Found problems: 85335
When rolling a certain unfair six-sided die with faces numbered $1, 2, 3, 4, 5$, and $6$, the probability of obtaining face $F$ is greater than $\frac{1}{6}$, the probability of obtaining the face opposite is less than $\frac{1}{6}$, the probability of obtaining any one of the other four faces is $\frac{1}{6}$, and the sum of the numbers on opposite faces is $7$. When two such dice are rolled, the probability of obtaining a sum of $7$ is $\frac{47}{288}$. Given that the probability of obtaining face $F$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.
Let $S=x_1x_2+x_3x_4+\cdots+x_{2015}x_{2016},$ where $x_1,x_2,\ldots,x_{2016}\in\{\sqrt{3}-\sqrt{2},\sqrt{3}+\sqrt{2}\}.$ Can $S$ be equal to $2016?$
[i]Cristian Lazăr[/i]
Let $S$ be the sum of all integers $b$ for which the polynomial $x^2+bx+2008b$ can be factored over the integers. Compute $|S|$.
Circle $ X$ has a radius of $ \pi$. Circle $ Y$ has a circumference of $ 8\pi$. Circle $ Z$ has an area of $ 9\pi$. List the circles in order from smallest to largest radius.
$ \textbf{(A)}\ X, Y, Z \qquad
\textbf{(B)}\ Z, X, Y \qquad
\textbf{(C)}\ Y, X, Z \qquad
\textbf{(D)}\ Z, Y, X \qquad
\textbf{(E)}\ X, Z, Y$
Prove that exprssion $A=\frac{25}{2}(n+2-\sqrt{2n+3})$, $(n\in\mathbb{N})$ is a perfect square of an integer if exprssion $A$ is an integer .
Peter and Basil together thought of ten quadratic trinomials. Then, Basil began calling consecutive natural numbers starting with some natural number. After each called number, Peter chose one of the ten polynomials at random and plugged in the called number. The results were recorded on the board. They eventually form a sequence. After they finished, their sequence was arithmetic. What is the greatest number of numbers that Basil could have called out?
Let $a$ and $b$ be positive real numbers satisfying
$$a -12b = 11 -\frac{100}{a} \,\,\,\,and \,\,\,\, a -\frac{12}{b}= 4 -\frac{100}{a}.$$
Then $a + b = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
How many ordered triples $(a, b, c)$, where $a$, $b$, and $c$ are from the set $\{ 1, 2, 3, \dots, 17 \}$, satisfy the equation
\[
a^3 + b^3 + c^3 + 2abc = a^2b + a^2c + b^2c + ab^2 + ac^2 + bc^2 \, ?
\]
Given a board of $3 \times 3$ you want to write the numbers $1, 2, 3, 4, 5, 6, 7, 8$ and a number in their boxes positive integer $M$, not necessarily different from the above. The goal is that the sum of the three numbers in each row be the same
$a)$ Find all the values of $M$ for which this is possible.
$b)$ For which of the values of $M$ found in $a)$ is it possible to arrange the numbers so that no only the three rows add the same but also the three columns add the same?
Let $BB_1$, $CC_1$ be the altitudes of triangle $ABC$, and $AD$ be the diameter of its circumcircle. The lines $BB_1$ and $DC_1$ meet at point $E$, the lines $CC_1$ and $DB_1$ meet at point $F$. Prove that $\angle CAE = \angle BAF$.
In a tennis tournament there are participants from $n$ different countries. Each team consists of a coach and a player whom should settle in a hotel. The rooms considered for the settlement of coaches are different from players' ones. Each player wants to be in a room whose roommates are [b][u]all[/u][/b] from countries which have a defense agreement with the player's country. Conversely, each coach wants to be in a room whose roommates are [b][u]all[/u][/b] from countries which don't have a defense agreement with the coach's country. Find the minimum number of the rooms such that we can [u][b]always[/b][/u] grant everyone's desire.
[i]proposed by Seyed Reza Hosseini and Mohammad Amin Ghiasi[/i]
Rectangle $ABCD$ is folded in half so that the vertices $D$ and $B$ coincide, creating the crease $\overline{EF}$, with $E$ on $\overline{AD}$ and $F$ on $\overline{BC}$. Let $O$ be the midpoint of $\overline{EF}$. If triangles $DOC$ and $DCF$ are congruent, what is the ratio $BC : CD$?
There is a square sheet of paper. How to get a rectangular sheet of paper with an aspect ratio equal to $\sqrt2$? (There are no tools, the sheet can only be bent.)
$n$ people (with names $1,2,\dots,n$) are around a table. Some of them are friends. At each step 2 friend can change their place. Find a necessary and sufficient condition for friendship relation between them that with these steps we can always reach to all of posiible permutations.
Circle $\Omega$ with center $O$ is the circumcircle of an acute triangle $\triangle ABC$ with $AB<BC$ and orthocenter $H$.
On the line $BO$ there is point $D$ such that $O$ is between $B$ and $D$ and $\angle ADC= \angle ABC$ . The semi-line starting at $H$ and parallel to $BO$ wich intersects segment $AC$ , intersects $\Omega$ at $E$. Prove that $BH=DE$.
Let $ P$ be a plane. Prove that there exists no function $ f: P\rightarrow P$ such that for every convex quadrilateral $ ABCD$, the points $ f(A),f(B),f(C),f(D)$ are the vertices of a concave quadrilateral.
