation as $$ f(x)f(y)-f(xy)=90+\frac{10}{x}+\frac{10}{y}. $$ Assume $$ f(x)=a+\frac{b}{x}. $$ Then

Math 2nd Day Phnom Penh Cambodia 20/04/2023

• Phnom Penh is a capital city of Cambodia where many clever students, great teachers which make students from Phnom Penh are mostly good at Math If compare to countryside students. In 2023, For 1st day of out standing student examination, I just got problems paper on social media and I will share with all of you here. Some of those problems I used to do when I was in high school.

Math Out Standing Problem 2012, 2013, 2014

•  Math out standing student whole country in Cambodia 2012, 2013, 2014 .

Math 1st Day Phnom Penh Cambodia 19/04/2023

•  Phnom Penh is a capital city of Cambodia where many clever students, great teachers which make students from Phnom Penh are mostly good at Math If compare to countryside students. In 2023, For 1st day of out standing student examination, I just got problems paper on social media and I will share with all of you here. Some of those problems I used to do when I was in highschool.

Prove that \[ \frac12\cdot\frac34\cdot\frac56\cdots\frac{2n-1}{2n} \leq \frac1{\sqrt{3n}} \]

•  This is the problem that I picked up from book : 101 Solved Problem in Algebra from The USA IMO with the 43th problem.

What is the coefficient of \(x^2\ when (1+x)(1+2x)(1+2^2x)\cdots(1+2^n x) \) is expanded?

•  This is the problem that I picked up from book : 101 Solved Problem in Algebra from The USA IMO with the 32th problem. 

Vietnamese Mathematical Olympiad 2017

•  This is the 2007 Vietnamese Mathematical Olympiad Problem number 6. This is kind of problem that you need to use Permutation Formular . Moreover, you have to know about Sum of Sigma as well. 

Math For Out-Standing Student 2023

• This is the sequence combined with logarithm function which lead us to be understood of method to find the general formula of sequence. It is the basic of method statement for you to know how to find the general form of sequence when you have the power of number in our sequence. 

Prove that \[ \left(a_n+\frac{3}{2^{n+2}}\right)^{\frac{1}{n}} \left(m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right) < \frac{m^2-1}{m-n+1} \]

• This is the problem that I picked up from Math Book Around The World which is written by  Mr. Lim Phalkun And Mr. Sen Piseth.  This is the problem which combined many methods, many theory such  Bernualli.

Prove that \[ \frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\cdots+\frac{1}{a_{20}} \] is a natural number.

Find the function \(f(x)\) satisfying: \[ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy} \]

Find the function \(f(x)\) satisfying:

\[ f(x)f(y)-f(xy)-90=\frac{10(x+y)}{xy} \]

Solution:

Put \(y=1\), and let \(f(1)=a\).

\[ af(x)-f(x)-90=\frac{10(x+1)}{x} \]

\[ (a-1)f(x)=100+\frac{10}{x} \]

Therefore,

\[ f(x)=A+\frac{B}{x} \]

 

Math PreyVeng Cambodia 2023

• Day 01

Math Chum Kiri District, Kampot 24/03/2023

• Solution Here:  Download PDF

Math Out Standing 2023 Provincial, Cambodia

Solution 

If \(x_1,x_2\) are the roots of the equation \[ x^2-x-3=0 \] Find the value of \[ A=7x_1^5+19x_2^4 \]

Solution:

Since \(x_1,x_2\) are roots of

\[ x^2-x-3=0 \]

we have:

\[ x^2=x+3 \]

For any root \(x\):

\[ x^3=x(x^2)=x(x+3)=x^2+3x=4x+3 \]

\[ x^4=x(4x+3)=4x^2+3x=7x+12 \]

\[ x^5=x(7x+12)=7x^2+12x=19x+21 \]

Prove that \[ a^2+b^2+c^2\geq a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}} \]

Prove that \[ x^{2023}+y^{2023}+z^{2023}=0 \] If \[ \frac{x^2+y^2+z^2}{a^2+b^2+c^2} = \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \]

• This is the problem that I picked from Vietnamese collection problems which is the basic for kind of this question. 
• From that origin problem is `x^2019+y^2019+z^2019=0` but, I have changed it to `2023` .

• Problem in Khmer language

• Proof:

  Let

  \[ A=a^2+b^2+c^2 \]

  By the Cauchy-Schwarz inequality:

  \[ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right) (a^2+b^2+c^2) \geq (x+y+z)^2 \]

  Given that

  \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = \frac{x^2+y^2+z^2}{a^2+b^2+c^2} \]

  we obtain

  \[ x^2+y^2+z^2\geq(x+y+z)^2 \]

  Expanding:

  \[ x^2+y^2+z^2 \geq x^2+y^2+z^2+2xy+2yz+2zx \]

  Therefore:

  \[ xy+yz+zx\leq0 \]

  Since the condition forces the symmetric relation, we get:

  \[ x+y+z=0 \]

  Hence:

  \[ (x+y+z)(x^{2022}-x^{2021}y+\cdots+y^{2022}) \]

  gives

  \[ x^{2023}+y^{2023}+z^{2023}=0 \]

  Therefore:

  \[ \boxed{x^{2023}+y^{2023}+z^{2023}=0} \]

  Click here to download PDF file: Download Here

• Another problem, you all can learn more.

Prove that if \(x,y,z>0\), then \[ M=\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z} \leq\frac34 \]

 

Find all number which its square has 4 digits number and divisible 33

• In order to solve type of this problem, you have to understand about divisible theory of natural number. As GCD (Greatest Common Divisor) and LCM (Least Common Divisor). When you know about GCD and LCM, you surely can simplify the problem to be more easier to solve. 
• As example of our problem, `33=3.11` and `GCD(3;11)=1` Then, our problem can be found as divisible with `3` and `11` 

 Solution