Found problems: 721
Find all prime numbers $p_1,…,p_n$ (not necessarily different) such that :
$$ \prod_{i=1}^n p_i=10 \sum_{i=1}^n p_i$$
Determine the least possible value of $f(1998),$ where $f:\Bbb{N}\to \Bbb{N}$ is a function such that for all $m,n\in {\Bbb N}$,
\[f\left( n^{2}f(m)\right) =m\left( f(n)\right) ^{2}. \]
Let $p$ be a prime number and the decimal notation of $\frac{1}{p}$ is periodical with a length of the period $4k$, $\frac{1}{p}=0,a_1 a_2…a_{4k} a_1 a_2…a_{4k}…$ .Prove that
$a_1+a_3+...+a_{4k-1}=a_2+a_4+...+a_{4k}$.
Take any 26 distinct numbers from {1, 2, ... , 100}. Show that there must be a non-empty subset of the $ 26$ whose product is a square.
[hide]
I think that the upper limit for such subset is 37.[/hide]
Suppose $a_1,a_2, \dots$ is an infinite strictly increasing sequence of positive integers and $p_1, p_2, \dots$ is a sequence of distinct primes such that $p_n \mid a_n$ for all $n \ge 1$. It turned out that $a_n-a_k=p_n-p_k$ for all $n,k \ge 1$. Prove that the sequence $(a_n)_n$ consists only of prime numbers.
Prove that $2005^2$ can be written in at least $4$ ways as the sum of 2 perfect (non-zero) squares.
Starting from 37, adding 5 before each previous term, forms the following sequence:
\[37,537,5537,55537,555537,...\]
How many prime numbers are there in this sequence?
Let $n\ge2$ be an integer. Prove that if $k^2+k+n$ is prime for all integers $k$ such that $0\le k\le\sqrt{n\over3}$, then $k^2+k+n$ is prime for all integers $k$ such that $0\le k\le n-2$.
The sequence $(a_n)_{n\in\mathbb{N}}$ is defined by $a_1=3$ and $$a_n=a_1a_2\cdots a_{n-1}-1$$ Show that there exist infinitely many prime number that divide at least one number in this sequences
Find all prime numbers $a,b,c$ and positive integers $k$ satisfying the equation \[a^2+b^2+16c^2 = 9k^2 + 1.\]
Proposed by Moldova
A positive integer is called [i]superprime[/i] if the difference between any two of its consecutive positive divisors is a prime number. Determine all superprime integers.
[i]Authored by Nikola Velov[/i]
is there a function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that
$i) \exists n\in \mathbb{N}:f(n)\neq n$
$ii)$ the number of divisors of $m$ is $f(n)$ if and only if the number of divisors of $f(m)$ is $n$
Find all pairs $(n, p)$ that satisfy the following condition, where $n$ is a positive integer and $p$ is a prime number.
[b]Condition)[/b] $2n-1$ is a divisor of $p-1$ and $p$ is a divisor of $4n^2+7$.
Two coprime positive integers $ a, b $ are given. Integer sequence $ \{ a_n \}, \{b_n \} $ satisties
\[ (a+b \sqrt2 )^{2n} = a_n + b_n \sqrt2 \]
Find all prime numbers $ p $ such that there exist positive integer $ n \le p $ satisfying $ p | b_n $.
Let $ p_1,\dots,p_k$ be prime numbers, and let $ S$ be the set of those integers whose all prime divisors are among $ p_1,\dots,p_k$. For a finite subset $ A$ of the integers let us denote by $ \mathcal G(A)$ the graph whose vertices are the elements of $ A$, and the edges are those pairs $ a,b\in A$ for which $ a \minus{} b\in S$. Does there exist for all $ m\geq 3$ an $ m$-element subset $ A$ of the integers such that
(i) $ \mathcal G(A)$ is complete?
(ii) $ \mathcal G(A)$ is connected, but all vertices have degree at most 2?
Find all the triples of positive integers $(a,b,c)$ for which the number
\[\frac{(a+b)^4}{c}+\frac{(b+c)^4}{a}+\frac{(c+a)^4}{b}\]
is an integer and $a+b+c$ is a prime.
Let $\mathbb{S}$ is the set of prime numbers that less or equal to 26. Is there any $a_1, a_2, a_3, a_4, a_5, a_6 \in \mathbb{N}$ such that $$ gcd(a_i,a_j) \in \mathbb{S} \qquad \text {for } 1\leq i \ne j \leq 6$$ and for every element $p$ of $\mathbb{S}$ there exists a pair of $ 1\leq k \ne l \leq 6$ such that $$s=gcd(a_k,a_l)?$$
Let $p \equiv 3 \,(\textrm{mod}\, 4)$ be a prime and $\theta$ some angle such that $\tan(\theta)$ is rational. Prove that $\tan((p+1)\theta)$ is a rational number with numerator divisible by $p$, that is, $\tan((p+1)\theta) = \frac{u}{v}$ with $u, v \in \mathbb{Z}, v >0, \textrm{mdc}(u, v) = 1$ and $u \equiv 0 \,(\textrm{mod}\,p) $.
