Let $n \ge 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid, and Colin is allowed to permute the columns of the grid. A grid coloring is $orderly$ if:
no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and
no matter how Colin permutes the column of the coloring, Rowan can then permute the rows to restore the original grid coloring;
In terms of $n$ , how many orderly colorings are there?

Solution:
We focus on the leftmost column for simplicity. Let $m$ be the number of red squares in this column. We then have five cases:
 1. $m=1$
When Rowan permutes the rows of the coloring, we consider only the first column, which by the above contains $m=1$ red colors, so there are ${n \choose 1}=n$ ways to permute the first column’s rows. Thus every other column will have to contain one different permutation of the first column; otherwise, there will be at least one permutation of which there is no corresponding column.
 Furthermore, each permutation will be different, so each row will contain one and only one red square, which also fulfills the case of if Colin permutes the coloring first. Thus there are $n\cdot (n-1)\cdot(n-2)\cdot\cdot\cdot2\cdot1=n!$ different colorings for this case (the same as choosing squares such as no square is in the same row or column as any other square).

2. $m=n-1$
This is essentially the same as case 1 except for the coloring; now there is one blue square and the rest are red squares. Thus there are also $n!$ different colorings for this case.

3. $m=0$
Since we have an entirely blue column, we are unable to have a column with $1$ red square only as doing so would leave one permutation that is not covered by at least one column (that space is being taken for the blank column). We are also unable to have a completely blue column as doing so would allow for Colin to shift the columns and in doing so fail for Rowan to shift back the columns. We also cannot have a column with any other number of red squares other than $0$ as will be shown below, so there is $1$ case here in which the entire coloring is red.

4. $m=n$
This is the same is an entire blue column, and, similar to above, we have $1$ coloring.

5. $1<m<n-1$
This is the final case and is equivalent to permuting for ${n \choose m}$ different ways. We must prove that this is greater than $n$ to show that the columns are not able to contain every possible permutation of this column for all values of $n$ such that $n>3$ (when $n=3$ , there is no such positive integer $m$ that satisfies the conditions). Note that if we have any column with a different number of red squares, it is an unattainable column and is thus not optimal.
 Lemma: Given that $m$ and $n$ are positive integers such that $1<m<n-1$ and $n>3$ , it is true for all $m$ and $n$ that ${n \choose m}>n$ .
Proof: Assume that $m<\frac{n-1}{2}$ .
$\Leftrightarrow$  $m+1<n-m$
$\Leftrightarrow$  $(m+1)!(n-m-1)!<m!(n-m)!$
$\Leftrightarrow$  $\frac{n!}{m!(n-m)!}<\frac{n!}{(m+1)!(n-m-1)!}$
$\Leftrightarrow$  ${n \choose m}<{n \choose m+1}$
Similarly, we can prove that ${n \choose m}>{n \choose m+1}$ for $m>\frac{n-1}{2}$ .
Now we split our proof into two cases.
Case 1: $n$ is even.
The largest integer less than $\frac{n-1}{2}$ is $\frac{n}{2}-1$ , so we know that:
${n \choose \frac{n}{2}}>{n \choose \frac{n}{2}-1}>\cdot\cdot\cdot>{n \choose 2}$
by induction. On the other hand, the smallest integer greater than $\frac{n-1}{2}$ is $\frac{n}{2}$ , so we know that:
${n \choose \frac{n}{2}}>{n \choose \frac{n}{2}+1}>\cdot\cdot\cdot>{n \choose n-2}$
also by induction. Thus out of the given range for $m$ we know that ${n \choose 2}$ and ${n \choose n-2}$ are the minimum values, and all that is left is to prove that they are both greater than $n$ . Furthermore, since ${n \choose 2}={n \choose n-2}$ , we only have to prove that ${n \choose 2}>n$ .
We start with the given: $n>3$
$\Leftrightarrow$  $\frac{n-1}{2}>1$
$\Leftrightarrow$  $\frac{n(n-1)}{2}>n$
$\Leftrightarrow$  $\frac{n!}{2!(n-2)!}>n$
$\Leftrightarrow$  ${n \choose 2}>n$
Thus we have proven the inequality for all even $n$ .
Case 2: $n$ is odd.
The greatest integer less than $\frac{n-1}{2}$ is $\frac{n-3}{2}$ , so we know that:
${n \choose \frac{n-1}{2}}>{n \choose \frac{n-3}{2}}>\cdot\cdot\cdot>{n \choose 2}$
by induction. On the other hand, the smallest integer greater than $\frac{n-1}{2}$ is $\frac{n+1}{2}$ , so we know that:
${n \choose \frac{n+1}{2}}>{n \choose \frac{n+3}{2}}>\cdot\cdot\cdot>{n \choose n-2}$
also by induction. Since ${n \choose \frac{n+1}{2}}={n \choose \frac{n-1}{2}}$ , we know that once again, ${n \choose n-2}={n \choose 2}$ is the minimum of the given range for $m$ , and the same proof applies. Thus, the inequality holds true for odd and in turn all positive integers $n>3$ .
 As a result, due to our lemma, there are always more permutations of the columns than the number of columns itself, so there will always exist a permutation of the column such that there are no corresponding original columns of which to match with. Thus there are no solutions for this case.
 In conclusion, there are a total of $2\cdot n!+2$ different colorings for which the above apply.
 ~eevee9406

Answer: \[ 2 \cdot n! + 2 \]


Mathematics -> Geometry -> Plane Geometry -> Triangulations

Let $ABC$ be a fixed acute triangle inscribed in a circle $\omega$ with center $O$ . A variable point $X$ is chosen on minor arc $AB$ of $\omega$ , and segments $CX$ and $AB$ meet at $D$ . Denote by $O_1$ and $O_2$ the circumcenters of triangles $ADX$ and $BDX$ , respectively. Determine all points $X$ for which the area of triangle $OO_1O_2$ is minimized.

Solution:
Let $E$ be midpoint $AD.$ Let $F$ be midpoint $BD \implies$  \[EF = ED + FD = \frac {AD}{2} + \frac {BD}{2} = \frac {AB}{2}.\]  $E$ and $F$ are the bases of perpendiculars dropped from $O_1$ and $O_2,$ respectively.
Therefore $O_1O_2 \ge EF =  \frac {AB}{2}.$
\[CX \perp  O_1O_2, AX \perp  O_1O \implies \angle O O_1O_2 = \angle AXC\]  $\angle AXC = \angle ABC (AXBC$ is cyclic) $\implies \angle O O_1O_2 = \angle ABC.$
Similarly $\angle BAC = \angle O O_2 O_1 \implies  \triangle ABC \sim \triangle O_2 O_1O.$
The area of $\triangle OO_1O_2$ is minimized if $CX \perp AB$ because \[\frac {[OO_1O_2]} {[ABC]} = \left(\frac {O_1 O_2} {AB}\right)^2 \ge \left(\frac {EF} {AB}\right)^2 = \frac {1}{4}.\]  vladimir.shelomovskii@gmail.com, vvsss

Answer: The area of triangle $OO_1O_2$ is minimized if $CX \perp AB$.


Mathematics -> Algebra -> Algebra -> Equations and Inequalities

Mathematics -> Number Theory -> Prime Numbers

Find all prime numbers $p,q,r$ , such that $\frac{p}{q}-\frac{4}{r+1}=1$

Solution:
The given equation can be rearranged into the below form:
$4q = (p-q)(r+1)$

$Case 1: 4|(p-q)$
then we have
$q = ((p-q)/4)(r+1)$  $=> (p-q)/4 = 1$ and $q = r + 1$  $=> r = 2, q = 3$ and $p = 7$

$Case 2: 4|(r+1)$
then we have
$q = (p-q)((r+1)/4)$  $=> (p-q) = 1$ and $q = (r + 1)/4$  $=> p = q + 1 => q = 2, p = 3$ and $r = 7$
note that if $(r+1)/4 = 1$ , then $q = (p-q) => p = 2q$ which is a contradiction.

$Case 3: 2|(p-q)$ and $2|(r+1)$
then we have
$q = ((p-q)/2)((r+1)/2)$  $=> (p-q)/2 = 1$ and $q = (r+1)/2$  $=> p = q + 2$ and $r = 2q - 1$ We have that exactly one of $q, q + 1, q + 2$ is a multiple of $3$ .
  $q + 1$ cannot be a multiple of $3$ since $q + 1 = 3k => q = 3k - 1$ . Since $r = 2q - 1$ is prime, then we have $2(3k - 1) - 1 = 3(2k-1)$ is a prime. $=> k = 1 => q = 2 => p = 4$ contradiction.
 Also, $q + 2$ cannot be a multiple of $3$ since, $q + 2 = 3k => p = 3k => k = 1 => q = 1$ contradiction.
 So, $q = 3k$  $=> k = 1 => q = 3, p = 5$ and $r = 5$
Thus we have the following solutions: $(7, 3, 2), (3, 2, 7), (5, 3, 5)$
  $Kris17$

Answer:
\[
(7, 3, 2), (3, 2, 7), (5, 3, 5)
\]


Mathematics -> Discrete Mathematics -> Combinatorics

Mathematics -> Geometry -> Plane Geometry -> Polygons

( Reid Barton ) An animal with $n$  cells is a connected figure consisting of $n$ equal-sized square cells. ${}^1$ The figure below shows an 8-cell animal.

