1) one group consists of n+1,n+2,...,3n. and the other group consists of other elements. each class has 2n many members and sum=$4n^2+n$. 4) $1006=\frac{1+2011}{2}> \sqrt(1*2011)$ $1006=\frac{2+2010}{2}> \sqrt(2*2010)$ ...... $1006=\frac{2011+1}{2}>\sqrt(2011*1)$ multiplying we get $1006^{2011} > 2011!$

2) Let there is n players. then total $\frac{n(n-1)}{2}$ matches were played. let x matches were draw. and y matches ends in a win and loss case. then $x+y=\frac{n(n-1)}{2}$ so for a draw total point obtained=+1+1=2 for a win and loss total point obtained=+3-1=2 so, total point scored in 1st round=2x+2y=90. $n(n-1)=90$ so $(n+9)(n-10)=90$ so $n=10$ 6) let all the primes upto p is following $2<3<5....<q<p$ then $2|\frac{p(p+1)}{2},3|\frac{p(p+1)}{2},......,q|\frac{p(p+1)}{2}$ since p is prime. that implies $2|\frac{(p+1)}{2},3|\frac{(p+1)}{2},......,q|\frac{(p+1)}{2}$ since $2<3<5....<q$ are prime that implies $2.3.5...q|\frac{(p+1)}{2}$ let $\frac{(p+1)}{2}=2.3.5...qK=L$ consider $L+1$ if $ p>3$ then $\frac{p+1}{2}+1=L+1<p$ also by construction 2,3,5,...q can't divide $L+1$ so, $L+1<p$ is a prime . so we get a prime less than p which is distinct from $\{2,3,5...q\}$ contradiction. so $p\leq 3$ now it is trivial that $p=3$ is only such prime. 9) The idea is to find a number of the form $10^n+1$ which has a square factor. then just smartly choosing a $n-1 digit $ no will serve our purpose. see $100000000001=11^2*23*4093*8779$ which is a 12 digit number. now see $10^2*23*4093*8779=82644628100$ is a 11 digit number. so, consider $n=82644628100$ take its repeat. $8264462810082644628100=82644628100*(100000000001) =11^2*10^2*23^2*4093^2*8779^2$ so we got one such number is $n=82644628100$

For problem $9$, there are no such numbers below $10000$(before the repeat operation). So checked till $1000010000$ EDIT:Yes, i too checked for non squarefree $10^n+1$, but gave up after it became unreasonably large. The factorisation used in your solution cannot possibly be expected in the exam. You could say that it is very easy to check that $11^2$ divides the number you found but proving that $10^n+1$ is squarefree for $n<11$ is quite the task!

Moonmathpi496 wrote: Higher Secondary: 2011 Problem 7: 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 $1$ friendship/enmity. 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 All $n$ who are perfect squares. With the properties we see there are 2 cliques who hate men of the other clique. Let their value be $F,E$ then $(F-E)^2=E+F=N$ after working out $\binom{F}{2}+\binom{E}{2}=EF$ problem 10: we have to find infinitely many solutions in integers for $3a^2+4=q^2$ or $6a^2+4=q^2$ No thing on consest for this one?

Solution to Problem 6: Let $2=p_1<p_1,...<p_n$ be the primes.Then $p_1...p_{n-1}p_n|1+2+...+p_n=\frac{p_n(p_n+1)}{2}$ $\implies p_1....p_{n-1}|\frac{p_n-1}{2}\implies p_n>p_1...p_{n-1}$ But if $n>4,p_n^2<p_1..p_{n-1}$ which contradicts the given fact.So $n=1,2,3,4$.Checking,the only value that satisfies the condition is $p=3$

Solution to Problem 4: I am proving this more generally. By A.m-G.m,$\frac {1+2+....+n}n>\sqrt[n] {1.2....n}\implies \frac {n(n+1)}{2n}>\sqrt[n]{n!}\implies (\frac {n+1}{2})^n>n!$ Now set $n=2001$ Thus $ (1006)^{2011} >2011!$

Solution to Problem 9: Let $N$ has $k$ digits.Then the repeat of $N,\overline{N}=(10^k+1)N$. Note that if $10^k+1$ is square-free,then we need $10^k+1$ to have a square factor. From the tricky lemma,$10^k+1$ is not square-free for $k=11$ Then $11^2|10^{11}+1$ Now note that $ 100000000001=11^{2}*23*4093*8779 $ $\overline{N}=11^{2}*23*4093*8779*N$ is a square.This suggests us to take a number divisible by $23*4093*8779$ which is a square.Now we need to check by hand to verify that $4093,8779$ are primes.Then a convenient choice for $ N=a^2*23*8779*4093$ for some $a$ But $N$ must be a $11$ digit number,as we derived.Check with some smaller values of $a$.We get $a=10$ because $23*4093*8779=826446281$ which is a $9$ digit number.Thus,$N=82644628100$ is such a number that satisfies the condition.