Given a rational number, write it as a fraction in lowest terms and calculate the product of the resulting numerator and denominator. For how many rational numbers between 0 and 1 will $ 20!$ be the resulting product?
Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?
[asy]path card=((0,0)--(0,3)--(2,3)--(2,0)--cycle);
draw(card, linewidth(1));
draw(shift(2.5,0)*card, linewidth(1));
draw(shift(5,0)*card, linewidth(1));
label("$44$", (1,1.5));
label("$59$", shift(2.5,0)*(1,1.5));
label("$38$", shift(5,0)*(1,1.5));[/asy]
$ \textbf{(A)}\ 13 \qquad
\textbf{(B)}\ 14 \qquad
\textbf{(C)}\ 15 \qquad
\textbf{(D)}\ 16 \qquad
\textbf{(E)}\ 17$
Let $n+1, n \geq 1$ positive integers be formed by taking the product of $n$ given prime numbers (a prime number can appear several times or also not appear at all in a product formed in this way.) Prove that among these $n+1$ one can find some numbers whose product is a perfect square.
Prove that these polynomials are irreducible in $\mathbb{Q}[x]:$
a) $\frac{{{x^p}}}{{p!}} + \frac{{{x^{p - 1}}}}{{(p - 1)!}} + ... + \frac{{{x^2}}}{2} + x + 1,$ with $p$ is a prime number.
b) $x^{2^n}+1,$ with $n$ is a positive integer.
In the diagram below, fill the $12$ circles with numbers from the following bank so that each number is used once. Two circles connected by a single line must contain relatively prime numbers. Two circles connected by a double line must contain numbers that are not relatively prime.
$$\text{Bank: } 20, 21, 22, 23, 24, 25, 27, 28, 30 ,32, 33 ,35$$
[asy]
real HRT3 = sqrt(3) / 2;
void drawCircle(real x, real y, real r) {
path p = circle((x,y), r);
draw(p);
fill(p, white);
}
void drawCell(int gx, int gy) {
real x = 0.5 * gx;
real y = HRT3 * gy;
drawCircle(x, y, 0.35);
}
void drawEdge(int gx1, int gy1, int gx2, int gy2, bool doubled) {
real x1 = 0.5 * gx1;
real y1 = HRT3 * gy1;
real x2 = 0.5 * gx2;
real y2 = HRT3 * gy2;
if (doubled) {
real dx = x2 - x1;
real dy = y2 - y1;
real ox = -0.035 * dy / sqrt(dx * dx + dy * dy);
real oy = 0.035 * dx / sqrt(dx * dx + dy * dy);
draw((x1+ox,y1+oy)--(x2+ox,y2+oy));
draw((x1-ox,y1-oy)--(x2-ox,y2-oy));
} else {
draw((x1,y1)--(x2,y2));
}
}
drawEdge(2, 0, 4, 0, true);
drawEdge(2, 0, 1, 1, true);
drawEdge(2, 0, 3, 1, true);
drawEdge(4, 0, 3, 1, false);
drawEdge(4, 0, 5, 1, false);
drawEdge(1, 1, 0, 2, false);
drawEdge(1, 1, 2, 2, false);
drawEdge(1, 1, 3, 1, false);
drawEdge(3, 1, 2, 2, true);
drawEdge(3, 1, 4, 2, true);
drawEdge(3, 1, 5, 1, false);
drawEdge(5, 1, 4, 2, true);
drawEdge(5, 1, 6, 2, false);
drawEdge(0, 2, 1, 3, false);
drawEdge(0, 2, 2, 2, false);
drawEdge(2, 2, 1, 3, false);
drawEdge(2, 2, 3, 3, true);
drawEdge(2, 2, 4, 2, false);
drawEdge(4, 2, 3, 3, false);
drawEdge(4, 2, 5, 3, false);
drawEdge(4, 2, 6, 2, false);
drawEdge(6, 2, 5, 3, true);
drawEdge(1, 3, 3, 3, true);
drawEdge(3, 3, 5, 3, false);
drawCell(2, 0);
drawCell(4, 0);
drawCell(1, 1);
drawCell(3, 1);
drawCell(5, 1);
drawCell(0, 2);
drawCell(2, 2);
drawCell(4, 2);
drawCell(6, 2);
drawCell(1, 3);
drawCell(3, 3);
drawCell(5, 3);
[/asy]
Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them $S$- and $Z$-tetrominoes, respectively.
Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove that no matter how we tile $P$ using only $S$- and $Z$-tetrominoes, we always use an even number of $Z$-tetrominoes.
[i]Proposed by Tamas Fleiner and Peter Pal Pach, Hungary[/i]
Let $f:\mathbb N\rightarrow\mathbb N$ be a non-decreasing function and let $n$ be an arbitrary natural number. Suppose that there are prime numbers $p_1,p_2,\dots,p_n$ and natural numbers $s_1,s_2,\dots,s_n$ such that for each $1\leq i\leq n$ the set $\{f(p_ir+s_i)|r=1,2,\dots\}$ is an infinite arithmetic progression. Prove that there is a natural number $a$ such that
\[f(a+1), f(a+2), \dots, f(a+n)\]
form an arithmetic progression.