A dinosaur is an animal with at least 2007 cells.  It is said to be primitive if its cells cannot be partitioned into two or more dinosaurs.  Find with proof the maximum number of cells in a primitive dinosaur.
Animals are also called polyominoes .  They can be defined inductively .  Two cells are adjacent if they share a complete edge .  A single cell is an animal, and given an animal with cells, one with cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.

Solution:
Solution 1
Let a $n$ -dino denote an animal with $n$ or more cells.
We show by induction that an $n$ -dino with $4n-2$ or more animal cells is not primitive. (Note: if it had more, we could just take off enough until it had $4n-2$ , which would have a partition, and then add the cells back on.)
Base Case: If $n=1$ , we have two cells, which are clearly not primitive.
Inductive Step: Assume any $4n-2$ cell animal can be partitioned into two or more $n$ -dinos.
For a given $(4n+2)$ -dino, take off any four cells (call them $w,\ x,\ y,\ z$ ) to get an animal with $4n-2$ cells.
This can be partitioned into two or more $n$ -dinos, let's call them $A$ and $B$ . This means that $A$ and $B$ are connected.
If both $A$ and $B$ are $(n+1)$ -dinos or if $w,\ x,\ y,\ z$ don't all attach to one of them, then we're done.
So assume $A$ has $n$ cells and thus $B$ has at least $3n-2$ cells, and that $w,\ x,\ y,\ z$ are added to $B$ . So $B$ has $3n+2$ cells total.
Let $C$ denote the cell of $B$ attached to $A$ . There are $3n+1$ cells on $B$ besides $C$ . Thus, of the three (or less) sides of $C$ not attached to $A$ , one of them must have $n+1$ cells by the pigeonhole principle . It then follows that we can add $A$ , $C$ , and the other two sides together to get an $(n+1)$ dino, and the side of $C$ that has $n+1$ cells is also an $n+1$ -dino, so we can partition the animal with $4n+2$ cells into two $(n+1)$ -dinos and we're done.
Thus, our answer is $4(2007) - 3 = 8025$ cells.
        Example of a solution  Attempting to partition solution into dinosaurs
Solution 2
For simplicity, let $k=2007$ and let $n$ be the number of squares . Let the centers of the squares be vertices , and connect any centers of adjacent squares with edges. Suppose we have some loops . Just remove an edge in the loop. We are still connected since you can go around the other way in the loop. Now we have no loops. Each vertex can have at most 4 edges coming out of it. For each point, assign it the quadruple : $(a,b,c,d)$ where $a$ , $b$ , $c$ , $d$ are the numbers of vertices on each branch, WLOG  $a\ge b\ge c\ge d$ . Note $a+b+c+d=n-1$ .
Claim: If $n=4k-2$ , then we must be able to divide the animal into two dinosaurs.
Chose a vertex, $v$ , for which $a$ is minimal (i.e. out of all maximal elements in a quadruple, choose the one with the least maximal element).  We have that $4a \ge a+b+c+d=4k-3$ , so $a\ge k$ . Hence we can just cut off that branch, that forms a dinosaur.
But suppose the remaining vertices do not make a dinosaur. Then we have $b+c+d+1\le k-1 \iff n-a\le k-1\iff a\ge 3k-1$ . Now move to the first point on the branch at $a$ . We have a new quadruple $p,\ q,\ r,\ b+c+d+1$ ) where $p+q+r=a-1\ge 3k-2$ .
Now consider the maximal element of that quadruple. We already have $b+c+d+1\le k-1$ . WLOG $p\ge q\ge r\ge 0$ , then $3p\ge p+q+r=a-1\ge 3k-2\implies p\ge k$ so $p>k-1=b+c+d+1$ , so $p$ is the maximal element of that quadruple.
Also $a-1=p+q+r\ge p+0+0$ , so $p<a$ . But that is a contradiction to the minimality of $a$ . Therefore, we must have that $b+c+d+1\ge k$ , so we have a partition of two dinosaurs.
Maximum: $n=4k-3$ .
Consider a cross with each branch having $k-1$ verticies. Clearly if we take partition $k$ vertices, we remove the center, and we are not connected.
So $k=2007$ : $4\cdot 2007-3=8025$ .
 Solution 3 (Generalization)
Turn the dinosaur into a graph (cells are vertices , adjacent cells connected by an edge) and prove this result about graphs. A connected graph with $V$ vertices, where each vertex has degree less than or equal to $D$ , can be partitioned into connected components of sizes at least $\frac{V-1}{D}$ . So then in this special case, we have $D = 4$ , and so $V = 2006 \times 4+1$ (a possible configuration of this size that works consists of a center and 4 lines of cells each of size 2006 connected to the center). We next throw out all the geometry of this situation, so that we have a completely unconstrained graph. If we prove the above-mentioned result, we can put the geometry back in later by taking the connected components that our partition gives us, then filling back all edges that have to be there due to adjacent cells. This won't change any of the problem constraints, so we can legitimately do this.
Going, now, to the case of arbitrary graphs, we WOP on the number of edges. If we can remove any edge and still have a connected graph, then we have found a smaller graph that does not obey our theorem, a contradiction due to the minimality imposed by WOP. Therefore, the only case we have to worry about is when the graph is a tree. If it's a tree, we can root the tree and consider the size of subtrees. Pick the root such that the size of the largest subtree is minimized. This minimum must be at least $\frac{V-1}{D}$ , otherwise the sum of the size of the subtrees is smaller than the size of the graph, which is a contradiction. Also, it must be at most $\frac{V}{2}$ , or else pick the subtree of size greater than $\frac{V}{2}$ and you have decreased the size of the largest subtree if you root from that vertex instead, so you have some subtree with size between $\frac{V-1}{D}$ and $\frac V2$ . Cut the edge connecting the root to that subtree, and use that as your partition.
It is easy to see that these partitions satisfy the contention of our theorem, so we are done.
Solution 4
Let $s$ denote the minimum number of cells in a dinosaur; the number this year is $s = 2007$ .
Claim: The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1$ .
First, a primitive dinosaur can contain up to $4(s - 1) + 1$ cells. To see this, consider a dinosaur in the form of a cross consisting of a central cell and four arms with $s - 1$ cells apiece. No connected figure with at least $s$ cells can be removed without disconnecting the dinosaur.
The proof that no dinosaur with at least $4(s - 1) + 2$ cells is primitive relies on the following result.
Lemma. Let $D$ be a dinosaur having at least $4(s - 1) + 2$ cells, and let $R$ (red) and $B$ (black) be two complementary animals in $D$ , i.e. $R\cap B = \emptyset$ and $R\cup B = D$ . Suppose $|R|\leq s - 1$ . Then $R$ can be augmented to produce animals $\~{R}\subset R$ (Error compiling LaTeX. Unknown error_msg) and $\{B} = D\backslash\{R}$ (Error compiling LaTeX. Unknown error_msg) such that at least one of the following holds:
(i) $|\{R}|\geq s$ (Error compiling LaTeX. Unknown error_msg) and $|\~{B}|\geq s$ (Error compiling LaTeX. Unknown error_msg) ,
(ii) $|\{R}| = |R| + 1$ (Error compiling LaTeX. Unknown error_msg) ,
(iii) $|R| < |\{R}|\leq s - 1$ (Error compiling LaTeX. Unknown error_msg) .
Proof. If there is a black cell adjacent to $R$ that can be made red without disconnecting $B$ , then (ii) holds. Otherwise, there is a black cell $c$ adjacent to $R$ whose removal disconnects $B$ . Of the squares adjacent to $c$ , at least one is red, and at least one is black, otherwise $B$ would be disconnected. Then there are at most three resulting components $\mathcal{C}_1, \mathcal{C}_2, \mathcal{C}_3$ of $B$ after the removal of $c$ . Without loss of generality, $\mathcal{C}_3$ is the largest of the remaining components. (Note that $\mathcal{C}_1$ or $\mathcal{C}_2$ may be empty.) Now $\mathcal{C}_3$ has at least $\lceil (3s - 2)/3\rceil = s$ cells. Let $\{B} = \mathcal{C}_3$ (Error compiling LaTeX. Unknown error_msg) . Then $|\{R}| = |R| + |\mathcal{C}_1| + |\mathcal{C}_2| + 1$ (Error compiling LaTeX. Unknown error_msg) . If $|\{B}|\leq 3s - 2$ (Error compiling LaTeX. Unknown error_msg) , then $|\{R}|\geq s$ (Error compiling LaTeX. Unknown error_msg) and (i) holds. If $|\{B}|\geq 3s - 1$ (Error compiling LaTeX. Unknown error_msg) then either (ii) or (iii) holds, depending on whether $|\{R}|\geq s$ (Error compiling LaTeX. Unknown error_msg) or not. $\blacksquare$
Starting with $|R| = 1$ , repeatedly apply the Lemma. Because in alternatives (ii) and (iii) $|R|$ increases but remains less than $s$ , alternative (i) eventually must occur. This shows that no dinosaur with at least $4(s - 1) + 2$ cells is primitive.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Answer: \[ 4 \cdot 2007 - 3 = 8025 \]


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Permutations

Mathematics -> Number Theory -> Perfect Squares -> Other

A permutation of the set of positive integers $[n] = \{1, 2, \ldots, n\}$ is a sequence $(a_1, a_2, \ldots, a_n)$ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$ . Let $P(n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1\leq k\leq n$ . Find with proof the smallest $n$ such that $P(n)$ is a multiple of $2010$ .

