Found problems: 85335
Zandrew Hao has $n^2$ dollars, where $n$ is an integer. He is a massive fan of the singer Pachary Zerry, and he wants to buy many copies of his $3$ albums, which cost $\$8$, $\$623$, and $\$835$ (two of them are very rare). Find the sum of the $3$ greatest values of $n$ such that Zandrew can't spend all of his money on albums.
Let $ x$ and $ y$ be positive integers such that $ 7x^5 \equal{} 11y^{13}$. The minimum possible value of $ x$ has a prime factorization $ a^cb^d$. What is $ a \plus{} b \plus{} c \plus{} d$?
$ \textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$
Determine the greatest positive integer $A{}$ with the following property: however we place the numbers $1,2,\ldots, 100$ on a $10\times 10$ board, each number appearing exactly once, we can find two numbers on the same row or column which differ by at least $A{}$.
Which of the following numbers has the largest prime factor?
$\text{(A)}\ 39 \qquad \text{(B)}\ 51 \qquad \text{(C)}\ 77 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 121$
Is it possible to divide a $6 \times 6$ chessboard into $18$ rectangles, each either $1 \times 2$ or $2 \times 1$, and to draw exactly one diagonal on each rectangle such that no two of these diagonals have a common endpoint?
(A Shapovalov)
Let $ n > 1, n \in \mathbb{Z}$ and $ B \equal{}\{1,2,\ldots, 2^n\}.$ A subset $ A$ of $ B$ is called weird if it contains exactly one of the distinct elements $ x,y \in B$ such that the sum of $ x$ and $ y$ is a power of two. How many weird subsets does $ B$ have?
Prove that if $a,b,c>0,$ then \[ \frac{ab}{3a+4b+5c}+\frac{bc}{3b+4c+5a}+\frac{ca}{3c+4a+5b}\le \frac{a+b+c}{12}. \]
[i] Nikolai Nikolov[/i]
The first number in the following sequence is $1$. It is followed by two $1$'s and two $2$'s. This is followed by three $1$'s, three $2$'s, and three $3$'s. The sequence continues in this fashion.
\[1,1,1,2,2,1,1,1,2,2,2,3,3,3,1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,\dots.\]
Find the $2014$th number in this sequence.
How many three-digit positive integers $x$ are there with the property that $x$ and $2x$ have only even digits? (One such number is $x = 220$, since $2x = 440$ and each of $x$ and $2x$ has only even digits.)
$ \mathrm{(A) \ } 16 \qquad \mathrm{(B) \ } 18 \qquad \mathrm {(C) \ } 64 \qquad \mathrm{(D) \ } 100 \qquad \mathrm{(E) \ } 125$
The unknown real numbers $x_1,x_2,\dots,x_n$ satisfy $x_1 < x_2 < \cdots < x_n,$ where $n \geq 3$. The numbers $s$, $t$ and $d_1,d_2,\dots,d_{n - 2}$ are given, such that \[ \begin{aligned} s & = \sum\limits_{i = 1}^nx_i, \\ t & = \sum\limits_{i = 1}^nx_i^2,\\ d_i & = x_{i + 2} - x_i,\ \ i = 1,2,\dots,n - 2. \end{aligned} \] For which $n$ is this information always sufficient to determine $x_1,x_2,\dots,x_n$ uniquely?
For how many ordered triples $ (a,b,c)$ of positive integers are the equations $ abc\plus{}9 \equal{} ab\plus{}bc\plus{}ca$ and $ a\plus{}b\plus{}c \equal{} 10$ satisfied?
Prove that for all odd $ n > 1$, we have $ 8n \plus{} 4|C^{4n}_{2n}$.
How many integers $n$ (with $1 \le n \le 2021$) have the property that $8n + 1$ is a perfect square?
The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers $ x, y,$ and $ r$ with $ |x| > |y|,$
\[ (x \plus{} y)^r \equal{} x^r \plus{} rx^{r \minus{} 1}y \plus{} \frac {r(r \minus{} 1)}2x^{r \minus{} 2}y^2 \plus{} \frac {r(r \minus{} 1)(r \minus{} 2)}{3!}x^{r \minus{} 3}y^3 \plus{} \cdots
\]
What are the first three digits to the right of the decimal point in the decimal representation of $ \left(10^{2002} \plus{} 1\right)^{10/7}?$
Given the positive integer $m \geq 2$, $n \geq 3$. Define the following set
$$S = \left\{(a, b) | a \in \{1, 2, \cdots, m\}, b \in \{1, 2, \cdots, n\} \right\}.$$
Let $A$ be a subset of $S$. If there does not exist positive integers $x_1, x_2, x_3, y_1, y_2, y_3$ such that $x_1 < x_2 < x_3, y_1 < y_2 < y_3$ and
$$(x_1, y_2), (x_2, y_1), (x_2, y_2), (x_2, y_3), (x_3, y_2) \in A.$$
Determine the largest possible number of elements in $A$.
Circles $\text{A}$ and $\text{B}$ are concentric, with the radius of $\text{A}$ being $\sqrt{17}$ times the radius of $B$. The largest line segment that can be draw in the region bounded by the two circles has length $32$. Compute the radius of circle $B$.
