[size=130][b]Higher Secondary: 2011[/b]
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Time: 4 Hours

[b]Problem 1:[/b]
Prove that for any non-negative integer $n$ the numbers $1, 2, 3, ..., 4n$ can be divided in tow mutually exclusive classes with equal number of members so that the sum of numbers of each class is equal.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=709

[b]Problem 2:[/b]
In the first round of a chess tournament, each player plays against every other player exactly once. A player gets $3, 1$ or $-1$ points respectively for winning, drawing or losing a match. After the end of the first round, it is found that the sum of the scores of all the players is $90$. How many players were there in the tournament?
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=708

[b]Problem 3:[/b]
$E$ is the midpoint of side $BC$ of rectangle $ABCD$. $A$ point $X$ is chosen on $BE$. $DX$ meets extended $AB$ at $P$. Find the position of $X$ so that the sum of the areas of  $\triangle BPX$ and  $\triangle DXC$ is maximum with proof.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=683

[b]Problem 4:[/b]
Which one is larger 2011! or, $(1006)^{2011}$? Justify your answer.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=707

[b]Problem 5:[/b]
In a scalene triangle $ABC$ with $\angle A = 90^{\circ}$, the tangent line at $A$ to its circumcircle meets line $BC$ at $M$ and the incircle touches $AC$ at $S$ and $AB$ at $R$. The lines $RS$ and $BC$ intersect at $N$ while the lines $AM$ and $SR$ intersect at $U$. Prove that the triangle $UMN$ is isosceles.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=706

[b]Problem 6:[/b]
$p$ is a prime and sum of the numbers from $1$ to $p$ is divisible by all primes less or equal to $p$. Find the value of $p$ with proof.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=693

[b]Problem 7:[/b]
Consider a group of $n > 1$ people. Any two people of this group are related by mutual friendship or mutual enmity. Any friend of a friend and any enemy of an enemy is a friend. If $A$ and $B$ are friends/enemies then we count it as $1$ [b]friendship/enmity[/b]. It is observed that the  number of friendships and number of enmities are equal in the group. Find all possible values of $n$.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=694

[b]Problem 8:[/b]
$ABC$ is a right angled triangle with $\angle A = 90^{\circ}$ and $D$ be the midpoint of $BC$. A point $F$ is chosen on $AB$. $CA$ and $DF$ meet at $G$ and $GB \parallel AD$. $CF$ and $AD$ meet at $O$ and $AF = FO$. $GO$ meets $BC$ at $R$. Find the sides of  $ABC$ if the area of $GDR$ is $\dfrac{2}{\sqrt{15}}$
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=704

[b]Problem 9:[/b]
The repeat of a natural number is obtained by writing it twice in a row (for example, the repeat of $123$ is $123123$). Find a positive integer (if any) whose repeat is a perfect square.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=703

[b]Problem 10:[/b]
Consider a square grid with $n$ rows and $n$ columns, where $n$ is odd (similar to a chessboard). Among the $n^2$ squares of the grid, $p$ are black and the others are white. The number of black squares is maximized while their arrangement is such that horizontally, vertically or diagonally neighboring black squares are separated by at least one white square between them. Show that there are infinitely many triplets of integers $(p, q, n)$ so that the number of white squares is $q^2$.
http://matholympiad.org.bd/forum/viewtopic.php?f=13&t=702

The problems of the Junior categories are available in [url=http://matholympiad.org.bd/forum/]BdMO Online forum[/url]: 
http://matholympiad.org.bd/forum/viewtopic.php?f=25&t=678
To solve the problem, we need to find a positive integer whose repeat is a perfect square. Let's denote the original number by \( n \). The repeat of \( n \) is obtained by writing it twice in a row, which can be represented as \( n \cdot 10^{k} + n \), where \( k \) is the number of digits in \( n \).

We need to find \( n \) such that \( n \cdot 10^{k} + n \) is a perfect square. Let's denote this number by \( m \), so we have:
\[ m = n \cdot 10^{k} + n = n (10^{k} + 1) \]

We need \( m \) to be a perfect square, so we can write:
\[ n (10^{k} + 1) = x^2 \]
for some integer \( x \).

This implies that \( n \) must be a factor of \( x^2 \). Since \( x^2 \) is a perfect square, \( n \) must also be a perfect square. Let's denote \( n = a^2 \) for some integer \( a \). Then we have:
\[ a^2 (10^{k} + 1) = x^2 \]

Dividing both sides by \( a^2 \), we get:
\[ 10^{k} + 1 = \left( \frac{x}{a} \right)^2 \]

Let \( \frac{x}{a} = b \), where \( b \) is an integer. Then we have:
\[ 10^{k} + 1 = b^2 \]

This implies:
\[ 10^{k} = b^2 - 1 \]
\[ 10^{k} = (b - 1)(b + 1) \]

Since \( 10^{k} \) is a power of 10, it can be written as \( 2^k \cdot 5^k \). Therefore, \( b - 1 \) and \( b + 1 \) must be powers of 2 and 5, respectively, or vice versa.

Let's consider the smallest possible values for \( k \):
- For \( k = 1 \), \( 10^1 = 10 \), which gives \( b - 1 = 2 \) and \( b + 1 = 5 \). This gives \( b = 3 \), but \( 10 \neq 3^2 - 1 \).
- For \( k = 2 \), \( 10^2 = 100 \), which gives \( b - 1 = 10 \) and \( b + 1 = 10 \). This gives \( b = 11 \), but \( 100 \neq 11^2 - 1 \).

We need to find a larger \( k \) such that \( 10^{k} = (b - 1)(b + 1) \). Let's try \( k = 55 \):
- For \( k = 55 \), \( 10^{55} = (b - 1)(b + 1) \).

We need to find \( b \) such that \( b - 1 \) and \( b + 1 \) are factors of \( 10^{55} \). Let's take \( b - 1 = 2 \cdot 10^{27} \) and \( b + 1 = 5 \cdot 10^{27} \):
- This gives \( b = 2 \cdot 10^{27} + 1 \).

Then we have:
\[ 10^{55} = (2 \cdot 10^{27})(5 \cdot 10^{27}) \]

Thus, \( n = 10^{27} \) and the repeat of \( n \) is \( 10^{55} + 1 \).

The final answer is \( \boxed{ 10^{55} + 1 } \).