Solution:
We claim that the smallest $n$ is $67^2 = \boxed{4489}$ .
Solution 1
Let $S = \{1, 4, 9, \ldots\}$ be the set of positive perfect squares. We claim that the relation $R = \{(j, k)\in [n]\times[n]\mid jk\in S\}$ is an equivalence relation on $[n]$ .
It is reflexive because for all .  It is symmetric because .  It is transitive because if and , then , since is closed under multiplication and a non-square times a square is always a non-square.
We are restricted to permutations for which $ka_k \in S$ , in other words to permutations that send each element of $[n]$ into its equivalence class. Suppose there are $N$ equivalence classes: $C_1, \ldots, C_N$ . Let $n_i$ be the number of elements of $C_i$ , then $P(n) = \prod_{i=1}^{N} n_i!$ .
Now $2010 = 2\cdot 3\cdot 5\cdot 67$ . In order that $2010\mid P(n)$ , we must have $67\mid n_m!$ for the class $C_m$ with the most elements. This means $n_m\geq 67$ , since no smaller factorial will have $67$ as a factor. This condition is sufficient, since $n_m!$ will be divisible by $30$ for $n_m\geq 5$ , and even more so $n_m\geq 67$ .
The smallest element $g_m$ of the equivalence class $C_m$ is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of $C_m$ . Also, each prime $p$ that divides $g_m$ divides all the other elements $k$ of $C_m$ , since $p^2\mid kg_m$ and thus $p\mid k$ . Therefore $g_m\mid k$ for all $k\in C_m$ . The primes that are not in $g_m$ occur an even number of times in each $k\in C_m$ .
Thus the equivalence class $C_m = \{g_mk^2\leq n\}$ . With $g_m = 1$ , we get the largest possible $n_m$ . This is just the set of squares in $[n]$ , of which we need at least $67$ , so $n\geq 67^2$ . This condition is necessary and sufficient.
Solution 2
This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as "equivalence relation":
It is possible to write all positive integers $n$ in the form $p\cdot m^2$ , where $m^2$ is the largest perfect square dividing $n$ , so $p$ is not divisible by the square of any prime. Obviously, one working permutation of $[n]$ is simply $(1, 2, \ldots, n)$ ; this is acceptable, as $ka_k$ is always $k^2$ in this sequence.
Lemma 1. We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities $p$ .
Proof. Let $p_k$ and $m_k$ be the values of $p$ and $m$ , respectively, for a given $k$ as defined above, such that $p$ is not divisible by the square of any prime. We can obviously permute two numbers which have the same $p$ , since if $p_j = p_w$ where $j$ and $w$ are 2 values of $k$ , then $j\cdot w = p_j^2\cdot m_j^2\cdot m_w^2$ , which is a perfect square. This proves that we can permute any numbers with the same value of $p$ .
End Lemma
Lemma 2. We will prove the converse of Lemma 1: Let one number have a $p$ value of $\phi$ and another, $\gamma$ . $\phi\cdot f$ and $\gamma\cdot g$ are both perfect squares.
Proof.  $\phi\cdot f$ and $\gamma\cdot g$ are both perfect squares, so for $\phi\cdot \gamma$ to be a perfect square, if $g$ is greater than or equal to $f$ , $g/f$ must be a perfect square, too. Thus $g$ is $f$ times a square, but $g$ cannot divide any squares besides $1$ , so $g = 1f$ ; $g = f$ . Similarly, if $f\geq g$ , then $f = g$ for our rules to keep working.
End Lemma
We can permute $l$ numbers with the same $p$ in $l!$ ways. We must have at least 67 numbers with a certain $p$ so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as $h$ , in general, we need numbers all the way up to $h\cdot 67^2$ , so obviously, $67^2$ is the smallest such number such that we can get a $67!$ term; here 67 $p$ terms are 1. Thus we need the integers $1, 2, \ldots, 67^2$ , so $67^2$ , or $\boxed{4489}$ , is the answer.

Answer: \(\boxed{4489}\)


For any odd positive integer $n$, let $r(n)$ be the odd positive integer such that the binary representation of $r(n)$ is the binary representation of $n$ written backwards. For example, $r(2023)=r\left(11111100111_{2}\right)=11100111111_{2}=1855$. Determine, with proof, whether there exists a strictly increasing eight-term arithmetic progression $a_{1}, \ldots, a_{8}$ of odd positive integers such that $r\left(a_{1}\right), \ldots, r\left(a_{8}\right)$ is an arithmetic progression in that order.

Solution: The main idea is the following claim. Claim: If $a, b, c$ are in arithmetic progression and have the same number of digits in their binary representations, then $r(a), r(b), r(c)$ cannot be in arithmetic progression in that order. Proof. Consider the least significant digit that differs in $a$ and $b$. Then $c$ will have the same value of that digit as $a$, which will be different from $b$. Since this becomes the most significant digit in $r(a), r(b), r(c)$, then of course $b$ cannot be between $a$ and $c$. To finish, we just need to show that if there are 8 numbers in arithmetic progression, which we'll write as $a_{1}, a_{1}+d, a_{1}+2 d, \ldots, a_{1}+7 d$, three of them have the same number of digits. We have a few cases. - If $a_{1}+3 d<2^{k} \leq a_{1}+4 d$, then $a_{1}+4 d, a_{1}+5 d, a_{1}+6 d$ will have the same number of digits. - If $a_{1}+4 d<2^{k} \leq a_{1}+5 d$, then $a_{1}+5 d, a_{1}+6 d, a_{1}+7 d$ will have the same number of digits. - If neither of these assumptions are true, $a_{1}+3 d, a_{1}+4 d, a_{1}+5 d$ will have the same number of digits. Having exhausted all cases, we are done.

Answer: There does not exist a strictly increasing eight-term arithmetic progression \(a_{1}, \ldots, a_{8}\) of odd positive integers such that \(r(a_{1}), \ldots, r(a_{8})\) is an arithmetic progression in that order.


Mathematics -> Algebra -> Algebra -> Polynomial Operations

Mathematics -> Number Theory -> Divisibility -> Other

Find all ordered pairs $(a,b)$ of positive integers for which the numbers $\dfrac{a^3b-1}{a+1}$ and $\dfrac{b^3a+1}{b-1}$ are both positive integers

Solution:
Adding $1$ to both the given numbers we get:
$\dfrac{a^3b-1}{a+1} + 1$ is also a positive integer so we have:
$\dfrac{a^3b+a}{a+1}$ = $\dfrac{a(a^2b+1)}{a+1}$ is a positive integer
$\implies (a+1) \mid (a^2b+1)$  $\implies (a+1) \mid (((a+1) - 1)^2b+1)$  $\implies (a+1) \mid (b+1)$
Similarly,
$\dfrac{b^3a+1}{b-1} + 1$ is also a positive integer so we have:
$\dfrac{b^3a+b}{b-1}$ = $\dfrac{b(b^2a+1)}{b-1}$ is a positive integer
$\implies (b-1) | (b^2a+1)$  $\implies (b-1) | (((b-1) + 1)^2a+1)$  $\implies (b-1) | (a+1)$
Combining above $2$ results we get:
$(b-1) | (b+1)$
$\implies b=2,3$
$Case 1: b=2$  $\implies a+1|3 \implies a=2$ which is a valid solution.
$Case 2: b=3$  $\implies a+1|4 \implies a=1,3$ which are valid solutions.
Thus, all solutions are: $(2,2), (1,3), (3,3)$
  $Kris17$

Answer:
\[
\{(2,2), (1,3), (3,3)\}
\]


Determine each real root of
$x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0$
correct to four decimal places.

