Let $a,b,c$ be distinct real numbers such that $a+b+c>0$. Let $M$ be the set of $3\times 3$ matrices with the property that each line and each column contain all given numbers $a,b,c$. Find $\{\max \{ \det A \mid A \in M \}$ and the number of matrices which realise the maximum value. Mircea Becheanu

Find all positive integers $A$ which can be represented in the form: \[ A = \left ( m - \dfrac 1n \right) \left( n - \dfrac 1p \right) \left( p - \dfrac 1m \right) \] where $m\geq n\geq p \geq 1$ are integer numbers. Ioan Bogdan

Click for solution Solution $\left ( m - \frac 1n \right) \left( n - \frac 1p \right) \left( p - \frac 1m \right) = \frac{(mn-1)(np-1)(pm-1)}{mnp}$ $\implies$ $p \lvert mn-1 \implies mn \equiv 1 \pmod p$ $m \lvert np-1 \implies np \equiv 1 \pmod m$ $n \lvert pm-1 \implies pm \equiv 1 \pmod n$ Let $p = \frac{mn-1}{k}$ $k<m$ to preserve the ordering of $m,n,p$ $n(mn-1) \equiv k \pmod m$ $m(mn-1) \equiv k \pmod n$ $n+k \equiv 0 \pmod m$ $m+k \equiv 0 \pmod n$ $m+k < 2n \implies m+k=n$ Therefore $n < 2m$ Therefore $n+k < 3k \implies n+k = 2m \implies m+k+k=2m \implies m=2k$ $m\frac{mn-1}{k} \equiv 1 \pmod n \implies 2(mn-1) \equiv 1 \pmod n \implies -2 \equiv 1 \pmod n \implies n=3$ or $n=1$ $m \le n \implies m=2$ or $m=1$ This leaves us with three ordered pairs $(m,n)$ $(1,1), (1,3), (2,3)$ $(m,n)=(1,1) \implies A = 0$ $(m,n)=(1,3) \implies p = 1$, contradicting the ordering. $(m,n)=(2,3) \implies p = 5$, and $A=21$ Therefore, the two possible values of $A$ are $0$ and $21$

Let $A$ be the set $A = \{ 1,2, \ldots, n\}$. Determine the maximum number of elements of a subset $B\subset A$ such that for all elements $x,y$ from $B$, $x+y$ cannot be divisible by $x-y$. Mircea Lascu, Dorel Mihet

Let $ P(X) = a_{n}X^{n} + a_{n - 1}X^{n - 1} + \ldots + a_{1}X + a_{0}$ be a real polynomial of degree $ n$. Suppose $ n$ is an even number and: a) $ a_{0} > 0$, $ a_{n} > 0$; b) $ a_{1}^{2} + a_{2}^{2} + \ldots + a_{n - 1}^{2}\leq\frac {4\min(a_{0}^{2} , a_{n}^{2})}{n - 1}$. Prove that $ P(x)\geq 0$ for all real values $ x$. Laurentiu Panaitopol

Let $A$ be the set $\{1,2,\ldots,n\}$, $n\geq 2$. Find the least number $n$ for which there exist permutations $\alpha$, $\beta$, $\gamma$, $\delta$ of the set $A$ with the property: \[ \sum_{i=1}^n \alpha(i) \beta (i) = \dfrac {19}{10} \sum^n_{i=1} \gamma(i)\delta(i) . \] Marcel Chirita

Click for solution Cute. For $n = 28$ it works: take $\alpha(i) = \beta(i) = \gamma(i) = i$ and $\delta(i) = 29 - i$, then we get \[ \sum_{i = 1}^{28}\alpha(i)\beta(i) = \sum_{i = 1}^{28}i^2 = 7714 \] and \[ \frac{19}{10} \cdot \sum_{i = 1}^{28}\gamma(i)\delta(i) = \frac{19}{10} \cdot \sum_{i = 1}^{28}i(29 - i) = \frac{19}{10} \cdot 4060 = 7714. \] However, it cannot work for $n \leq 27$. Indeed, by the rearrangement inequality we have \[ \frac{19}{10} \cdot \sum_{i = 1}^n\gamma(i)\delta(i) \geq \frac{19}{10} \cdot \sum_{i = 1}^ni(n + 1 - i) = \frac{19}{10} \cdot \frac{n(n + 1)(n + 2)}{6} \] and \[ \sum_{i = 1}^n\alpha(i)\beta(i) \leq \sum_{i = 1}^ni^2 = \frac{n(n + 1)(2n + 1)}{6}. \] Now, since $n \leq 27$, \[ \frac{19}{10} \cdot (n + 2) > 2n + 1 \] (since this is equivalent to $n < 28$), so the right-hand side will always be "too big". So, the solution is $n = 28$.

The plane is covered with network of regular congruent disjoint hexagons. Prove that there cannot exist a square which has its four vertices in the vertices of the hexagons. Gabriel Nagy

Click for solution We can even assume the lattice is triangular: for each hexagon, add a point in its center and connect it to the six vertices of the hexagon. The conclusion holds for this larger lattice as well: Assume WLOG that we can find points $ A,B $ belonging to this lattice s.t. if $ O $ is the origin of the plane, $ OB $ is obtained from $ OA $ through a rotation of angle $ \frac\pi 2 $ around $ O $. This is equivalent to finding integers $ a,b,c,d $ s.t. $ \frac{a+b\varepsilon}{c+d\varepsilon}=i $, where $ \varepsilon $ is a solution to $ x^2+x+1=0 $ ($ a+b\varepsilon $ represents the point $ A $, while $ c+d\varepsilon $ - the point $ B $). In turn, this implies that we can find rationals $ u,v $ with $ u+v\varepsilon=i $. This is claerly absurd, since eliminating $ u $ between $ u+v\varepsilon=i,u+v\bar\varepsilon=-i $ gives an irrational value for $ v $.

