Physicists at Princeton are trying to analyze atom entanglement using the following experiment. Originally there is one atom in the space and it starts splitting according to the following procedure. If after $n$ minutes there are atoms $a_1, \dots, a_N$, in the following minute every atom $a_i$ splits into four new atoms, $a_i^{(1)},a_i^{(2)},a_i^{(3)},a_i^{(4)}$. Atoms $a_i^{(j)}$ and $a_k^{(j)}$ are entangled if and only the atoms $a_i$ and $a_k$ were entangled after $n$ minutes. Moreover, atoms $a_i^{(j)}$ and $a_k^{(j+1)}$ are entangled for all $1 \le i$, $k \le N$ and $j = 1$, $2$, $3$. Therefore, after one minute there is $4$ atoms, after two minutes there are $16$ atoms and so on. Physicists are now interested in the number of unordered quadruplets of atoms $\{b_1, b_2, b_3, b_4\}$ among which there is an odd number of entanglements. What is the number of such quadruplets after $3$ minutes? [i]Remark[/i]. Note that atom entanglement is not transitive. In other words, if atoms $a_i$, $a_j$ are entangled and if $a_j$, $a_k$ are entangled, this does not necessarily mean that $a_i$ and $a_k$ are entangled.

For a finite non empty set of primes $P$, let $m(P)$ denote the largest possible number of consecutive positive integers, each of which is divisible by at least one member of $P$. (i) Show that $|P|\le m(P)$, with equality if and only if $\min(P)>|P|$. (ii) Show that $m(P)<(|P|+1)(2^{|P|}-1)$. (The number $|P|$ is the size of set $P$) [i]Dan Schwarz, Romania[/i]

Point $B$ is on $\overline{AC}$ with $AB = 9$ and $BC = 21$. Point $D$ is not on $\overline{AC}$ so that $AD = CD$, and $AD$ and $BD$ are integers. Let $s$ be the sum of all possible perimeters of $\triangle ACD$. Find $s$.

Let $F(x_1,..., x_n)$ be a polynomial with real coefficients in $ n > 1$ “indeterminate” variables $x_1,..., x_n$. We say that $F$ is $n$-[i]alternating [/i]if for all integers $1 \le i < j \le n$, $$F(x_1,..., x_i,..., x_j,..., x_n) = - F(x_1,..., x_j,..., x_i,..., x_n),$$ i.e. swapping the order of indeterminates $x_i, x_j$ flips the sign of the polynomial. For example, $x^2_1x_2 - x^2_2x_1$ is $2$-alternating, whereas $x_1x_2x_3 +2x_2x_3$ is not $3$-alternating. [i]Note: two polynomials $P(x_1,..., x_n)$ and $Q(x_1,..., x_n)$ are considered equal if and only if each monomial constituent $ax^{k_1}_1... x^{k_n}_n$ of $P$ appears in $Q$ with the same coefficient $a$, and vice versa. This is equivalent to saying that $P(x_1,..., x_n) = 0$ if and only if every possible monomial constituent of $P$ has coefficient $0$. [/i] (1) Compute a $3$-alternating polynomial of degree $3$. (2) Prove that the degree of any nonzero $n$-alternating polynomial is at least ${n \choose 2}$.

Determine the smallest positive real number $ k$ with the following property. Let $ ABCD$ be a convex quadrilateral, and let points $ A_1$, $ B_1$, $ C_1$, and $ D_1$ lie on sides $ AB$, $ BC$, $ CD$, and $ DA$, respectively. Consider the areas of triangles $ AA_1D_1$, $ BB_1A_1$, $ CC_1B_1$ and $ DD_1C_1$; let $ S$ be the sum of the two smallest ones, and let $ S_1$ be the area of quadrilateral $ A_1B_1C_1D_1$. Then we always have $ kS_1\ge S$. [i]Author: Zuming Feng and Oleg Golberg, USA[/i]

For odd positive integers $n$, define $f(n)$ to be the smallest odd integer greater than $n$ that is not relatively prime to $n$. Compute the smallest $n$ such that $f(f(n))$ is not divisible by $3$.

Evaluate $(2-\sec^2{1^\circ})(2-\sec^2{2^\circ})(2-\sec^2{3^\circ})\cdots(2-\sec^2{89^\circ}).$

The triangle table is constructed according to the rule: You put the natural number $a>1$ in the upper row, and then you write under the number $k$ from the left side $k^2$, and from the right side -- $(k+1)$. For example, if $a = 2$, you get the table on the picture. Prove that all the numbers on each particular line are different. 2 / \ / \ 4 3 / \ / \ 16 5 9 4 / \ / \ /\ / \

There are $n\ge 2$ line segments in the plane such that every two segments cross and no three segments meet at a point. Geoff has to choose an endpoint of each segment and place a frog on it facing the other endpoint. Then he will clap his hands $n-1$ times. Every time he claps,each frog will immediately jump forward to the next intersection point on its segment. Frogs never change the direction of their jumps. Geoff wishes to place the frogs in such a way that no two of them will ever occupy the same intersection point at the same time. (a) Prove that Geoff can always fulfill his wish if $n$ is odd. (b) Prove that Geoff can never fulfill his wish if $n$ is even.

Let $M,N$ be the midpoints of the sides $AD,BC$ respectively of the convex quadrilateral $ABCD$, $K=AN \cap BM$, $L=CM \cap DN$. Find the smallest possible $c\in R$ such that $S(MKNL)<c \cdot S(ABCD)$ for any convex quadrilateral $ABCD$. I. Voronovich

If $a$, $b$ and $c$ are sides of triangle which perimeter equals $1$, prove that: $a^2+b^2+c^2+4abc<\frac{1}{2}$

If $n$ is a positive integer, let $\phi(n)$ be the number of positive integers less than or equal to $n$ that are relatively prime to $n$. Compute the value of the infinite sum \[ \sum_{n=1}^\infty \frac{\phi(n) 2^n}{9^n - 2^n} \, . \]

Let $k$ be a positive integer. Compute $$\sum_{n_1=1}^\infty\sum_{n_2=1}^\infty\cdots\sum_{n_k=1}^\infty\frac1{n_1n_2\cdots n_k(n_1+n_2+\ldots+n_k+1)}.$$

Find the number of positive integers $x$ less than $100$ for which $$3^x + 5^x + 7^x + 11^x + 13^x + 17^x + 19^x$$ is prime.

Let $d \geq 3$ be a positive integer. The binary strings of length $d$ are splitted into $2^{d-1}$ pairs, such that the strings in each pair differ in exactly one position. Show that there exists an $\textit{alternating cycle}$ of length at most $2d-2$, i.e. at most $2d-2$ binary strings that can be arranged on a circle so that any pair of adjacent strings differ in exactly one position and exactly half of the pairs of adjacent strings are pairs in the split.

For $x>0,$ show that $e^x < (1+x)^{1+x}.$