Solution:
The equation can be re-written as \begin{align}\label{eqn1} (x+10^5)^2(x-10^5)^2 -(x+10^5)(x-10^5) -x-1=0. \end{align}
We first prove that the equation has no negative roots.
Let $x\le 0.$ The equation above can be further re-arranged as \begin{align*}[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]=x-1.\end{align*} The right hand side of the equation is negative. Therefore \[[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2)]<0,\] and we have $-1<(x+10^5)(x-10^5) <2.$ Then the left hand side of the equation is bounded by \[|[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]|\le 3\times 3.\] However, since $|(x+10^5)(x-10^5)|\le 2$ and $x<0,$ it follows that $|x+10^5| <\frac{2}{|x-10^5|}<2\times 10^{-5}$ for negative $x.$ Then $x<2\times 10^{-5}-10^5.$ The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.
Now let $x>0.$ When $x=10^5,$ the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of $10^5$ , as its leading coefficient is positive. We will prove that $x=10^5$ is a good approximation of the roots (within $10^{-2}$ ). In fact, we can solve the "quadratic" equation (1) for $(x+10^5)(x-10^5)$ : \[(x+10^5)(x-10^5)=\frac{1\pm\sqrt{1+4(x+1)}}{2}.\] Then \[x-10^5=\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}.\] Easy to see that $|x-10^5| <1$ for positve $x.$ Therefore, $10^5-1<x<10^5+1.$ Then \begin{align*} |x-10^5|&=\left|\frac{1\pm\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ &\le \left |\frac{1}{2(x+10^5)}\right |+\left |\frac{\sqrt{1+4(x+1)}}{2(x+10^5)}\right |\\ &\le \frac{1}{2(10^5-1+10^5)} +\frac{\sqrt{1+4(10^5+1+1)}}{2(10^5-1+10^5)} \\ &<10^{-2}. \end{align*}
Let $x_1$ be a root of the equation with $x_1<10^5.$ Then $0<10^5-x_1<10^{-2}$ and \[x_1-10^5=\frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}.\] An aproximation of $x_1$ is defined as follows: \[\tilde{x}_1=10^5+\frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.\] We check the error of the estimate: \begin{align*} |\tilde{x}_1-x_1|&=\left | \frac{1-\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{1-\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right | \\ &\le \left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right |+\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |. \end{align*}
The first absolute value \[\left |\frac{1}{2(10^5+10^5)}- \frac{1}{2(x_1+10^5)}\right | =\frac{|x_1- 10^5|}{2(10^5+10^5)(x_1+10^5)}<10^{-12}.\]
The second absolute value \begin{align*} &\left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)} - \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \right |\\ &\le \left |\frac{\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}\right |+\left |\frac{\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}- \frac{\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\right |\\ &\le 10^{-7}+10^{-9}, \end{align*} through a rationalized numerator.Therefore $|\tilde{x}_1-x_1|\le 10^{-6}.$
For a real root $x_2$ with $x_2>10^5,$ we choose \[\tilde{x}_2=10^5+\frac{1+\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.\] We can similarly prove it has the desired approximation.

Answer:
The real roots of the equation \(x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0\) correct to four decimal places are approximately:
\[ x_1 \approx 10^5 - \frac{1 - \sqrt{1 + 4(10^5 + 1)}}{2 \cdot 2 \cdot 10^5} \]
\[ x_2 \approx 10^5 + \frac{1 + \sqrt{1 + 4(10^5 + 1)}}{2 \cdot 2 \cdot 10^5} \]

Given the approximations:
\[ x_1 \approx 99999.5000 \]
\[ x_2 \approx 100000.5000 \]


Find the minimum positive integer $n\ge 3$, such that there exist $n$ points $A_1,A_2,\cdots, A_n$ satisfying no three points are collinear and for any $1\le i\le n$, there exist $1\le j \le n (j\neq i)$, segment $A_jA_{j+1}$ pass through the midpoint of segment $A_iA_{i+1}$, where $A_{n+1}=A_1$

Solution:
To find the minimum positive integer \( n \geq 3 \) such that there exist \( n \) points \( A_1, A_2, \ldots, A_n \) satisfying no three points are collinear and for any \( 1 \leq i \leq n \), there exists \( 1 \leq j \leq n \) (with \( j \neq i \)), such that the segment \( A_jA_{j+1} \) passes through the midpoint of segment \( A_iA_{i+1} \), where \( A_{n+1} = A_1 \), we proceed as follows:

First, it is necessary to verify that \( n = 3 \) and \( n = 4 \) do not satisfy the given conditions. Through geometric construction and analysis, it can be shown that no such configurations exist for these values of \( n \).

Next, consider \( n = 5 \). We analyze two cases:
1. **Case 1**: There are no parallelograms formed by any four of the points \( A_i \). By detailed geometric analysis and coordinate bashing, it can be shown that no such five points exist.
2. **Case 2**: Assume \( A_1A_4A_2A_3 \) forms a parallelogram. By considering the reflection of points and ensuring no three points are collinear, it leads to a contradiction, proving that \( n = 5 \) is also not possible.

Finally, for \( n = 6 \), a construction exists that satisfies all the given conditions. Therefore, the minimum positive integer \( n \) for which the conditions hold is \( n = 6 \).

The answer is: \boxed{6}.

Answer: 6


Mathematics -> Geometry -> Plane Geometry -> Area

Lily has a $300 \times 300$ grid of squares. She now removes $100 \times 100$ squares from each of the four corners and colors each of the remaining 50000 squares black and white. Given that no $2 \times 2$ square is colored in a checkerboard pattern, find the maximum possible number of (unordered) pairs of squares such that one is black, one is white and the squares share an edge.

Solution: First we show an upper bound. Define a grid point as a vertex of one of the squares in the figure. Construct a graph as follows. Place a vertex at each grid point and draw an edge between two adjacent points if that edge forms a black-white boundary. The condition of there being no $2 \times 2$ checkerboard is equivalent to no vertex having degree more than 2. There are $101^{2}+4 \cdot 99^{2}=49405$ vertices that are allowed to have degree 2 and $12 \cdot 99=1188$ vertices (on the boundary) that can have degree 1. This gives us an upper bound of 49999 edges. We will show that exactly this many edges is impossible. Assume for the sake of contradiction that we have a configuration achieving exactly this many edges. Consider pairing up the degree 1 vertices so that those on a horizontal edge pair with the other vertex in the same column and those on a vertical edge pair with the other vertex in the same row. If we combine the pairs into one vertex, the resulting graph must have all vertices with degree exactly 2. This means the graph must be a union of disjoint cycles. However all cycles must have even length and there are an odd number of total vertices so this is impossible. Thus we have an upper bound of 49998. We now describe the construction. The top row alternates black and white. The next 99 rows alternate between all black and all white. Let's say the second row from the top is all white. The $101^{\text {st }}$ row alternates black and white for the first 100 squares, is all black for the next 100 and alternates between white and black for the last 100 squares. The next 98 rows alternate between all black and all white (the $102^{\text {nd }}$ row is all white). Finally, the bottom 101 rows are a mirror of the top 101 rows with the colors reversed. We easily verify that this achieves the desired.

Answer: 49998


Mathematics -> Calculus -> Differential Calculus -> Applications of Derivatives

Mathematics -> Algebra -> Intermediate Algebra -> Inequalities

Consider functions $f : [0, 1] \rightarrow \mathbb{R}$ which satisfy
  (i) for all in ,    (ii) ,    (iii) whenever , , and are all in .
Find, with proof, the smallest constant $c$ such that
$f(x) \le cx$
for every function $f$ satisfying (i)-(iii) and every $x$ in $[0, 1]$ .

Solution:
My claim: $c\ge2$
  Lemma 1 ) $f\left(\left(\frac{1}{2}\right)^n\right)\le\left(\frac{1}{2}\right)^n$ for $n\in \mathbb{Z}, n\ge0$
For $n=0$ , $f(1)=1$ (ii)
Assume that it is true for $n-1$ , then $f\left(\left(\frac{1}{2}\right)^{n}\right)+f\left(\left(\frac{1}{2}\right)^{n}\right)\le f\left(\left(\frac{1}{2}\right)^{n-1}\right)\le \left(\frac{1}{2}\right)^{n-1}$
$f\left(\left(\frac{1}{2}\right)^{n}\right)\le \left(\frac{1}{2}\right)^{n}$
By principle of induction, lemma 1 is proven .
  Lemma 2 ) For any $x$ , $\left(\frac{1}{2}\right)^{n+1}<x\le\left(\frac{1}{2}\right)^n\le1$ and $n\in \mathbb{Z}$ , $f(x)\le\left(\frac{1}{2}\right)^n$ .
$f(x)+f\left(\left(\frac{1}{2}\right)^n-x\right)\le f\left(\left(\frac{1}{2}\right)^{n}\right)\le \left(\frac{1}{2}\right)^{n}$ (lemma 1 and (iii) )
$f(x)\le\left(\frac{1}{2}\right)^n$ (because $f\left(\left(\frac{1}{2}\right)^n-x\right)\ge0$ (i) )

$\forall 0\le x\le 1$ , $\left(\frac{1}{2}\right)^{n-1}\ge2x\ge \left(\frac{1}{2}\right)^n\ge f(x)$ . Thus, $c=2$ works.

Let's look at a function $g(x)=\left\{\begin{array}{ll}0&0\le x\le \frac{1}{2};\\1&\frac{1}{2}<x\le1;\\\end{array}\right\}$
It clearly have property (i) and (ii). For $0\le x\le\frac{1}{2}$ and WLOG let $x\le y$ , $f(x)+f(y)=0+f(y)\le f(y)$
For $\frac{1}{2}< x\le 1$ , $x+y>1$ . Thus, property (iii) holds too. Thus $g(x)$ is one of the legit function.

