Six distinct positive integers $a,b,c.d,e, f$ are given. Jack and Jill calculated the sums of each pair of these numbers. Jack claims that he has $10$ prime numbers while Jill claims that she has $9$ prime numbers among the sums. Who has the correct claim?

In Wonderland, the government of each country consists of exactly $a$ men and $b$ women, where $a$ and $b$ are fixed natural numbers and $b > 1$. For improving of relationships between countries, all possible working groups consisting of exactly one government member from each country, at least $n$ among whom are women, are formed (where $n$ is a fixed non-negative integer). The same person may belong to many working groups. Find all possibilities how many countries can be in Wonderland, given that the number of all working groups is prime.

The natural numbers $a_1,a_2,...,a_n, n \ge 12$, are smaller than $9n^2$ and pairwise coprime. Show that at least one of these numbers is prime.

Consider a prime number $p\geqslant 11$. We call a triple $a,b,c$ of natural numbers [i]suitable[/i] if they give non-zero, pairwise distinct residues modulo $p{}$. Further, for any natural numbers $a,b,c,k$ we define \[f_k(a,b,c)=a(b-c)^{p-k}+b(c-a)^{p-k}+c(a-b)^{p-k}.\]Prove that there exist suitable $a,b,c$ for which $p\mid f_2(a,b,c)$. Furthermore, for each such triple, prove that there exists $k\geqslant 3$ for which $p\nmid f_k(a,b,c)$ and determine the minimal $k{}$ with this property. [i]Călin Popescu and Marian Andronache[/i]

For which $ p$ prime numbers, there is an integer root of the polynominal $ 1 \plus{} p \plus{} Q(x^1)\cdot\ Q(x^2)\ldots\ Q(x^{2p \minus{} 2})$ such that $ Q(x)$ is a polynominal with integer coefficients?

Let $p_1<p_2<p_3<p_4$ and $q_1<q_2<q_3<q_4$ be two sets of prime numbers, such that $p_4 - p_1 = 8$ and $q_4 - q_1= 8$. Suppose $p_1 > 5$ and $q_1>5$. Prove that $30$ divides $p_1 - q_1$.

We write $\{a,b,c\}$ for the set of three different positive integers $a, b$, and $c$. By choosing some or all of the numbers a, b and c, we can form seven nonempty subsets of $\{a,b,c\}$. We can then calculate the sum of the elements of each subset. For example, for the set $\{4,7,42\}$ we will find sums of $4, 7, 42,11, 46, 49$, and $53$ for its seven subsets. Since $7, 11$, and $53$ are prime, the set $\{4,7,42\}$ has exactly three subsets whose sums are prime. (Recall that prime numbers are numbers with exactly two different factors, $1$ and themselves. In particular, the number $1$ is not prime.) What is the largest possible number of subsets with prime sums that a set of three different positive integers can have? Give an example of a set $\{a,b,c\}$ that has that number of subsets with prime sums, and explain why no other three-element set could have more.

For every integer $n\geq 3$, let $s_n$ be the sum of all primes (strictly) less than $n$. Are there infinitely many integers $n\geq 3$ such that $s_n$ is coprime to $n$? [i]Russian Competition[/i]

Find all natural $ x $ for which $ 3x+1 $ and $ 6x-2 $ are perfect squares, and the number $ 6x^2-1 $ is prime.

There are 168 primes below 1000. Then sum of all primes below 1000 is [list=1] [*] 11555 [*] 76127 [*] 57298 [*] 81722 [/list]

For two different prime numbers $p, q$, define $S_{p,q} = \{p,q,pq\}$. If two elements in $S_{p,q}$ are numbers in the form of $x^2 + 2005y^2, (x, y \in Z)$, prove that all three elements in $S_{p,q}$ are in such form.

Let $p\geq 3$ a prime number, $a$ and $b$ integers such that $\gcd(a, b)=1$. Let $n$ a natural number such that $p$ divides $a^{2^n}+b^{2^n}$, prove that $2^{n+1}$ divides $p-1$.

Let $m,n$ be positive integers and $p$ be a prime number. Show that if $\frac{7^m + p \cdot 2^n}{7^m - p \cdot 2^n}$ is an integer, then it is a prime number.

Find the greatest possible value of $pq + r$, where p, q, and r are (not necessarily distinct) prime numbers satisfying $pq + qr + rp = 2016$.

Let $p$ be a prime number. Find all polynomials $P$ with integer coefficients with the following properties: $(a)$ $P(x)>x$ for all positive integers $x$. $(b)$ The sequence defined by $p_0:=p$, $p_{n+1}:=P(p_n)$ for all positive integers $n$, satisfies the property that for all positive integers $m$ there exists some $l\geq 0$ such that $m\mid p_l$.