Found problems: 583
How many ordered pairs of positive integers $(m, n)$ satisfy the system
\begin{align*}
\gcd (m^3, n^2) & = 2^2 \cdot 3^2,
\\ \text{LCM} [m^2, n^3] & = 2^4 \cdot 3^4 \cdot 5^6,
\end{align*}
where $\gcd(a, b)$ and $\text{LCM}[a, b]$ denote the greatest common divisor and least common multiple of $a$ and $b$, respectively?
If positive integers, $a,b,c$ are pair-wise co-prime, and, \[\ a^2|(b^3+c^3), b^2|(a^3+c^3), c^2|(a^3+b^3) \] find $a,b,$ and $c$
Ten positive integers are arranged around a circle. Each number is one more than the greatest common divisor of its two neighbors. What is the sum of the ten numbers?
Let $A$ be the set of four-digit natural numbers having exactly two distinct digits, none of which is zero. Interchanging the two digits of $n\in A$ yields a number $f (n) \in A$ (for instance, $f (3111) = 1333$). Find those $n \in A$ with $n > f (n)$ for which $gcd(n, f (n))$ is the largest possible.
[b](a)[/b] Each of Peter and Basil thinks of three positive integers. For each pair of his numbers, Peter writes down the greatest common divisor of the two numbers. For each pair of his numbers, Basil writes down the least common multiple of the two numbers. If both Peter and Basil write down the same three numbers, prove that these three numbers are equal to each other.
[b](b)[/b] Can the analogous result be proved if each of Peter and Basil thinks of four positive integers instead?
Prove that if $m$ and $s$ are integers with $ms=2000^{2001}$, then the equation $mx^2-sy^2=3$ has no integer solutions.
Find five positive integers such that the greatest common divisor of each pair is equal to the difference between them.
(SI Tokarev)
Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$. Which of the following must be a divisor of $a$?
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$
$33$ horsemen are riding in the same direction along a circular road. Their speeds are constant and pairwise distinct. There is a single point on the road where the horsemen can surpass one another. Can they ride in this fashion for arbitrarily long time ?
For a positive integer number $n$ we denote $d(n)$ as the greatest common divisor of the binomial coefficients $\dbinom{n+1}{n} , \dbinom{n+2}{n} ,..., \dbinom{2n}{n}$.
Find all possible values of $d(n)$
The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base 10, where digit $a$ is not zero. Find the three-digit number $abc$.
For two positive integers a and b, which are relatively prime, find all integer that can be the great common divisor of $a+b$ and $\frac{a^{2005}+b^{2005}}{a+b}$.
$100$ distinct natural numbers $a_1, a_2, a_3, \ldots, a_{100}$ are written on the board. Then, under each number $a_i$, someone wrote a number $b_i$, such that $b_i$ is the sum of $a_i$ and the greatest common factor of the other $99$ numbers. What is the least possible number of distinct natural numbers that can be among $b_1, b_2, b_3, \ldots, b_{100}$?
On a blackboard there are three positive integers. In each step the three numbers on the board are denoted as $a, b, c$ such that $a >gcd(b, c)$, then $a$ gets replaced by $ a-gcd(b, c)$. The game ends if there is no way to denote the numbers such that $a >gcd(b, c)$.
Prove that the game always ends and that the last three numbers on the blackboard only depend on the starting numbers.
(Theresia Eisenkölbl)
(a) Let $\gcd(m, k) = 1$. Prove that there exist integers $a_1, a_2, . . . , a_m$ and $b_1, b_2, . . . , b_k$ such that each product $a_ib_j$ ($i = 1, 2, \cdots ,m; \ j = 1, 2, \cdots, k$) gives a different residue when divided by $mk.$
(b) Let $\gcd(m, k) > 1$. Prove that for any integers $a_1, a_2, . . . , a_m$ and $b_1, b_2, . . . , b_k$ there must be two products $a_ib_j$ and $a_sb_t$ ($(i, j) \neq (s, t)$) that give the same residue when divided by $mk.$
[i]Proposed by Hungary.[/i]
Let $m,n$ be integers. It is known that there are integers $a,b$ such that $am+bn=1$ if, and only if, the greatest common divisor of $m,n$ is 1. [i]You are not required to prove this[/i].