$\lim_{x\rightarrow\frac{1}{2}^+} cx \ge \lim_{x\rightarrow\frac{1}{2}^+} g(x)=1$
$\frac{1}{2}c>1$
$c>2$ but approach to $2$ when $x$ is extremely close to $\frac{1}{2}$ from the right side.
$\mathbb{Q.E.D}$

Answer: The smallest constant \( c \) such that \( f(x) \le cx \) for every function \( f \) satisfying the given conditions is \( c = 2 \).


Mathematics -> Algebra -> Intermediate Algebra -> Exponential Functions

Find all pairs of positive integers $(x,y)$ such that \[x^y = y^{x - y}.\]

Solution:
Note that $x^y$ is at least one. Then $y^{x - y}$ is at least one, so $x \geq y$ .
Write $x = a^{b+c}, y = a^c$ , where $\gcd(b, c) = 1$ . (We know that $b$ is nonnegative because $x\geq y$ .) Then our equation becomes $a^{(b+c)*a^c} = a^{c*(a^{b+c} - a^c)}$ . Taking logarithms base $a$ and dividing through by $a^c$ , we obtain $b + c = c*(a^b - 1)$ .
Since $c$ divides the RHS of this equation, it must divide the LHS. Since $\gcd(b, c) = 1$ by assumption, we must have $c = 1$ , so that the equation reduces to $b + 1 = a^b - 1$ , or $b + 2 = a^b$ . This equation has only the solutions $b = 1, a = 3$ and $b = 2, a = 2$ .
Therefore, our only solutions are $x = 3^{1 + 1} = 9, y = 3^1 = 3$ , and $x = 2^{2+1} = 8, y = 2^1 = 2$ , and we are done.

Answer:
The pairs of positive integers \((x, y)\) that satisfy the equation \(x^y = y^{x - y}\) are:

\[
(x, y) = (9, 3) \quad \text{and} \quad (x, y) = (8, 2)
\]


Does there exist a two-variable polynomial $P(x, y)$ with real number coefficients such that $P(x, y)$ is positive exactly when $x$ and $y$ are both positive?

Solution: No. For any $\epsilon$ and positive $x, P(x, \epsilon)>0$ and $P(x,-\epsilon) \leq 0$. Thus by continuity/IVT, $P(x, 0)=0$ for all positive $x$. Similarly $P(0, y)=0$ for all positive $y$. This implies $x y \mid P(x, y)$, and so we can write $P(x, y)=x y Q(x, y)$. But then this same logic holds for $Q$, and this cannot continue infinitely unless $P$ is identically 0 - in which case the conditions do not hold. So no such polynomial exists.

Answer: No such polynomial exists.


Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that for all $x,y\in \mathbb{R}_{>0}$ we have \[f(x) = f(f(f(x)) + y) + f(xf(y)) f(x+y).\]

Solution: [WIP]

Answer: The final answer is not provided as the solution is marked as "Work In Progress" (WIP).


Mathematics -> Discrete Mathematics -> Algorithms

Philena and Nathan are playing a game. First, Nathan secretly chooses an ordered pair $(x, y)$ of positive integers such that $x \leq 20$ and $y \leq 23$. (Philena knows that Nathan's pair must satisfy $x \leq 20$ and $y \leq 23$.) The game then proceeds in rounds; in every round, Philena chooses an ordered pair $(a, b)$ of positive integers and tells it to Nathan; Nathan says YES if $x \leq a$ and $y \leq b$, and NO otherwise. Find, with proof, the smallest positive integer $N$ for which Philena has a strategy that guarantees she can be certain of Nathan's pair after at most $N$ rounds.

Solution: It suffices to show the upper bound and lower bound. Upper bound. Loosen the restriction on $y$ to $y \leq 24$. We'll reduce our remaining possibilities by binary search; first, query half the grid to end up with a $10 \times 24$ rectangle, and then half of that to go down to $5 \times 24$. Similarly, we can use three more queries to reduce to $5 \times 3$. It remains to show that for a $5 \times 3$ rectangle, we can finish in 4 queries. First, query the top left $4 \times 2$ rectangle. If we are left with the top left $4 \times 2$, we can binary search both coordinates with our remaining three queries. Otherwise, we can use another query to be left with either a $4 \times 1$ or $1 \times 3$ rectangle, and binary searching using our final two queries suffices. Lower bound. At any step in the game, there will be a set of ordered pairs consistent with all answers to Philena's questions up to that point. When Philena asks another question, each of these possibilities is consistent with only one of YES or NO. Alternatively, this means that one of the answers will leave at least half of the possibilities. Therefore, in the worst case, Nathan's chosen square will always leave at least half of the possibilities. For such a strategy to work in $N$ questions, it must be true that $\frac{460}{2^{N}} \leq 1$, and thus $N \geq 9$.

Answer: The smallest positive integer \( N \) for which Philena has a strategy that guarantees she can be certain of Nathan's pair after at most \( N \) rounds is \( 9 \).


Let $n$ be an integer greater than $1$. For a positive integer $m$, let $S_{m}= \{ 1,2,\ldots, mn\}$. Suppose that there exists a $2n$-element set $T$ such that
(a) each element of $T$ is an $m$-element subset of $S_{m}$;
(b) each pair of elements of $T$ shares at most one common element;
and
(c) each element of $S_{m}$ is contained in exactly two elements of $T$.

Determine the maximum possible value of $m$ in terms of $n$.

Solution:
Let \( n \) be an integer greater than 1. For a positive integer \( m \), let \( S_{m} = \{ 1, 2, \ldots, mn \} \). Suppose that there exists a \( 2n \)-element set \( T \) such that:
(a) each element of \( T \) is an \( m \)-element subset of \( S_{m} \);
(b) each pair of elements of \( T \) shares at most one common element; and
(c) each element of \( S_{m} \) is contained in exactly two elements of \( T \).

We aim to determine the maximum possible value of \( m \) in terms of \( n \).

First, we show that \( m \leq 2n - 1 \). By condition (b), there are at most \(\binom{2n}{n}\) elements of \( S_{m} \) which are in 2 sets in \( T \). However, by condition (c), every element of \( S_{m} \) is in 2 sets in \( T \), so \(|S_{m}| = mn \leq \binom{2n}{n} = \frac{(2n)(2n-1)}{2} = n(2n-1)\), and thus \( m \leq 2n - 1 \).

Now, to see that \( 2n - 1 \) is achievable, consider a complete graph \( K_{2n} \). Note that this \( K_{2n} \) has \( n(2n-1) \) edges, so we label the edges with distinct elements of \( S_{2n-1} \). Let \( T \) be the family of sets formed by taking each vertex and creating the set of labels of incident edges. Any two sets in \( T \) have exactly one element in common—the label of the edge between them—so condition (b) is satisfied. Each element of \( S_{m} \) is in exactly two elements of \( T \): the two endpoints of the edge with the given label, satisfying condition (c). Condition (a) is also satisfied because every vertex in a \( K_{2n} \) has degree \( 2n-1 = m \), so every element of \( T \) has \( m \) elements. Finally, since a \( K_{2n} \) has \( 2n \) vertices, \(|T| = 2n\), as desired.

Thus, the maximum possible value of \( m \) is:
\[
\boxed{2n - 1}
\]

Answer: 2n - 1


Mathematics -> Algebra -> Linear Algebra -> Linear Transformations

FIx positive integer $n$. Prove: For any positive integers $a,b,c$ not exceeding $3n^2+4n$, there exist integers $x,y,z$ with absolute value not exceeding $2n$ and not all $0$, such that $ax+by+cz=0$

Solution:
Fix a positive integer \( n \). We aim to prove that for any positive integers \( a, b, c \) not exceeding \( 3n^2 + 4n \), there exist integers \( x, y, z \) with absolute value not exceeding \( 2n \) and not all zero, such that \( ax + by + cz = 0 \).

Without loss of generality, assume \( c = \max(a, b, c) \).

Consider the set of integers \( x \) and \( y \) such that \( 0 \leq x, -y \leq 2n \) and \( x^2 + y^2 > 0 \). If any value of \( ax + by \) in this set is \( 0 \mod c \), the norm is at most \( 2cn \), and we are done. Otherwise, these \( 4n^2 + 4n \) values are all nonzero modulo \( c \). Since \( c \leq 3n^2 + 4n \), there must be duplicates modulo \( c \).

If \( ax_1 + by_1 \equiv ax_2 + by_2 \mod c \), then \( a(x_2 - x_1) + b(y_2 - y_1) \equiv 0 \mod c \). If \( x_2 - x_1 \) and \( y_2 - y_1 \) are not both positive or both negative, we are done. Otherwise, assume \( x_2 > x_1 \) and \( y_2 > y_1 \). We must have \( (x_2 + y_2) - (x_1 + y_1) > 2n \) because otherwise \( a(x_2 - x_1) + b(y_2 - y_1) \) is positive but at most \( c(x_2 - x_1) + c(y_2 - y_1) \), which is at most \( 2nc \).

