Let $ A$ be a set of natural numbers in which if $ a$ , $ b$ belong to $ A$ ($ a>b$) then either $ a+b$ or $ a-b$ belong to $ A$ ( both cases may be posible at the same time). Decide wheter there is or not a set $ A$ consisting on exactly $ 100$ elements which has four elements $ x$, $ y$ , $ z$ , $ w$ ( not necesarilly distinct) that satisfy $ x-y=512$ and $ z-w=460$ Daniel

Given 365 cards, in which distinct numbers are written. We may ask for any three cards, the order of numbers written in them. Is it always possible to find out the order of all 365 cards by 2000 such questions?

In a circumference with center $ O$ we draw two equal chord $ AB=CD$ and if $ AB \cap CD =L$ then $ AL>BL$ and $ DL>CL$ We consider $ M \in AL$ and $ N \in DL$ such that $ \widehat {ALC} =2 \widehat {MON}$ Prove that the chord determined by extending $ MN$ has the same as length as both $ AB$ and $ CD$

Find all integers solutions for $ xy+yz+zx-xyz=2$

Let $p$ be a prime with $p>5$, and let $S=\{p-n^2 \vert n \in \mathbb{N}, {n}^{2}<p \}$. Prove that $S$ contains two elements $a$ and $b$ such that $a \vert b$ and $1<a<b$.

Click for solution We show that the smallest possible $a$ works. Let $n= \lfloor \sqrt p \rfloor$. If $n^2+1=p$, then we set $a=p-(n-1)^2=n^2+1-n^2+2n-1=2n$ and $b=p-1^2=n^2$. This works, because surely $p$ is odd, so $2|n$ giving $a=2n|n^2=b$. In all other cases we can set $a=p-n^2>1$. There are three cases: a) $n^2+n > p$: Then we set $c=n^2+n-p>0$. Additionally $n^2 < p$ gives $c=n^2-p+n<n$. b) $n^2+n=p$: this implies $n|p$, impossible. c) $n^2+n < p$: Then we set $c=p-n^2-n>0$. Additionally $p < (n+1)^2$ gives $c=p-n^2-n=p-(n+1)^2+n+1<n+1$. If $c=n$, then we would have $n=p-n^2-n$ giving $n|p$, impossible. So again $c<n$. In all possible cases, we now set $b=p-c^2$ and get $b=p-c^2 \equiv p- (\pm n)^2 \equiv p-n^2 \equiv 0 \mod (p-n^2)$, being the result.

Let $ n$ be a natural number, and we consider the sequence $ a_1, a_2 \ldots , a_{2n}$ where $ a_i \in (-1,0,1)$ If we make the sum of consecutive members of the sequence, starting from one with an odd index and finishing in one with and even index, the result is $ \le 2$ and $ \ge -2$ How many sequence are there satisfying this conditions?