Solution to Problem 3. Let $AB=DC=l, AD=BC=w, BX=x, BP=y$. By Opposite Angle Theorem, $\angle DXC=\angle PXB$ and clearly $\angle XCD = \angle XBP= 90^{\circ}$ $\implies \triangle DXC \sim \triangle PBX\implies \frac{y}{x}=\frac{l}{w-x}\implies y=\frac{xl}{w-x}$. It is known that $x\ge 0, x\le\frac{w}{2}\Longleftrightarrow w-2x\ge 0, w>x\Longleftrightarrow w-x\ge 0\Longleftrightarrow \frac{1}{w-x}\ge 0, l\ge 0$, so multiplying them together yields $0\le \frac{xl(w-2x)}{w-x}=x\left(l-\frac{xl}{w-x}\right)=x(l-y)$ $\Longleftrightarrow x(y-l)\le 0 \Longleftrightarrow xy + wl - xl \le wl \Longleftrightarrow \frac{xy}{2}+\frac{(w-x)l}{2}\le \frac{wl}{2}$ $\Longleftrightarrow [BXP]+[CXD]\le \frac{[ABCD]}{2}$, with equality holding and if and only if $x=0,\frac{w}{2}$.

mathmdmb wrote: Solution to Problem 9: Let $N$ has $k$ digits.Then the repeat of $N,\overline{N}=(10^k+1)N$. Note that if $10^k+1$ is square-free,then we need $10^k+1$ to have a square factor. From the tricky lemma,$10^k+1$ is not square-free for $k=11$ Then $11^2|10^{11}+1$ Now note that $ 100000000001=11^{2}*23*4093*8779 $ $\overline{N}=11^{2}*23*4093*8779*N$ is a square.This suggests us to take a number divisible by $23*4093*8779$ which is a square.Now we need to check by hand to verify that $4093,8779$ are primes.Then a convenient choice for $ N=a^2*23*8779*4093$ for some $a$ But $N$ must be a $11$ digit number,as we derived.Check with some smaller values of $a$.We get $a=10$ because $23*4093*8779=826446281$ which is a $9$ digit number.Thus,$N=82644628100$ is such a number that satisfies the condition. Why do we have to check that thing in red? All we needed was to find a prime whose square divided our number. On a different line; we don't have to find the least such number; any will do. Therefore we need to find a prime $p$ such that $p^2 \mid 10^{2k}-1$, but $p \nmid 10{k} -1$ (so we are having $p^2 \mid 10^k + 1$). It is convenient to work with $p=11$, since $10 \equiv -1 \pmod{11}$ has an easy to handle residue modulo $11$. Now, $\varphi(11^2) = 110$, so try $k=55$. All is fine, since $10^{55} \equiv (-1)^{55} = -1 \not \equiv 1 \pmod{11}$. Thus $p=11$, $k=55$ will do. All we need now is to take the $55$-digit number $10^2 \cdot (10^{55} + 1)/121$.

Solution to Problem 5. Let $\angle ACB=c$. Let the inradius be $r$ and circumradius be $R$, as usual. Let the incenter of $\triangle ABC$ be $I$. Note that $IS\perp AC$ and $IR\perp AB$, which, combined with $IS=IR=r$ and $\angle A=90^{\circ}$, implies that $ARIS$ is a square. Then $\angle ARS=\angle ASR=45^{\circ}$. By the Opposite Angle Theorem, $\angle NRB=\angle ARS=45^{\circ}$. Note that $\angle RBN=180^{\circ}-\angle RBC=c+90^{\circ}$. By the Interior Angle Theorem, $\angle UNM=\angle RNB= 180^{\circ}-\angle NRB-\angle RBN=45^{\circ}-c$. Let the circumcenter of $\triangle ABC$ be $X$. Since $XA=XC=R$, $\angle XAC=\angle XCA=c$. Note that $\angle XAM=90^{\circ}\implies \angle SAU=90^{\circ}+c$. Remember that $\angle ASR=45^{\circ}$. By the Opposite and Interior Angle Theorems, $\angle NUM=\angle AUS=180^{\circ}-\angle UAS-\angle ASU=45^{\circ}-c$. Thus, since $\angle UNM=\angle NUM$, $\triangle NUM$ is isosceles. The hardest part was drawing the diagram LOL. After that, the $\angle A=90^{\circ}$ simplified the whole angle chasing.

mavropnevma wrote: Thus $p=11$, $k=55$ will do. All we need now is to take the $55$-digit number $10^2 \cdot (\boxed{10^55 + 1})/121$. I think it will be $10^{55}+1$