Day 1
On a table there is a pile with $ T$ tokens which incrementally shall be converted into piles with three tokens each. Each step is constituted of selecting one pile removing one of its tokens. And then the remaining pile is separated into two piles. Is there a sequence of steps that can accomplish this process? a.) $ T = 1000$ (Cono Sur) b.) $ T = 2001$ (BWM)
Click for solution Let n, m be the number of disks and the number of piles at a moment. Each movement will rduce n by one, but increase m by one. So n+m = 1000 + 1 = 1001 Since there are no empty piles, when only 3-disk piles exist: 3m + m = 1001 4m = 1001 Which is contradictory. Similarly, since 2002 is not a multiple of 4, point b) is solved.
Consider a circle with centre $O$, and $3$ points on it, $A,B$ and $C$, such that $\angle {AOB}< \angle {BOC}$. Let $D$ be the midpoint on the arc $AC$ that contains the point $B$. Consider a point $K$ on $BC$ such that $DK \perp BC$. Prove that $AB+BK=KC$.
Find the number of elements that a set $B$ can have, contained in $(1, 2, ... , n)$, according to the following property: For any elements $a$ and $b$ on $B$ ($a \ne b$), $(a-b) \not| (a+b)$.
Day 2
On a chess board ($8\times 8$) there are written the numbers $1$ to $64$: on the first line, from left to right, there are the numbers $1, 2, 3, ... , 8$; on the second line, from left to right, there are the numbers $9, 10, 11, ... , 16$;etc. The ''$ + $'' and ''$- $'' signs are put to each number such that, in each line and in each column, there are $4$ ''$+$'' signs and $4$ ''$-$'' signs. Then, the $64$ numbers are added. Find all the possible values of this sum.
Prove that there exists a succession $a_1, a_2, ... , a_k, ...$, where each $a_i$ is a digit ($a_i \in (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)$ ) and $a_0=6$, such that, for each positive integrer $n$, the number $x_n=a_0+10a_1+100a_2+...+10^{n-1}a_{n-1}$ verify that $x_n^2-x_n$ is divisible by $10^n$.
Prove that, given a positive integrer $n$, there exists a positive integrer $k_n$ with the following property: Given any $k_n$ points in the space, $4$ by $4$ non-coplanar, and associated integrer numbers between $1$ and $n$ to each sharp edge that meets $2$ of this points, there's necessairly a triangle determined by $3$ of them, whose sharp edges have associated the same number.