Determine all triples of real numbers $(a,b,c)$ such that \begin{eqnarray*} xyz &=& 8 \\ x^2y + y^2z + z^2x &=& 73 \\ x(y-z)^2 + y(z-x)^2 + z(x-y)^2 &=& 98 . \end{eqnarray*}

Click for solution $(x-y)(y-z)(z-x)=xy^2+yz^2+zx^2-x^2y-y^2z-z^2x$ so $x=y$ or $y=z$ or $z=x$

A flea jumps in a straight numbered line. It jumps first from point $0$ to point $1$. Afterwards, if its last jump was from $A$ to $B$, then the next jump is from $B$ to one of the points $B + (B - A) - 1$, $B + (B - A)$, $B + (B-A) + 1$. Prove that if the flea arrived twice at the point $n$, $n$ positive integer, then it performed at least $\lceil 2\sqrt n\rceil$ jumps.

Click for solution My solution: We say that at step $i$, we move $d_i$ units, where $d_i$ is an integer. From the condition we get that $|d_{i+1}-d{i}|$ is less than or equal to $1$ Suppose we fall two times on point $n$, there is a point to the right of $n$, or $n$ itself, such that the Flea makes a jump from it onto it (so it can go back). Let $m$ be this point. Take a coordinate system, where we will draw rectangles of base the segment between $i$ and $i+1$ and altitude $d_i$. Cleraly the sum (signed sum) of this rectangles from $0$ to $j$ is where the Flea is before the $j$ step. So suppose our Flea arrives to $m$, in $x$ steps for the last time, $d_x$ is $1$ at most, because the next step has distance $0$. so the Flea takes at most $x+1$ steps on steping back in $n$. Now draw the lines from $(0,1)$ with slope $1$, and from $(x,1)$ with slope $-1$, now traslade the altitude of the rectangles up, until they touch one of these lines with their leftmost vertex. The new area is bigger than or equal to the area of the rectangles, so in fact it is bigger than $m$ and $n$. But it is easy to calculate this area and from here obtain the bound we want for $x$

Let $p > 3$ be a prime. Prove that if $$\sum_{i=1 }^{p-1}{1\over i^p} = {n\over m}, $$with $ gcd\,\, (n,m) = 1$, then $p^3$ divides $n$.

Click for solution Well, sometimes I think it's easier to put up a solution than searching... I suspected that it was well known too. But I solved it in a slightly different way. I used the extended binomial expansion, that is, \[ (a + b)^\alpha = \sum_{k=0}^\infty{\alpha \choose k}a^{\alpha-k}b^k \] for $\alpha$ real. We define ${\alpha \choose k}$ for $\alpha$ real and $k$ nonnegative integer as ${\alpha \choose 0} = 1$ and \[ {\alpha\choose k} = \dfrac{\alpha(\alpha-1)(\alpha-2)\ldots (\alpha - k + 1)}{k!} \] for $k > 0$. So \[ a^{-p} + (p-a)^{-p} = a^{-p} + \sum_{k=0}^\infty {-p\choose k}(-a)^{-p-k}p^k \equiv a^{-p} - a^{-p} + {-p\choose 1}a^{-p-1}p - {-p\choose 2}a^{-p-2}p^2 \pmod{p^3} \] It's not hard to see that $p$ divides ${-p\choose 2}$, so \[ a^{-p} + (p-a)^{-p} \equiv -a^{-p-1}p^2\pmod{p^3} \] OK, it looks like "cheating" but it's not. You can check by hand that, indeed, \[ (p-a)^p \cdot (-a^{-p}-a^{-p-1}p^2) \equiv 1 \pmod{p^3} \] In fact, using the "regular" binomial expansion, \[ (p-a)^p \equiv -a^p + {p\choose 1}a^{p-1}p \pmod p \] and \[ (p-a)^p \cdot (-a^{-p}-a^{-p-1}p^2) \equiv (-a^p + a^{p-1}p^2)\cdot (-a^{-p}-a^{-p-1}p^2) \equiv 1 + a^{-1}p^2 -a^{-1}p^2\equiv 1 \pmod{p^3} \] Thus we only need to prove that \[ \sum_{k=1}^{p-1}a^{-p-1}\equiv 0\pmod p \] or, recalling that $a^{-p-1} \equiv (a^{-1})^{p+1} \equiv (a^{-1})^2\pmod p$, \[ \sum_{k=1}^{p-1}(a^{-1})^2\equiv 0\pmod p \] This is a standard problem in number theory. Let $S = \sum_{k=1}^{p-1}(a^{-1})^2\bmod p$. Thus, since $\mathbb{Z}/p\mathbb{Z}^* = \{1,2,\ldots, p-1\}$ is a reduced residue system, $2^{-2}\mathbb{Z}/p\mathbb{Z}^* = \mathbb{Z}/p\mathbb{Z}^*$ and \[ 2^{-2}S \equiv S\pmod p\iff S \equiv 0 \pmod p, \] since $p \neq 3$ and we are done.

Denote by $a \bmod b$ the remainder of the euclidean division of $a$ by $b$. Determine all pairs of positive integers $(a,p)$ such that $p$ is prime and \[ a \bmod p + a\bmod 2p + a\bmod 3p + a\bmod 4p = a + p. \]

Click for solution modulo $p$ we get $3a=0$ so $p=3$ or $a=0$ modp With $p=3$ we need to check cases, since our expression is at most $2+5+8+11=26$ so a is at most $23$ (I checked them manually actually) With $a=0$ mod p, we can use that $ap$ mod $bp$=$p(a$mod$b)$. And it also reduces to check some cases. The solutions $(p,3p)$ and $(3,1)$, $(3,17)$

Let $O$ be the circumcenter of acutangle triangle $ABC$ and let $A_1$ be some point in the smallest arc $BC$ of the circumcircle of $ABC$. Let $A_2$ and $A_3$ points on sides $AB$ and $AC$, respectively, such that $\angle BA_1A_2 = \angle OAC$ and $\angle CA_1A_3 = \angle OAB$. Prove that the line $A_2A_3$ passes through the orthocenter of $ABC$.

Click for solution Let the altitudes from $B$ respectively $C$ intersect the circumcircle at B' respectively C'. Then points $(C',A_2,A_1)$ and $(A_1,A_3,B')$ are colliniar. Hence by Pascal's theorem apllied to $ABB'A_1C'C$ we conclude that points $A_1,H,A_3$ are collinar.

Let $n$ be a fixed positive integer. The points $A_1$, $A_2$, $\ldots$, $A_{2n}$ are on a straight line. Color each point blue or red according to the following procedure: draw $n$ pairwise disjoint circumferences, each with diameter $A_iA_j$ for some $i \neq j$ and such that every point $A_k$ belongs to exactly one circumference. Points in the same circumference must be of the same color. Determine the number of ways of coloring these $2n$ points when we vary the $n$ circumferences and the distribution of the colors.

Click for solution This was an easy number six. First, every circle goes from an odd point to an even point, so there is aan equal number of odd blue points and even blue points and the same for red. There are $\binom{2n}{n}$ permutations satisfying that. Now we prove all of them can be obtained by the coloring process. We make induction on $n$ Clearly there are two consecutive points of the same color, putting a circle trought them, and removing them leaves a nice scenario for $n-1$ case So we are done