Now suppose that $p,q$ are different odd primes. In each case determine if there are integers $a,b$ such that $ap+bq=1$ so that the given condition is satisfied:
[list]
a. $p$ divides $b$ and $q$ divides $a$;
b. $p$ divides $a$ and $q$ divides $b$;
c. $p$ does not divide $a$ and $q$ does not divide $b$.
[/list]
Find all nonnegative integers $x,y,z$ such that $(x+1)^{y+1} + 1= (x+2)^{z+1}$.
[asy]
/*Using regular asymptote, this diagram would take 30 min to make. Using cse5, this takes 5 minutes. Conclusion? CSE5 IS THE BEST PACKAGE EVER CREATED!!!!*/
size(100);
import cse5;
pathpen=black;
anglefontpen=black;
pointpen=black;
anglepen=black;
dotfactor=3;
pair A=(0,0),B=(0.5,0.5*sqrt(3)),C=(3,0),D=(1.7,0),EE;
EE=(B+C)/2;
D(MP("$A$",A,W)--MP("$B$",B,N)--MP("$C$",C,E)--cycle);
D(MP("$E$",EE,N)--MP("$D$",D,S));
D(D);D(EE);
MA("80^\circ",8,D,EE,C,0.1);
MA("20^\circ",8,EE,C,D,0.3,2,shift(1,3)*C);
draw(arc(shift(-0.1,0.05)*C,0.25,100,180),arrow =ArcArrow());
MA("100^\circ",8,A,B,C,0.1,0);
MA("60^\circ",8,C,A,B,0.1,0);
//Credit to TheMaskedMagician for the diagram
[/asy]
In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. If the length of $AC$ is $1$ and $\measuredangle BAC = 60^\circ$, $\measuredangle ABC = 100^\circ$, $\measuredangle ACB = 20^\circ$ and $\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals
$\textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad\textbf{(B) }\frac{\sqrt{3}}{8}\qquad\textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad\textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad\textbf{(E) }\frac{1}{8}$
Let $m$ and $n$ be relatively prime positive integers. Determine all possible values of \[\gcd(2^m - 2^n, 2^{m^2+mn+n^2}- 1).\]
For $x, y$ positive integers, $x^2-4y+1$ is a multiple of $(x-2y)(1-2y)$. Prove that $|x-2y|$ is a square number.
Find all functions $f:\mathbb{N}\setminus\{1\} \rightarrow\mathbb{N}$ such that for all distinct $x,y\in \mathbb{N}$ with $y\ge 2018$, $$\gcd(f(x),y)\cdot \mathrm{lcm}(x,f(y))=f(x)f(y).$$
Prove that equation $ p^{4}\plus{}q^{4}\equal{}r^{4}$ does not have solution in set of prime numbers.
Positive integer $n$ is such that number $n^2-9$ has exactly $6$ positive divisors. Prove that GCD $(n-3, n+3)=1$
Let $P$ be a polynomial with integer coefficients such that $P(0)=0$ and
\[\gcd(P(0), P(1), P(2), \ldots ) = 1.\]
Show there are infinitely many $n$ such that
\[\gcd(P(n)- P(0), P(n+1)-P(1), P(n+2)-P(2), \ldots) = n.\]
A set $X$ of positive integers is said to be [i]iberic[/i] if $X$ is a subset of $\{2, 3, \dots, 2018\}$, and whenever $m, n$ are both in $X$, $\gcd(m, n)$ is also in $X$. An iberic set is said to be [i]olympic[/i] if it is not properly contained in any other iberic set. Find all olympic iberic sets that contain the number $33$.