Found problems: 5923
Source: 2018 Canadian Open Math Challenge Part A Problem 4
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In the sequence of positive integers, starting with $2018, 121, 16, ...$ each term is the square of the sum of digits of the previous term. What is the $2018^{\text{th}}$ term of the sequence?
For any odd prime $p$ and any integer $n,$ let $d_p (n) \in \{ 0,1, \dots, p-1 \}$ denote the remainder when $n$ is divided by $p.$ We say that $(a_0, a_1, a_2, \dots)$ is a [i]p-sequence[/i], if $a_0$ is a positive integer coprime to $p,$ and $a_{n+1} =a_n + d_p (a_n)$ for $n \geqslant 0.$
(a) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $(a_0, a_1, a_2, \dots)$ and $(b_0, b_1, b_2, \dots)$ such that $a_n >b_n$ for infinitely many $n,$ and $b_n > a_n$ for infinitely many $n?$
(b) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $(a_0, a_1, a_2, \dots)$ and $(b_0, b_1, b_2, \dots)$ such that $a_0 <b_0,$ but $a_n >b_n$ for all $n \geqslant 1?$
[I]United Kingdom[/i]
Suppose $P$ is a polynomial with integer coefficients such that for every positive integer $n$, the sum of the decimal digits of $|P(n)|$ is not a Fibonacci number. Must $P$ be constant?
(A [i]Fibonacci number[/i] is an element of the sequence $F_0, F_1, \dots$ defined recursively by $F_0=0, F_1=1,$ and $F_{k+2} = F_{k+1}+F_k$ for $k\ge 0$.)
[i]Nikolai Beluhov[/i]
Let $ABCD$ be a convex quadrilateral with precisely one pair of parallel sides.
$(a)$ Show that the lengths of its sides $AB,BC,CD, DA$ (in this order) do not form an arithmetic progression.
$(b)$ Show that there is such a quadrilateral for which the lengths of its sides $AB ,BC,CD,DA$ form an arithmetic progression after the order of the lengths is changed.
Given $3n$ cards, each of them will be written with a number from the following sequence:
$$2, 3, ..., n, n + 1, n + 3, n + 4, ..., 2n + 1, 2n + 2, 2n + 4, ..., 3n + 3$$
with each number used exactly once. Then every card is arranged from left to right in random order. Determine the probability such that for every $i$ with $1\le i \le 3n$, the number written on the $i$-th card, counted from the left, is greater than or equal to $i$.
In the sequence $ 2001, 2002, 2003, \ldots$, each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is $ 2001 \plus{} 2002 \minus{} 2003 \equal{} 2000$. What is the $ 2004^\text{th}$ term in this sequence?
$ \textbf{(A)} \minus{} \! 2004 \qquad \textbf{(B)} \minus{} \! 2 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 4003 \qquad \textbf{(E)}\ 6007$
Let $\{a_n \}_n$ be a sequence of real numbers such there there are countably infinite distinct subsequences converging to the same point. We call two subsequences distinct if they do not have a common term. Which of the following statements always holds:
(A) $\{a_n \}_n$ is bounded
(B) $\{a_n \}_n$ is unbounded
(C) The set of convergent subsequence $\{a_n \}_n$ is countable
(D) None of these
Let $a$ and $b$ be non-negative integers. Consider a sequence $s_1$, $s_2$, $s_3$, $. . .$ such that $s_1 = a$, $s_2 = b$, and $s_{i+1} = |s_i - s_{i-1}|$ for $i \ge 2$. Prove that there is some $i$ for which $s_i = 0$.
Let $m, n$ be positive integers. Let $S(n,m)$ be the number of sequences of length $n$ and consisting of $0$ and $1$ in which there exists a $0$ in any consecutive $m$ digits. Prove that
\[S(2015n,n).S(2015m,m)\ge S(2015n,m).S(2015m,n)\]
Rational numbers are written in the following sequence: $\frac{1}{1},\frac{2}{1},\frac{1}{2},\frac{3}{1},\frac{2}{2},\frac{1}{3},\frac{4}{1},\frac{3}{2},\frac{2}{3},\frac{1}{4}, . . .$
In which position of this sequence is $\frac{2005}{2004}$ ?
We have $10$ points on a line $A_1,A_2\ldots A_{10}$ in that order. Initially there are $n$ chips on point $A_1$. Now we are allowed to perform two types of moves. Take two chips on $A_i$, remove them and place one chip on $A_{i+1}$, or take two chips on $A_{i+1}$, remove them, and place a chip on $A_{i+2}$ and $A_i$ . Find the minimum possible value of $n$ such that it is possible to get a chip on $A_{10}$ through a sequence of moves.
A maths teacher has $10$ cards with the numbers $1$ to $10$ on them, one number per card. She places these cards in some order in a line next to each other on the table. The students come to the table, one at a time. The student whose turn it is goes once through the line of cards from left to right and removes every card she encounters that is (at that moment) the lowest card on the table. This continues till all cards are removed from the table. For example, if the line is in order $3$, $1$, $4$, $5$, $8,$ $6$, $9$, $10$, $2$, $7$ from left to right, the first student takes cards $1$ and $2$. Then the second student comes who, in our example, takes the cards $3$, $4$, $5$, $6$, and $7$. The third student then takes the cards $8$, $9$, and $10$.