[center][img]https://cdn.artofproblemsolving.com/attachments/7/4/6bc4aed9842cdfbeb95853d508a22b61a10c9c.png[/img][/center]
Find all possible values of $$\frac{(a+b-c)^2}{(a-c)(b-c)}+\frac{(b+c-a)^2}{(b-a)(c-a)}+\frac{(c+a-b)^2}{(c-b)(a-b)}$$
Two players, the builder and the destroyer, plays the following game. Builder starts and players chooses alternatively different elements from the set $\{0,1,\ldots,10\}.$ Builder wins if some four integer of those six integer he chose forms an arithmetic sequence. Destroyer wins if he can prevent to form such an arithmetic four-tuple. Which one has a winning strategy?
Let $a,b,c$ be side lengths of a triangle with $a+b+c = 1$. Prove that $a^2 +b^2 +c^2 +4abc \le \frac12$ .
The Baltic Sea has $2016$ harbours. There are two-way ferry connections between some of them. It is impossible to make a sequence of direct voyages $C_1 - C_2 - ... - C_{1062}$ where all the harbours $C_1, . . . , C_{1062}$ are distinct. Prove that there exist two disjoint sets $A$ and $B$ of $477$ harbours each, such that there is no harbour in $A$ with a direct ferry connection to a harbour in $B.$
Let a,b,c be three distinct positive numbers.
Consider the quadratic polynomial $P (x) =\frac{c(x - a)(x - b)}{(c -a)(c -b)}+\frac{a(x - b)(x - c)}{(a - b)(a - c)}+\frac{b(x -c)(x - a)}{(b - c)(b - a)}+ 1$.
The value of $P (2017)$ is
(A): $2015$ (B): $2016$ (C): $2017$ (D): $2018$ (E): None of the above.
In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled.
[i]Proposed by Jerry Xu[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{21}$
Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows:
Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane.
[asy]
import geometry;
size(8cm);
pair A = (0,sqrt(3));
pair B = (-1,0);
pair C = (1,0);
pair M = (0,0);
for (int i = -1; i <= 2; ++i) {
draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3)));
draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3)));
draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3)));
}
draw(A--B--C--A, red);
dot(M);
label("$A$",A+(0,0.25),N);
label("$B$",B-(0.25,0),SW);
label("$C$",C+(0.25,0),SE);
label("$M$",M,S);
[/asy]
It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases:
$1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that
$$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that
$$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case.
By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution
$$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$).
We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution).
From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are
\begin{align*}
(x,y)&=(0,1),(4,7),(56,97) \\
\longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97)
\end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex).
For $k=3$, using the multiplicative principle we get two new smaller solutions
\begin{align*}
(x,y)&=(0,3),(12,21) \\
\longrightarrow (n,m,l)&=(0,1,3),(12,1,21)
\end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$.
For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$.
Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$.
$2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that
$$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that
$$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions.
Our final answer is thus $1+7+13=\boxed{21}$.
[/hide]
Let $\Gamma$ be a circle and let $d$ be a line such that $\Gamma$ and $d$ have no common points. Further, let $AB$ be a diameter of the circle $\Gamma$; assume that this diameter $AB$ is perpendicular to the line $d$, and the point $B$ is nearer to the line $d$ than the point $A$. Let $C$ be an arbitrary point on the circle $\Gamma$, different from the points $A$ and $B$. Let $D$ be the point of intersection of the lines $AC$ and $d$. One of the two tangents from the point $D$ to the circle $\Gamma$ touches this circle $\Gamma$ at a point $E$; hereby, we assume that the points $B$ and $E$ lie in the same halfplane with respect to the line $AC$. Denote by $F$ the point of intersection of the lines $BE$ and $d$. Let the line $AF$ intersect the circle $\Gamma$ at a point $G$, different from $A$.
Prove that the reflection of the point $G$ in the line $AB$ lies on the line $CF$.
Let $ABCDEF$ be a convex hexagon, and let $P= AB \cap CD$, $Q = CD \cap EF$, $R = EF \cap AB$, $S = BC \cap DE$, $T = DE \cap FA$, $U = FA \cap BC$. Prove that
$\frac{PQ}{CD} = \frac{QR}{EF} = \frac{RP}{AB}$ if and only if $\frac{ST}{DE} = \frac{TU}{FA} = \frac{US}{BC}$
Can ten distinct numbers $a_1, a_2, b_1, b_2, b_3, c_1, c_2, d_1, d_2, d_3$ be chosen from $\{0, 1, 2, \ldots, 14\}$, so that the $14$ differences $|a_1 - b_1|$, $|a_1 - b_2|$, $|a_1 - b_3|$, $|a_2 - b_1|$, $|a_2 - b_2|$, $|a_2 - b_3|$, $|c_1 - d_1|$, $|c_1 - d_2|$, $|c_1 - d_3|$, $|c_2 - d_1|$, $|c_2 - d_2|$, $|c_2 - d_3|$, $|a_1 - c_1|$, and $|a_2 - c_2|$ are all distinct?