Wednesday, March 31, 2021

1979 IMO Problems And Solutions

Problem 01

If $p$ and $q$ are natural numbers so that $\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319},$ Prove that $p$ is divisible with $1979$ .

Solution

1963 IMO Problems And Solutions

Problem 01

Find all real roots of the equation

$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$ ,

where $p$ is a real parameter.

Solution

Problem 01

Real numbers $a_1, a_2, \dots , a_n$ are given. For each $i$ ( $1\le i\le n$ ) define

$d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}$

and let

$d=\max\{d_i:1\le i\le n\}$ .

(a) Prove that, for any real numbers $x_1\le x_2\le \cdots\le x_n$ ,

$\max\{|x_i-a_i|:1\le i\le n\}\ge \dfrac{d}{2} (*)$

(b) Show that there are real numbers $x_1\le x_2\le x_n$ such that equality holds in (*)

Solution

Problem 01

Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$ . Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$ . Find all sets $A$ of four distinct positive integers which achieve the largest possible value of $n_A$

Solution

Problem 01

Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a$ and $b$ , $f(2a) + 2f(b) = f(f(a + b)).$

Solutions

Solution 1

Let us substitute $0$ in for $a$ to get $f(0) + 2f(b) = f(f(b)).$

Now, since the domain and range of $f$ are the same, we can let $x = f(b)$ and $f(0)$ equal some constant $c$ to get $c + 2x = f(x).$

Problem 01

Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$ .

Solutions

Solution 1

Let $N = 100a + 10b+c$ for some digits $a,b,$ and $c$ . Then $100a + 10b+c = 11m$ for some $m$ . We also have $m=a^2+b^2+c^2$ . Substituting this into the first equation and simplification, we get

Problem 01 (1959 IMO)

Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number.

Solution 01 (Euclidean Algorithm)

Cambodian Olympiad Math

Cambodian Olympiad Math is written by Lim Sovanvichet . This book is shared on social media by many students which is very important document for all Khmer students. In this book, there are many International Math Problems which he translated from other foreign books and wrote them in to Khmer language.

Download This Book Here:

Vietnames's Sequence Math

This is a document that I research from the internet and rewrite it to upload at my website, I think It can help you do Math with different language which makes you to be better at Math. Why Vietnames students can challenge with other students around the world.

Here you can check out my collection.

Monday, March 22, 2021

When is $n^2+2021n$ a perfect square?

Solution

Let $m,n\in\mathbb{N}$ such that $n^2+2021n=m^2$. Since $2021=43\times47$, we have

\[ 4n^2+4\times2021n=4m^2 \]

Adding $2021^2$ to both sides,

\[ (2n+2021)^2-4m^2=2021^2 \]

Therefore,

\[ (2n-2m+2021)(2n+2m+2021)=2021^2. \]

Case 1: $2021^2=1\times2021^2$.

Then $2n-2m+2021=1$ and $2n+2m+2021=2021^2$. Adding both equations,

\[ 4n+2\times2021=2021^2+1\Rightarrow4n=(2021-1)^2, \]

so $n=1020100$.

Case 2: $2021^2=43\times(43\times47)^2$.

Then $2n-2m+2021=43$ and $2n+2m+2021=(43\times47)^2$. Solving similarly, we get $n=22747$.

Hence, $n^2+2021n$ is a perfect square when

\[ \boxed{n=1020100\ \text{or}\ n=22747}. \]

Here you can download the PDF file (Solution by Thin Sokkean)

Sunday, March 14, 2021

Irish Math Olympiad 2009

Find all positive integers $n$ such that $n^8+n+1$ is a prime.

Irish Math Olympiad 2009

Solution

Let $f(x)=x^8+x+1$. If $w$ is the third root of unity $(w=e^{2i\pi/3})$, then $w^2+w+1=0$ and $w^3=1$, so $w^8=w^2$. Therefore, $f(w)=w^8+w+1=w^2+w+1=0$. Hence, $w$ is a root of both $f(x)=x^8+x+1$ and $x^2+x+1$, so

\[ f(x)=(x^2+x+1)g(x). \]

We have

\[ x^8+x+1=(x^2+x+1)(x^6-x^5+x^3-x^2+1). \]

Since $f(1)=1^8+1+1=3$, $n=1$ gives a prime number.

For $n>1$,

\[ n^8+n+1=(n^2+n+1)(n^6-n^5+n^3-n^2+1). \]

Both factors are greater than $1$, so $n^8+n+1$ is composite for all $n>1$.

Therefore, the only positive integer satisfying the condition is

\[ \boxed{n=1}. \]

Solution by: Thin Sokkean

Check PDF file to download here

Monday, March 8, 2021

2001 Dutch Math Olympiad

Suppose that for all $x,y\in\mathbb{R}$, we have $f(x+y)=f(x)+f(y)+xy$ and $f(4)=10$. Find the value of $f(2001)$.

Solution

Prove that $\frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640}=1$.

Solution

1. We observe that

\[ \frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640} =\frac12+\frac1{20}+\frac15+\frac18+\frac1{11}+\frac1{110}+\frac1{40}+\frac1{1640} \] \[ =\frac{22}{40}+\frac{13}{40}+\frac{11}{110}+\frac{41}{1640} =\frac{22}{40}+\frac{13}{40}+\frac4{40}+\frac1{40} =\frac{40}{40}=1. \]

Hence, $\frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640}=1$.

2. Find the minimum value of $A=|x-1|+|x-2|+\cdots+|x-100|$.

Using $|a|+|b|\geq|a-b|$, we have

\[ |x-k|+|x-(101-k)|\geq|101-2k|,\quad k=1,2,\ldots,50. \]

Therefore,

\[ |x-1|+|x-100|\geq99,\ |x-2|+|x-99|\geq97,\ldots,\ |x-50|+|x-51|\geq1. \]

Adding all inequalities,

\[ A\geq99+97+\cdots+1=\frac{50}{2}(1+99)=2500. \]

The equality holds when $x\in\bigcap_{k=1}^{50}[k,101-k]=[50,51]$.

Hence,

\[ \boxed{A_{\min}=2500\quad\text{when }x\in[50,51].} \]

Monday, February 1, 2021

Sequence Book For Grade 12

This book is related to Sequence Problem, the tips to solve Sequence problem, which is written in Khmer language. I hope this book is used for all of you.

IMO 2019 in South Africa

Let Z be the set of integers. Determine all function `f: Z \rightarrow Z` such that, for all integers `a` and `b`

`f(2a)+2f(b)=f(f(a+b))` `(1)`

Answer: The solution are `f(n)=0` and `f(n)=2n+k` for any constant `k\in Z`

Substituting `a =0, b= n+1 ` gives `f(f(n+1))=f(0)+2f(n+1)`.

Substituting `a =1, b= n` gives `f(f(n+1))=f(2)+2f(n)` .

In particular, `f(0)+2f(n+1)=f(2)+2f(n)`, and so `f(n+1)-f(n)=1/2(f(2)-f(0))` .

Labels

Wednesday, March 31, 2021

Problem 01

Solution

Monday, March 29, 2021

Problem 01

Solution

Problem 01

Solution

Problem 01

Solution

Problem 01

Solutions

Solution 1

Problem 01

Solutions

Solution 1

Problem 01 (1959 IMO)

Solution 01 (Euclidean Algorithm)

Thursday, March 25, 2021

Monday, March 22, 2021

Solution

Sunday, March 14, 2021

Solution

Monday, March 8, 2021

Sunday, March 7, 2021

Solution

Monday, February 1, 2021

Saturday, January 2, 2021