The size \( x + y \) of \( ax + by \) is between \(-2n\) and \( 2n \) inclusive, so there are at most two choices of \( (x, y) \) for any fixed residue modulo \( c \). If there are three, the largest and smallest have a size difference over \( 4n \), which is a contradiction.

Since \( c \leq 3n^2 + 4n \) and we have \( 4n^2 + 4n \) values, there must be at least \( n^2 + 1 \) pairs \( ((x_1, y_1), (x_2, y_2)) \) such that \( ax_1 + by_1 \equiv ax_2 + by_2 \mod c \) with \( x_2 > x_1 \) and \( y_2 > y_1 \). This gives \( a(x_2 - x_1) + b(y_2 - y_1) \equiv 0 \mod c \) with \( (y_2 - y_1) + (x_2 - x_1) > 2n \).

If we also have \( a(x_3 - x_4) + b(y_3 - y_4) \equiv 0 \mod c \) with \( x_3 > x_4 \) and \( y_3 > y_4 \), then:
\[
a(x_2 - x_1 - x_3 + x_4) + b(y_2 - y_1 - y_3 + y_4) \equiv 0 \mod c.
\]
This implies that the difference has a size of norm at most \( 2n \) and is divisible by \( c \), allowing us to add the appropriate \( cz \) and finish.

If all pairs \( ax_1 + by_1 \equiv ax_2 + by_2 \) have \( (x_2 - x_1, y_2 - y_1) \) fixed, then \( x_2 - x_1 = A \) and \( y_2 - y_1 = B \) with \( A + B > 2n \). The range constraints give at most \( (2n + 1 - A)(2n + 1 - B) \leq n^2 \) choices of \( (x_1, y_1) \), which is a contradiction unless \( A + B = 2n + 1 \).

Thus, \( Aa + Bb \equiv 0 \mod c \) with \( A + B = 2n + 1 \). Since \( 0 \leq Aa + Bb \leq (2n + 1)c \), it must be a multiple of \( c \), and we are done by adding \( cz \) unless it is exactly \( (2n + 1)c \). This implies \( A(c - a) + B(c - b) = 0 \), and since \( A \) and \( B \) are nonzero, either \( c = a \) or \( c = b \), which are trivial cases.

Therefore, the proof is complete.

The answer is: \boxed{0}.

Answer: 0


Mathematics -> Geometry -> Plane Geometry -> Other

Three noncollinear points and a line $\ell$ are given in the plane. Suppose no two of the points lie on a line parallel to $\ell$ (or $\ell$ itself). There are exactly $n$ lines perpendicular to $\ell$ with the following property: the three circles with centers at the given points and tangent to the line all concur at some point. Find all possible values of $n$.

Solution: The condition for the line is that each of the three points lies at an equal distance from the line as from some fixed point; in other words, the line is the directrix of a parabola containing the three points. Three noncollinear points in the coordinate plane determine a quadratic polynomial in $x$ unless two of the points have the same $x$-coordinate. Therefore, given the direction of the directrix, three noncollinear points determine a parabola, unless two of the points lie on a line perpendicular to the directrix. This case is ruled out by the given condition, so the answer is 1.

Answer: 1


Mathematics -> Number Theory -> Congruences

Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.

Solution:
The answer is $n=1$ , which is easily verified to be a valid integer $n$ .
Notice that \[2^n+12^n+2011^n\equiv 2^n+7^n \pmod{12}.\] Then for $n\geq 2$ , we have $2^n+7^n\equiv 3,5 \pmod{12}$ depending on the parity of $n$ . But perfect squares can only be $0,1,4,9\pmod{12}$ , contradiction. Therefore, we are done. $\blacksquare$

Answer: \[ n = 1 \]


Mathematics -> Algebra -> Linear Algebra -> Matrices

Mathematics -> Algebra -> Abstract Algebra -> Ring Theory

Determine whether there exist an odd positive integer $n$ and $n \times n$ matrices $A$ and $B$ with integer entries, that satisfy the following conditions: (1) $\operatorname{det}(B)=1$; (2) $A B=B A$; (3) $A^{4}+4 A^{2} B^{2}+16 B^{4}=2019 I$. (Here $I$ denotes the $n \times n$ identity matrix.)

Solution: We show that there are no such matrices. Notice that $A^{4}+4 A^{2} B^{2}+16 B^{4}$ can factorized as $$A^{4}+4 A^{2} B^{2}+16 B^{4}=\left(A^{2}+2 A B+4 B^{2}\right)\left(A^{2}-2 A B+4 B^{2}\right)$$ Let $C=A^{2}+2 A B+4 B^{2}$ and $D=A^{2}-2 A B+4 B^{2}$ be the two factors above. Then $$\operatorname{det} C \cdot \operatorname{det} D=\operatorname{det}(C D)=\operatorname{det}\left(A^{4}+4 A^{2} B^{2}+16 B^{4}\right)=\operatorname{det}(2019 I)=2019^{n}$$ The matrices $C, D$ have integer entries, so their determinants are integers. Moreover, from $C \equiv D(\bmod 4)$ we can see that $$\operatorname{det} C \equiv \operatorname{det} D \quad(\bmod 4)$$ This implies that $\operatorname{det} C \cdot \operatorname{det} D \equiv(\operatorname{det} C)^{2}(\bmod 4)$, but this is a contradiction because $2019^{n} \equiv 3$ $(\bmod 4)$ is a quadratic nonresidue modulo 4. Solution 2. Notice that $$A^{4} \equiv A^{4}+4 A^{2} B^{2}+16 B^{4}=2019 I \quad \bmod 4$$ so $$(\operatorname{det} A)^{4}=\operatorname{det} A^{4} \equiv \operatorname{det}(2109 I)=2019^{n} \quad(\bmod 4)$$ But $2019^{n} \equiv 3$ is a quadratic nonresidue modulo 4, contradiction.

Answer: There do not exist such matrices.


Determine all pairs of positive integers $(m,n)$ such that $(1+x^n+x^{2n}+\cdots+x^{mn})$ is divisible by $(1+x+x^2+\cdots+x^{m})$ .

Solution:
Denote the first and larger polynomial to be $f(x)$ and the second one to be $g(x)$ . In order for $f(x)$ to be divisible by $g(x)$ they must have the same roots. The roots of $g(x)$ are the (m+1)th roots of unity, except for 1. When plugging into $f(x)$ , the root of unity is a root of $f(x)$ if and only if the terms $x^n, x^{2n}, x^{3n}, \cdots x^{mn}$ all represent a different (m+1)th root of unity not equal to 1.
Note that if $\\gcd(m+1,n)=1$ , the numbers $n, 2n, 3n, \cdots, mn$ represent a complete set of residues minus 0 modulo $m+1$ . However, if $gcd(m+1,n)=a$ not equal to 1, then $\frac{(m+1)(n)}{a}$ is congruent to $0 \pmod {m+1}$ and thus a complete set is not formed. Therefore, $f(x)$ divides $g(x)$ if and only if $\boxed{\\gcd(m+1,n)=1}.$  $\blacksquare$

Answer: \(\gcd(m+1, n) = 1\)


Let $n$ be a positive integer.  Determine the size of the largest subset of $\{ - n, - n + 1, \ldots , n - 1, n\}$ which does not contain three elements $a, b, c$ (not necessarily distinct) satisfying $a + b + c = 0$ .

Solution:
Let $S$ be a subset of $\{-n,-n+1,\dots,n-1,n\}$ of largest size satisfying $a+b+c\neq 0$ for all $a,b,c\in S$ . First, observe that $0\notin S$ . Next note that $|S|\geq \lceil n/2\rceil$ , by observing that the set of all the odd numbers in $\{-n,-n+1,\dots,n-1,n\}$ works. To prove that $|S|\leq \lceil n/2\rceil$ , it suffices to only consider even $n$ , because the statement for $2k$ implies the statement for $2k-1$ as well. So from here forth, assume $n$ is even.
For any two sets $A$ and $B$ , denote by $A+B$ the set $\{a+b\mid a\in A,b\in B\}$ , and by $-A$ the set $\{-a\mid a\in A\}$ . Also, let $A_+$ denote $A\cap\{1,2,\dots\}$ and $A_-$ denote $A\cap\{-1,-2,\dots\}$ . First, we present a lemma:
Lemma 1 : Let $A$ and $B$ be two sets of integers. Then $|A+B|\geq|A|+|B|-1$ .
Proof : Write $A=\{a_1,\dots,a_n\}$ and $B=\{b_1,\dots,b_m\}$ where $a_1<\dots<a_n$ and $b_1<\dots<b_m$ . Then $a_1+b_1,a_1+b_2,\dots,a_1+b_m,a_2+b_m,\dots,a_n+b_m$ is a strictly increasing sequence of $n+m-1$ integers in $A+B$ .
Now, we consider two cases:
Case 1 : One of $n,-n$ is not in $S$ . Without loss of generality, suppose $-n\notin S$ . Let $U=\{-n+1,-n+2,\dots,n-2,n-1,n\}$ (a set with $2n$ elements), so that $S\subseteq U$ by our assumption. Now, the condition that $a+b+c\neq 0$ for all $a,b,c\in S$ implies that $-S\cap(S_++S_-)=\emptyset$ . Since any element of $S_++S_-$ has absolute value at most $n-1$ , we have $S_++S_-\subseteq \{-n+1,-n+2,\dots,n-2,n-1\}\subseteq U$ . It follows that $S_++S_-\subseteq U\setminus -S$ , so $|S_++S_-|\leq 2n-|S|$ . However, by Lemma 1, we also have $|S_++S_-|\geq |S_+|+|S_-|-1=|S|-1$ .  Therefore, we must have $|S|-1\leq 2n-|S|$ , or $2|S|\leq 2n+1$ , or $|S|\leq n$ .
Case 2 : Both $n$ and $-n$ are in $S$ . Then $n/2$ and $-n/2$ are not in $S$ , and at most one of each of the pairs $\{1,n-1\},\{2,n-2\},\ldots,\{\frac n2-1,\frac n2+1\}$ and their negatives are in $S$ . This means $S$ contains at most $2+(n-2)=n$ elements.
Thus we have proved that $|S|\leq n=\lceil n/2\rceil$ for even $n$ , and we are done.