Let $A$ be the number of sequences of cards that the teacher can choose so that exactly nine students get a turn to pick cards. Let $B$ be the number of sequences of cards that the teacher can choose so that exactly two students get a turn to pick cards. Prove that $A = B$.
For a positive integer $a$, define $F_1 ^{(a)}=1$, $F_2 ^{(a)}=a$ and for $n>2$, $F_n ^{(a)}=F_{n-1} ^{(a)}+F_{n-2} ^{(a)}$. A positive integer is fibonatic when it is equal to $F_n ^{(a)}$ for a positive integer $a$ and $n>3$. Prove that there are infintely many not fibonatic integers.
Let $q_{0},q_{1},...$ be a sequence of integers such that
a) for any $m>n$ we have $m-n\mid q_{m}-q_{n}$, and
b) $|q_{n}|\leq n^{10}, \ \forall n\geq 0$.
Prove there exists a polynomial $Q$ such that $q_{n}=Q(n), \ \forall n\geq 0$.
A finite sequence of natural numbers $a_1, a_2, \dots, a_n$ is given. A sub-sequence $a_{k+1}, a_{k+2}, \dots, a_l$ will be called a [i]repetition[/i] if there exists a natural number $p\leq \frac{l-k}2$ such that $a_i=a_{i+p}$ for $k+1\leq i\leq l-p$, but $a_i\neq a_{i+p}$ for $i=k$ (if $k>0$) and $i=l-p+1$ (if $l<n$).
Show that the sequence contains less than $n$ repetitions.
Let $A$ and $B$ be two sets of real numbers. Suppose that the elements of the set $AB = \{ab: a\in A, b\in B\}$ form a finite arithmetic progression. Prove that one of these sets contains no more than three elements
Find the number of sequences of 10 letters where all the letters are either $A$ or $B$, the first letter is $A$, the last letter is $B$, and the sequence contains no three consecutive letters reading $ABA$. For example, count $AAABBABBAB$ and $ABBBBBBBAB$ but not $AABBAABABB$ or $AAAABBBBBA$.
An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?
[asy]
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fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white);
fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white);
fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white);
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label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S);
label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S);
[/asy]
$\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}$
Starting from the number $ 1$ we write down a sequence of numbers where the next number in the sequence is obtained from the previous one either by doubling it, or by rearranging its digits (not allowing the first digit of the rearranged number to be $0$). For instance we might begin:
$$1, 2, 4, 8, 16, 61, 122, 212, 424,...$$
Is it possible to construct such a sequence that ends with the number $1,000,000,000$? Is it possible to construct one that ends with the number $9,876,543,210$?
Two vertices of a cube are $A,O$ such that $AO$ is the diagonal of one its faces. A $n-$run is a sequence of $n+1$ vertices of the cube such that each $2$ consecutive vertices in the sequence are $2$ ends of one side of the cube. Is the $1386-$runs from $O$ to itself less than $1386-$runs from $O$ to $A$ or more than it?
The Fibonacci sequence is given by $a_1 = 1, a_2 = 2$ and $a_{n+1} = a_n +a_{n-1}$ for $n > 1$. Express $a_{2n}$ in terms of only $a_{n-1},a_n,a_{n+1}$.
The sequence $ \{ a_n \} _ { n \ge 0 } $ is defined by $ a_0 = 2 , a_1 = 4 $ and
\[ a_{n+1} = \frac{a_n a_{n-1}}{2} + a_n + a_{n-1} \]
for all positive integers $ n $. Determine all prime numbers $ p $ for which there exists a positive integer $ m $ such that $ p $ divides the number $ a_m - 1 $.
Define sequences $\{a_n\},\ \{b_n\}$ by
\[a_n=\int_{-\frac {\pi}6}^{\frac{\pi}6} e^{n\sin \theta}d\theta,\ b_n=\int_{-\frac {\pi}6}^{\frac{\pi}6} e^{n\sin \theta}\cos \theta d\theta\ (n=1,\ 2,\ 3,\ \cdots).\]
(1) Find $b_n$.
(2) Prove that for each $n$, $b_n\leq a_n\leq \frac 2{\sqrt{3}}b_n.$
(3) Find $\lim_{n\to\infty} \frac 1{n}\ln (na_n).$
Let $x_0,\dots,x_{2017}$ are positive integers and $x_{2017}\geq\dots\geq x_0=1$ such that $A=\{x_1,\dots,x_{2017}\}$ consists of exactly $25$ different numbers. Prove that $\sum_{i=2}^{2017}(x_i-x_{i-2})x_i\geq 623$, and find the number of sequences that holds the case of equality.
Let $n \geq 3$ be an integer. A sequence $P_1, P_2, \ldots, P_n$ of distinct points in the plane is called [i]good[/i] if no three of them are collinear, the polyline $P_1P_2 \ldots P_n$ is non-self-intersecting and the triangle $P_iP_{i + 1}P_{i + 2}$ is oriented counterclockwise for every $i = 1, 2, \ldots, n - 2$.
For every integer $n \geq 3$ determine the greatest possible integer $k$ with the following property: there exist $n$ distinct points $A_1, A_2, \ldots, A_n$ in the plane for which there are $k$ distinct permutations $\sigma : \{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}$ such that $A_{\sigma(1)}, A_{\sigma(2)}, \ldots, A_{\sigma(n)}$ is good.
(A polyline $P_1P_2 \ldots P_n$ consists of the segments $P_1P_2, P_2P_3, \ldots, P_{n - 1}P_n$.)