[i]Dinu Şerbănescu[/i]
Find the greatest $n$ such that $(z+1)^n = z^n + 1$ has all its non-zero roots in the unitary circumference, e.g. $(\alpha+1)^n = \alpha^n + 1, \alpha \neq 0$ implies $|\alpha| = 1.$
In how many ways can the set $N ={1, 2, \cdots, n}$ be partitioned in the form $p(N) = A_{1}\cup A_{2}\cup \cdots \cup A_{k},$ where $A_{i}$ consists of elements forming arithmetic progressions, all with the same common positive difference $d_{p}$ and of length at least one? At least two?
[hide="Solution"]
[b]Part 1[/b]
Claim: There are $2^{n}-2$ ways of performing the desired partitioning.
Let $d(k)$ equal the number of ways $N$ can be partitioned as above with common difference $k.$ We are thus trying to evaluate
$\sum_{i=1}^{n-1}d(i)$
[b]Lemma: $d(i) = 2^{n-i}$[/b]
We may divide $N$ into various rows where the first term of each row denotes a residue $\bmod{i}.$ The only exception is the last row, as no row starts with $0$; the last row will start with $i.$ We display the rows as such:
$1, 1+i, 1+2i, 1+3i, \cdots$
$2, 2+i, 2+2i, 2+3i, \cdots$
$\cdots$
$i, 2i, 3i, \cdots$
Since all numbers have one lowest remainder $\bmod{i}$ and we have covered all possible remainders, all elements of $N$ occur exactly once in these rows.
Now, we can take $k$ line segments and partition a given row above; all entries within two segments would belong to the same set. For example, we can have:
$1| 1+i, 1+2i, 1+3i | 1+4i | 1+5i, 1+6i, 1+7i, 1+8i,$
which would result in the various subsets: ${1},{1+i, 1+2i, 1+3i},{1+4i},{1+5i, 1+6i, 1+7i, 1+8i}.$ For any given row with $k$ elements, we can have at most $k-1$ segments. Thus, we can arrange any number of segments where the number lies between $0$ and $k-1$, inclusive, in:
$\binom{k-1}{0}+\binom{k-1}{1}+\cdots+\binom{k-1}{k-1}= 2^{k-1}$
ways. Applying the same principle to the other rows and multiplying, we see that the result always gives us $2^{n-i},$ as desired.
We now proceed to the original proof.
Since $d(i) = 2^{n-i}$ by the above lemma, we have:
$\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}= 2^{n}-2$
Thus, there are $2^{n}-2$ ways of partitioning the set as desired.
[b]Part 2[/b]
Everything is the same as above, except the lemma slightly changes to $d(i) = 2^{n-i}-i.$ Evaluating the earlier sum gives us:
$\sum_{i=1}^{n-1}d(i) = \sum_{i=1}^{n-1}2^{n-i}-i = 2^{n}-\frac{n(n-1)}{2}-2$
[/hide]
Triangle $ABC$ has sidelengths $AB=14,AC=13,$ and $BC=15.$ Point $D$ is chosen in the interior of $\overline{AB}$ and point $E$ is selected uniformly at random from $\overline{AD}.$ Point $F$ is then defined to be the intersection point of the perpendicular to $\overline{AB}$ at $E$ and the union of segments $\overline{AC}$ and $\overline{BC}.$ Suppose that $D$ is chosen such that the expected value of the length of $\overline{EF}$ is maximized. Find $AD.$
Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(x)f(y)+f(x+y)=xy$ for all real numbers $x$ and $y$.
We call a positive integer [i] good[/i] if the sum of the reciprocals of all its natural divisors are integers. Prove that if
$ m $ is a [i]good [/i] number, and $ p> m $ is a prime number, then $ pm $ is not [i]good[/i].
Let $k$ be a positive integer. Find all positive integers $n$, such that it is possible to mark $n$ points on the sides of a triangle (different from its vertices) and connect some of them with a line in such a way that the following conditions are satisfied:
1) there is at least $1$ marked point on each side,
2) for each pair of points $X$ and $Y$ marked on different sides, on the third side there exist exactly $k$ marked points which are connected to both $X$ and $Y$ and exactly k points which are connected to neither $X$ nor $Y$
A binary string is a word containing only $0$s and $1$s. In a binary string, a $1-$run is a non extendable substring containing only $1$s. Given a positive integer $n$, let $B(n)$ be the number of $1-$runs in the binary representation of $n$. For example, $B(107)=3$ since $107$ in binary is $1101011$ which has exactly three $1-$runs. What is the following expression equal to? $$B(1)+B(2)+B(3)+ \dots + B(255)$$
Let $f\in C^3(\mathbb{R})$ nonnegative function with $f(0)=f'(0)=0, f''(0)>0$. Define $g(x)$ as follows:
\[ \{ \begin{array}{ccc}g(x)= (\frac{\sqrt{f(x)}}{f'(x)})' &\text{for}& x\not=0 \\ g(x)=0 &\text{for}& x=0\end{array} \]
(a) Show that $g$ is bounded in some neighbourhood of $0$.
(b) Is the above true for $f\in C^2(\mathbb{R})$?
Let $G$ be a graph on $n\geq 6$ vertices and every vertex is of degree at least 3. If $C_{1}, C_{2}, \dots, C_{k}$ are all the cycles in $G$, determine all possible values of $\gcd(|C_{1}|, |C_{2}|, \dots, |C_{k}|)$ where $|C|$ denotes the number of vertices in the cycle $C$.