Determine all positive integers $n$ such that $n$ divides $3^n - 2^n$.

Click for solution Let $p$ be the smallest prime divisor of $n$. Then $p\mid 3^n-2^n$ and $p \mid 3^{p-1}-2^{p-1}$, so $p \mid 3^{\gcd (n,p-1)}-2^{\gcd (n,p-1)}$ because $a_n = 3^n-2^n$ is a Mersenne sequence (well-known) $= 3-2 = 1$, so the only $n$ is 1.

Let $ABCD$ be a square and $a$ be the length of his edges. The segments $AE$ and $CF$ are perpendicular on the square's plane in the same half-space and they have the length $AE=a$, $CF=b$ where $a<b<a\sqrt 3$. If $K$ denoted the set of the interior points of the square $ABCD$ determine $\min_{M\in K} \left( \max ( EM, FM ) \right) $ and $\max_{M\in K} \left( \min (EM,FM) \right)$. Octavian Stanasila

Prove that for all real numbers $\alpha_1,\alpha_2,\ldots,\alpha_n$ we have \[ \sum_{i=1}^n \sum_{j=1}^n ij \cos {(\alpha_i - \alpha_j )} \geq 0. \] Octavian Stanasila

Let $a,b,c$ be integer numbers such that $(a+b+c) \mid (a^{2}+b^{2}+c^{2})$. Show that there exist infinitely many positive integers $n$ such that $(a+b+c) \mid (a^{n}+b^{n}+c^{n})$. Laurentiu Panaitopol

Click for solution Consider the equation $x^3 - (a+b+c) x^2 + (ab + bc + ca) x - abc = 0$. This equation is the characteristic equation of the recurrence relation $a_n = (a+b+c)a_{n-1} - (ab + bc + ca)a_{n-2} + (abc)a_{n-3}$. This is equivalent to $a_n = a_1a_{n-1} - \frac{{a_1}^2-a_2}{2} a_{n-2} + (abc) a_{n-3}$. We note that the formula for $a_n$ is $a_n = a^n + b^n + c^n$. Thus $a_1 = a + b + c, a_2 = a^2 + b^2 + c^2, a_3 = a^3 + b^3 + c^3$. If $\frac{{a_1}^2-a_2}{2} \equiv 0 \pmod{a_1}$, then $a_n \equiv (abc)a_{n-3} \pmod{a_1} \Rightarrow a_i \equiv 0 \pmod{a_1} \forall i \equiv 1$ or $2 \pmod{3}, i \in \mathbb{Z}^+$, since $a_1 \equiv a_2 \equiv 0 \pmod{a_1}$, and we are done. Likewise, if $a_1$ is odd, then ${a_1}^2-a_2 \equiv 0 \pmod{a_1} \Rightarrow \frac{{a_1}^2-a_2}{2} \equiv 0 \pmod{a_1}$ (Note that we are guaranteed that $\frac{{a_1}^2-a_2}{2} \in \mathbb{Z}$ since $\frac{{a_1}^2-a_2}{2} = ab + bc + ca \in \mathbb{Z}$) This means that we once again have the similar case of $a_n \equiv (abc)a_{n-3} \pmod{a_1}$ and we are done. Thus we can assume that $a_1$ is even and that $\frac{{a_1}^2-a_2}{2} \equiv \frac{a_1}{2} \pmod{a_1} \Rightarrow a_n \equiv \frac{a_1}{2}a_{n-2} + (abc)a_{n-3} \pmod{a_1}$. Note that since $a_n = a^n + b^n + c^n$, we have $a_i$ will have the same parity throughout $\forall i \in \mathbb{Z}^+$. Since we have assumed $a_1$ is even, we have $a_i$ is even $\forall i \in \mathbb{Z}^+$. Suppose that $\not\exists$ infinitely many $n$ such that $a_1 \mid a_n$, say $M$ is the largest integer such that $a_1 \mid a_M$. Now consider $a_{M+3}$. We have $a_{M+3} \equiv \frac{a_1}{2}a_{M+1} + (abc)a_{M} \equiv 0 \pmod{a_1}$, since $a_{M+1}$ is even. (Contradiction) $\therefore \exists$ infinitely many $n$ such that $a + b + c \mid a^n + b^n + c^n$.

Let $P(X,Y)=X^2+2aXY+Y^2$ be a real polynomial where $|a|\geq 1$. For a given positive integer $n$, $n\geq 2$ consider the system of equations: \[ P(x_1,x_2) = P(x_2,x_3) = \ldots = P(x_{n-1},x_n) = P(x_n,x_1) = 0 . \] We call two solutions $(x_1,x_2,\ldots,x_n)$ and $(y_1,y_2,\ldots,y_n)$ of the system to be equivalent if there exists a real number $\lambda \neq 0$, $x_1=\lambda y_1$, $\ldots$, $x_n= \lambda y_n$. How many nonequivalent solutions does the system have? Mircea Becheanu