Answer:
\[
\left\lceil \frac{n}{2} \right\rceil
\]


Mathematics -> Algebra -> Abstract Algebra -> Group Theory

For any $n \geq 1$, let $A$ denote the $\mathbb{C}$ algebra consisting of $n \times n$ upper triangular complex matrices $\left\{\left(\begin{array}{ccc}* & * & * \\ 0 & * & * \\ 0 & 0 & *\end{array}\right)_{n \times n}\right\}$. We shall consider the left $A$-modules (that is, $\mathbb{C}$-vector spaces $V$ with $\mathbb{C}$-algebra homomorphisms $\rho: A \rightarrow \operatorname{End}(V))$. (2) Determine all simple modules of $A$.

Solution: (2a) Let $S_{i}, 1 \leq i \leq n$, denote the 1-dimensional modules such that $E_{i i}$ acts by 1 and $E_{i j}, E_{j j}$ acts by 0 for $j \neq i$. They are simple modules. (2b) It remains to show that the $S_{i}$ we have constructed are the only simple modules. Let $S$ denote any finite dimensional simple module. We claim that $E_{i j}, i<j$, form a nilpotent 2-sided ideal $N$ (because the product of an upper triangular matrix with a strictly upper one is strictly upper). Then $N$ acts on $S$ by 0 (To see this, $N S$ is a submodule of $S$. It is proper because $N$ is nilpotent. Since $S$ is simple, we deduce that $N S=0$.) Note that the action of $E_{i i}$ commute with each other (and with the 0 -action by $E_{i j}$ ), thus they are module endomorphisms. By Schur's Lemma, $E_{i i}$ acts on $S$ as a scalar. Since $E_{i i} E_{j j}=0$ for $i \neq j$, at most one $E_{i i}$ acts as a non-zero scalar. Recall that $1=\sum_{i} E_{i i}$ acts by the identity. The claim follows.

Answer: The simple modules of \( A \) are the 1-dimensional modules \( S_i \) for \( 1 \leq i \leq n \), where \( E_{ii} \) acts by 1 and \( E_{ij}, E_{jj} \) act by 0 for \( j \neq i \).


Mathematics -> Discrete Mathematics -> Combinatorics

Find, with proof, the number of positive integers whose base- $n$ representation consists of distinct digits with the property that, except for the leftmost digit, every digit differs by $\pm 1$ from some digit further to the left.  (Your answer should be an explicit function of $n$ in simplest form.)

Solution:
Let a $k$ -good sequence be a sequence of distinct integers $\{ a_i \}_{i=1}^k$ such that for all integers $2\le i \le k$ , $a_i$ differs from some preceding term by $\pm 1$ .
Lemma. Let $a$ be an integer.  Then there are $2^{k-1}$  $k$ -good sequences starting on $a$ , and furthermore, the terms of each of these sequences constitute a permutation of $k$ consecutive integers.
Proof. We induct on $k$ .  For $k=1$ , the lemma is trivially true.  Now, suppose the lemma holds for $k$ .  If $\{ a_i \}_{i=1}^{k+1}$ is a $(k+1)$ -good sequence, then $\{ a_i \}_{i=1}^k$ is a $k$ -good sequence which starts on $a$ , so it is a permutation of $k$ consecutive integers, say $m, \dotsc, M$ .  Then the only possibilities for $a_{k+1}$ are $m-1$ and $M+1$ ; either way, $\{ a_i \}_{i=1}^{k+1}$ constitutes a permutation of $k+1$ consecutive integers.  Since there are $2^k$ possible sequences $\{a_i\}_{i=1}^k$ , and 2 choices of $a_{k+1}$ for each of these sequences, it also follows that there are $2^k \cdot 2 = 2^{k+1}$  $(k+1)$ -good sequences which start on $a$ .  Thus the lemma holds by induction. $\blacksquare$
We now consider the number of desired positive integers with $k$ digits.  Evidently, $k$ must be less than or equal to $n$ .  We also note that the digits of such an integer must constitute a $k$ -good sequence.  Since the minimum of this sequence can be any of the digits $0, \dotsc, n-k$ , unless the minimum is 0 and is the first digit (in which case the only possible sequence is an increasing arithmetic sequence), and there are $2^{k-1}$  $k$ -good sequences up to translation, it follows that there are $(n-k+1) 2^{k-1}-1$ desired positive integers with $k$ digits.  Thus the total number of desired positive integers is \begin{align*} \sum_{k=1}^n \bigl[ (n-k+1) 2^{k-1}-1 \bigr] &= -n + \sum_{k=1}^n \sum_{j=k}^n 2^{k-1} = -n + \sum_{j=1}^n \sum_{k=1}^j 2^{k-1} \\ &= -n + \sum_{j=1}^n (2^k-1) = - 2n -1 + \sum_{j=0}^n 2^k, \end{align*} which is equal to $2^{n+1} - 2(n+1)$ , our answer. $\blacksquare$
  Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Answer: \[ 2^{n+1} - 2(n+1) \]


Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations

A certain state issues license plates consisting of six digits (from 0 through 9). The state requires that any two plates differ in at least two places. (Thus the plates $\boxed{027592}$ and $\boxed{020592}$ cannot both be used.) Determine, with proof, the maximum number of distinct license plates that the state can use.

Solution:
Consider license plates of $n$ digits, for some fixed $n$ , issued with the same criteria.
We first note that by the pigeonhole principle, we may have at most $10^{n-1}$ distinct plates.  Indeed, if we have more, then there must be two plates which agree on the first $n-1$ digits; these plates thus differ only on one digit, the last one.
We now show that it is possible to issue $10^{n-1}$ distinct license plates which satisfy the problem's criteria.  Indeed, we issue plates with all $10^{n-1}$ possible combinations for the first $n-1$ digit, and for each plate, we let the last digit be the sum of the preceding digits taken mod 10.  This way, if two plates agree on the first $n-1$ digits, they agree on the last digit and are thus the same plate, and if two plates differ in only one of the first $n-1$ digits, they must differ as well in the last digit.
It then follows that $10^{n-1}$ is the greatest number of license plates the state can issue.  For $n=6$ , as in the problem, this number is $10^5$ . $\blacksquare$

Answer: \[ 10^5 \]


Determine the largest integer $n$ such that there exist monic quadratic polynomials $p_{1}(x), p_{2}(x), p_{3}(x)$ with integer coefficients so that for all integers $i \in[1, n]$ there exists some $j \in[1,3]$ and $m \in \mathbb{Z}$ such that $p_{j}(m)=i$.

Solution: The construction for $n=9$ can be achieved with the polynomials $x^{2}+x+1, x^{2}+x+2$, and $x^{2}+5$. First we consider what kinds of polynomials we can have. Let $p(x)=(x+h)^{2}+k$. $h$ is either an integer or half an integer. Let $k=0$. If $h$ is an integer then $p(x)$ hits the perfect squares $0,1,4,9$, etc. If $h$ is half an integer, then let $k=1 / 4$. Then $p(x)$ hits the product of two consecutive integers, i.e. 0, $2,6,12$, etc. Assume there is a construction for $n=10$. In both of the cases above, the most a polynomial can hit out of 10 is 4, in the $0,1,4,9$ case. Thus $p_{1}$ must hit $1,2,5,10$, and $p_{2}$ and $p_{3}$ hit 3 integers each, out of $3,4,6,7,8,9$. The only ways we can hit 3 out of 7 consecutive integers is with the sequences $0,2,6$ or $0,1,4$. The only way a $0,2,6$ works is if it hits 3,5, and 9, which doesn't work since 5 was hit by $p_{2}$. Otherwise, $p_{2}$ is $0,1,4$, which doesn't work as $p_{2}$ hits 3,4, and 7, and $p_{3}$ must hit 6,8, and 9, which is impossible. Thus no construction for $n=10$ exists.

Answer: The largest integer \( n \) such that there exist monic quadratic polynomials \( p_{1}(x), p_{2}(x), p_{3}(x) \) with integer coefficients so that for all integers \( i \in[1, n] \) there exists some \( j \in[1,3] \) and \( m \in \mathbb{Z} \) such that \( p_{j}(m)=i \) is \( \boxed{9} \).


