Found problems: 56
An $n$-term sequence $(x_1, x_2, \ldots, x_n)$ in which each term is either 0 or 1 is called a [i]binary sequence of length [/i]$n$. Let $a_n$ be the number of binary sequences of length $n$ containing no three consecutive terms equal to 0, 1, 0 in that order. Let $b_n$ be the number of binary sequences of length $n$ that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that $b_{n+1} = 2a_n$ for all positive integers $n$.
Positive integers $x_1, x_2, \dots, x_n$ ($n \ge 4$) are arranged in a circle such that each $x_i$ divides the sum of the neighbors; that is \[ \frac{x_{i-1}+x_{i+1}}{x_i} = k_i \] is an integer for each $i$, where $x_0 = x_n$, $x_{n+1} = x_1$. Prove that \[ 2n \le k_1 + k_2 + \dots + k_n < 3n. \]
Let $n$ be a fixed positive integer. Initially, $n$ 1's are written on a blackboard. Every minute, David picks two numbers $x$ and $y$ written on the blackboard, erases them, and writes the number $(x+y)^4$ on the blackboard. Show that after $n-1$ minutes, the number written on the blackboard is at least $2^{\frac{4n^2-4}{3}}$.
[i]Proposed by Calvin Deng[/i]
Let $S$ be a set of rational numbers with the following properties:
(a) $\frac12$ is an element in $S$,
(b) if $x$ is in $S$, then both $\frac{1}{x+1}$ and $\frac{x}{x+1}$ are in $S$.
Prove that $S$ contains all rational numbers in the interval $(0, 1)$.
For any positive integer $n$ let $f(n)$ be the number of divisors of $n$ ending with $1$ or $9$ in base $10$ and let $g(n)$ be the number of divisors of $n$ ending with digit $3$ or $7$ in base $10$. Prove that $f(n)\geqslant g(n)$ for all nonnegative integers $n$.
[i](Swiss Mathematical Olympiad 2011, Final round, problem 9)[/i]
Determine all positive integers $n$, $n\ge2$, such that the following statement is true:
If $(a_1,a_2,...,a_n)$ is a sequence of positive integers with $a_1+a_2+\cdots+a_n=2n-1$, then there is block of (at least two) consecutive terms in the sequence with their (arithmetic) mean being an integer.