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.

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Given that \[ x^3+\frac{1}{x^3}=4, \] find the value of \[ P=x^6+\frac{4}{x^3}. \]

Solve the Algebraic Problem

Given that

\[ x^3+\frac{1}{x^3}=4, \]

find the value of

\[ P=x^6+\frac{4}{x^3}. \]

---

Solution

Step 1. Introduce a New Variable

Let

\[ a=x^3. \]

Then the given equation becomes

\[ a+\frac{1}{a}=4. \]

Step 2. Form a Quadratic Equation

Multiply both sides by \(a\):

\[ a^2+1=4a. \]

Hence,

\[ a^2-4a+1=0. \]

Therefore,

\[ a^2=4a-1. \]

Step 3. Evaluate the Required Expression

Since

\[ P=x^6+\frac{4}{x^3}=a^2+\frac{4}{a}, \]

and

\[ \frac{1}{a}=4-a, \]

we obtain

\[ \begin{aligned} P &=a^2+4(4-a)\\ &=a^2-4a+16. \end{aligned} \]

Substitute \(a^2=4a-1\):

\[ \begin{aligned} P &=(4a-1)-4a+16\\ &=15. \end{aligned} \]

---

Final Answer

\[ \boxed{15} \]

Solve the Equation

Solve the following equation:

\[ \frac{x^3+6x^2+15x+13}{x^2+4x+7} = \sqrt{x+9} - \frac{1}{x+12}. \]

---

Solution

Step 1. Determine the Domain

Since the square root is defined only when

\[ x+9\ge0, \]

we must have

\[ x\ge-9. \]

Next, rewrite the numerator:

\[ x^3+6x^2+15x+13 = (x+2)(x^2+4x+7)-1. \]

Therefore,

\[ \frac{x^3+6x^2+15x+13}{x^2+4x+7} = (x+2)-\frac{1}{x^2+4x+7}. \]

Notice that

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

and similarly,

\[ x+12=(\sqrt{x+9})^2+3. \]

Hence the equation becomes

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. 

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,

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

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

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

 

Cambodia National Math 2019, 22/04/2019 Day 02

 

Cambodia National Math 2019, 22/04/2019 Day 02

Math Cambodia 2019 Day 02

• This was the problem that released for Out Standing Student in Cambodia in 2019 for 2nd day of testing.
• There were two days of the testing. This is the first day of exam.
• You all can Click here to download

Cambodia Grade 12, 22/04/2019 Day 01

Cambodia National Math 2019, 22/04/2019

• This was the problem that released for Out Standing Student in Cambodia in 2019.
• There were two days of the testing. This is the first day of exam.

• You can watch solution here:

Prove that the polynomial \[ x^{9999}+x^{8888}+x^{7777}+\cdots+x^{1111}+1 \] is divisible by \[ x^9+x^8+x^7+\cdots+x+1 \]

 Problem: 01

Prove that the polynomial `x^9999+x^8888+x^7777+...+x^1111+1` is divisible by `x^9+x^8+x^7+....+x+1`

Solution

Find all functions \(f(x)\) satisfying: \[ (x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2) \]

 

Find all functions \(f(x)\) satisfying:

\[ (x-y)f(x+y)-(x+y)f(x-y)=4xy(x^2-y^2) \]

Solution

Vietnamese Mathematical Olympiad 2022 Let \(a\), \(b\), and \(c\) be the roots of the equation $$ x^3-x^2-1=0. $$ Find the value of $$ \frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}. $$

Let \(a\), \(b\), and \(c\) be nonzero real numbers satisfying

$$ a+b+c=2022, $$ and $$ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2022}. $$

Find the value of

$$ \frac{1}{a^{2023}}+\frac{1}{b^{2023}}+\frac{1}{c^{2023}}. $$

Solution

Find all polynomials \(P(x)\) satisfying $$ P(x-1)\,P(x+1)=P(x^2-1) \qquad\text{for all }x\in\mathbb{R}. $$

Solution

Suppose that \(\alpha\) is a root of \(P(x)\). Then

$$ P(\alpha)=0. $$

Substituting \(x=\alpha+1\) into the given equation,

$$ P((\alpha+1)-1)\,P((\alpha+1)+1) = P((\alpha+1)^2-1). $$

Since \(P(\alpha)=0\), it follows that

$$ 0\cdot P(\alpha+2)=P(\alpha^2+2\alpha), $$

and hence

$$ P(\alpha^2+2\alpha)=0. $$

Find the last two digits of $$ N=\left(1!+2!+3!+\cdots+101!\right)^{101}. $$

Solution

Finding the last two digits of \(N\) is equivalent to finding

$$ N \pmod{100}. $$

Since

$$ 10!=2^8\cdot3^4\cdot5^2\cdot7, $$

we have

$$ 10!\equiv0\pmod{100}. $$

Hence, every factorial \(n!\) with \(n\ge10\) is also divisible by \(100\). Therefore,

$$ N\equiv\left(1!+2!+3!+\cdots+9!\right)^{101}\pmod{100}. $$

Now,

$$ \begin{aligned} 1!+2!+\cdots+9! &=1+2+6+24+120+720+5040+40320+362880\\ &\equiv1+2+6+24+20+20+40+20+80\\ &\equiv13\pmod{100}. \end{aligned} $$

Thus,

$$ N\equiv13^{101}\pmod{100}. $$

Observe that

$$ \begin{aligned} 13^2&=169\equiv69\pmod{100},\\ 13^4&\equiv69^2=4761\equiv61\pmod{100},\\ 13^5&\equiv61\cdot13=793\equiv93\pmod{100}. \end{aligned} $$

Hence,

$$ 13^{20}=(13^5)^4\equiv93^4\equiv1\pmod{100}. $$

Therefore,

$$ \begin{aligned} N &\equiv13^{101}\\ &=13(13^{20})^5\\ &\equiv13\cdot1^5\\ &\equiv13\pmod{100}. \end{aligned} $$

Answer

$$ \boxed{13} $$

Therefore, the last two digits of $$ \left(1!+2!+\cdots+101!\right)^{101} $$ are 13.

Prove that \(A_n=3^{\,n+3}-4^{\,4n+2}\) is divisible by \(11\) for every positive integer \(n\).

We will prove by Mathematical Induction.

For \(n=0\), \(A_0=3^3-4^2=27-16=11\), which is divisible by \(11\).

Assume that \(A_k\) is divisible by \(11\), i.e., \(11\mid A_k\).

We shall prove that \(A_{k+1}\) is also divisible by \(11\).

\[ \begin{aligned} A_{k+1} &=3^{k+4}-4^{4k+6}\\ &=3\cdot3^{k+3}-4^4\cdot4^{4k+2}\\ &=3\cdot3^{k+3}-256\cdot4^{4k+2}\\ &=3\cdot3^{k+3}-3\cdot4^{4k+2}-253\cdot4^{4k+2}\\ &=3\left(3^{k+3}-4^{4k+2}\right)-253\cdot4^{4k+2}\\ &=3A_k-253\cdot4^{4k+2}. \end{aligned} \]

Since \(11\mid A_k\) by the induction hypothesis and \(253=11\times23\), both terms on the right-hand side are divisible by \(11\). Hence \(11\mid A_{k+1}\).

Therefore, by the Principle of Mathematical Induction, \(11\mid\left(3^{n+3}-4^{4n+2}\right)\) for every nonnegative integer \(n\).

Solution by: Thin Sokkean

Prove that \(\forall x\in\mathbb{R},\ |a\cos x+b\sin x|\le\sqrt{a^2+b^2}.\)

a. Prove that \(\forall x\in\mathbb{R}\), \(|a\cos x+b\sin x|\leq\sqrt{a^2+b^2}\).

b. Find the maximum and minimum of $$ f(x)=20\cos x+21\sin x+27. $$

Solution

a. Prove that \(\forall x\in\mathbb{R}\), \(|a\cos x+b\sin x|\leq\sqrt{a^2+b^2}\).

We have:

$$ a\cos x+b\sin x =\sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}}\cos x+ \frac{b}{\sqrt{a^2+b^2}}\sin x\right) $$

Let

$$ \sin\alpha=\frac{a}{\sqrt{a^2+b^2}} $$

\(X_1,X_2\) are the roots of \(X^2-(2\cos t+3\sin t)X-11\sin^2t=0\). Find the minimum value of \(A=X_1^2+X_1X_2+X_2^2\).

Solution

\(X_1,X_2\) are the roots of \(X^2-(2\cos t+3\sin t)X-11\sin^2t=0\). By Vieta's formulas, we have \(X_1+X_2=2\cos t+3\sin t\) and \(X_1X_2=-11\sin^2t\).

We have \(A=X_1^2+X_1X_2+X_2^2=(X_1+X_2)^2-X_1X_2\).