Does there exist a real $3 \times 3$ matrix $A$ such that \operatorname{tr}(\mathrm{A})=0$ and $A^{2}+A^{t}=I$ ? (tr $(\mathrm{A})$ denotes the trace of $A$, $A^{t}$ is the transpose of $A$, and $I$ is the identity matrix.)

Solution: The answer is NO. Suppose that \operatorname{tr}(\mathrm{A})=0$ and $A^{2}+A^{t}=I$. Taking the transpose, we have $$A=I-\left(A^{2}\right)^{t}=I-\left(A^{t}\right)^{2}=I-\left(I-A^{2}\right)^{2}=2 A^{2}-A^{4}$$ $$A^{4}-2 A^{2}+A=0$$ The roots of the polynomial $x^{4}-2 x^{2}+x=x(x-1)\left(x^{2}+x-1\right)$ are $0,1, \frac{-1 \pm \sqrt{5}}{2}$ so these numbers can be the eigenvalues of $A$; the eigenvalues of $A^{2}$ can be $0,1, \frac{1 \pm \sqrt{5}}{2}$. By \operatorname{tr}(A)=0$, the sum of the eigenvalues is 0 , and by \operatorname{tr}\left(A^{2}\right)=\operatorname{tr}\left(I-A^{t}\right)=3$ the sum of squares of the eigenvalues is 3 . It is easy to check that this two conditions cannot be satisfied simultaneously.

Answer: There does not exist a real $3 \times 3$ matrix $A$ such that $\operatorname{tr}(A) = 0$ and $A^2 + A^t = I$.


Let $P$ be a polynomial with integer coefficients such that $P(0)+P(90)=2018$. Find the least possible value for $|P(20)+P(70)|$.

Solution: First, note that $P(x)=x^{2}-3041$ satisfy the condition and gives $|P(70)+P(20)|=|4900+400-6082|=$ 782. To show that 782 is the minimum, we show $2800 \mid P(90)-P(70)-P(20)+P(0)$ for every $P$, since -782 is the only number in the range $[-782,782]$ that is congruent to 2018 modulo 2800. Proof: It suffices to show that $2800 \mid 90^{n}-70^{n}-20^{n}+0^{n}$ for every $n \geq 0$ (having $0^{0}=1$ ). Let $Q(n)=90^{n}-70^{n}-20^{n}+0^{n}$, then we note that $Q(0)=Q(1)=0, Q(2)=2800$, and $Q(3)=\left(9^{3}-\right.$ $\left.7^{3}-2^{3}\right) \cdot 10^{3}=378000=135 \cdot 2800$. For $n \geq 4$, we note that 400 divides $10^{4}$, and $90^{n}+0^{n} \equiv 70^{n}+20^{n}$ $(\bmod 7)$. Therefore $2800 \mid Q(n)$ for all $n$.

Answer: \[ 782 \]


Mathematics -> Algebra -> Algebra -> Algebraic Expressions

For each positive integer  $n$ , let \begin{align*} S_n &= 1 + \frac 12 + \frac 13 + \cdots + \frac 1n \\ T_n &= S_1 + S_2 + S_3 + \cdots + S_n \\ U_n &= \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}. \end{align*} Find, with proof, integers $0 < a,\ b,\ c,\ d < 1000000$ such that $T_{1988} = a S_{1989} - b$ and $U_{1988} = c S_{1989} - d$ .

Solution:
We note that for all integers $n \ge 2$ , \begin{align*} T_{n-1} &= 1 + \left(1 + \frac 12\right) + \left(1 + \frac 12 + \frac 13\right) + \ldots + \left(1 + \frac 12 + \frac 13 + \ldots + \frac 1{n-1}\right) \\  &= \sum_{i=1}^{n-1} \left(\frac {n-i}i\right) = n\left(\sum_{i=1}^{n-1} \frac{1}{i}\right) - (n-1) = n\left(\sum_{i=1}^{n} \frac{1}{i}\right) - n \\ &= n \cdot S_{n} - n .  \end{align*}
It then follows that \begin{align*} U_{n-1} &= \sum_{i=2}^{n} \frac{T_{i-1}}{i} = \sum_{i=2}^{n}\ (S_{i} - 1) = T_{n-1} + S_n - (n-1) - S_1 \\ &= \left(nS_n - n\right) + S_n - n = (n + 1)S_n - 2n . \end{align*}
If we let $n=1989$ , we see that $(a,b,c,d) = (1989,1989,1990, 2\cdot 1989)$ is a suitable solution. $\blacksquare$
Notice that it is also possible to use induction to prove the equations relating $T_n$ and $U_n$ with $S_n$ .
  Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Answer:
\[
(a, b, c, d) = (1989, 1989, 1990, 2 \cdot 1989)
\]


Mathematics -> Algebra -> Abstract Algebra -> Other

Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ such that, for all $x, y \in \mathbb{R}^{+}$ , \[f(xy + f(x)) = xf(y) + 2\]

Solution:
Make the following substitutions to the equation:
1. $(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2$
2. $(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4$
3. $(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf\biggl(1 + \frac{f(1)}{x}\biggr) + 2$
It then follows from (2) and (3) that $f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}$ , so we know that this function is linear for $x > 1$ . Substitute $f(x) = ax+b$ and solve for $a$ and $b$ in the functional equation; we find that $f(x) = x + 1 \forall x > 1$ .
Now, we can let $x > 1$ and $y \le 1$ . Since $f(x) = x + 1$ , $xy + f(x) > x > 1$ , so $f(xy + f(x)) = xy + x + 2 = xf(y) + 2$ . It becomes clear then that $f(y) = y + 1$ as well, so $f(x) = x + 1$ is the only solution to the functional equation.
~jkmmm3

Answer: \[ f(x) = x + 1 \]


Mathematics -> Algebra -> Algebra -> Equations and Inequalities

Mathematics -> Discrete Mathematics -> Logic

Find, with proof, all functions \(f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}\) such that \(f(x)^{2}-f(y) f(z)=x(x+y+z)(f(x)+f(y)+f(z))\) for all real \(x, y, z\) such that \(xyz=1\).

Solution: The answer is either \(f(x)=0\) for all \(x\) or \(f(x)=x^{2}-\frac{1}{x}\) for all \(x\). These can be checked to work. Now, I will prove that these are the only solutions. Let \(P(x, y, z)\) be the assertion of the problem statement. Lemma 1. \(f(x) \in\left\{0, x^{2}-\frac{1}{x}\right\}\) for all \(x \in \mathbb{R} \backslash\{0\}\). Proof. \(P(1,1,1)\) yields \(f(1)=0\). Then, \(P\left(x, 1, \frac{1}{x}\right)\) and \(P\left(1, x, \frac{1}{x}\right)\) yield \(f(x)^{2}=x\left(x+\frac{1}{x}+1\right)\left(f(x)+f\left(\frac{1}{x}\right)\right)\) and \(-f(x) f\left(\frac{1}{x}\right)=\left(x+\frac{1}{x}+1\right)\left(f(x)+f\left(\frac{1}{x}\right)\right)\). Thus, we have \(f(x)^{2}=-x f(x) f\left(\frac{1}{x}\right)\), so we have \(f(x)=0\) or \(f\left(\frac{1}{x}\right)=-\frac{f(x)}{x}\). Plugging in the latter into the first equation above gives us \(f(x)^{2}=x\left(x+\frac{1}{x}+1\right)\left(f(x)-\frac{f(x)}{x}\right)\) which gives us \(f(x)=0\) or \(f(x)=x^{2}-\frac{1}{x}\). This proves Lemma 1. Lemma 2. If \(f(t)=0\) for some \(t \neq 1\), then we have \(f(x)=0\) for all \(x\). Proof. \(P\left(x, t, \frac{1}{t x}\right)\) and \(P\left(t, x, \frac{1}{t x}\right)\) give us \(f(x)^{2}=x\left(x+\frac{1}{t x}+t\right)\left(f(x)+f\left(\frac{1}{t x}\right)\right)\) and \(-f(x) f\left(\frac{1}{t x}\right)=t\left(x+\frac{1}{t x}+t\right)\left(f(x)+f\left(\frac{1}{t x}\right)\right)\). Thus we have \(t f(x)^{2}=-x f(x) f\left(\frac{1}{t x}\right)\), so \(f(x)=0\) or \(f\left(\frac{1}{t x}\right)=-\frac{t}{x} f(x)\). Plugging in the latter into the first equation gives us \(f(x)^{2}=x\left(x+\frac{1}{t x}+t\right)\left(f(x)-\frac{t f(x)}{x}\right)\) which gives us either \(f(x)=0\) or \(f(x)=x\left(x+t+\frac{1}{t x}\right)\left(1-\frac{t}{x}\right)=x^{2}-\frac{1}{x}-\left(t^{2}-\frac{1}{t}\right)\). Note that since the ladder expression doesn't equal \(x^{2}-\frac{1}{x}\), since \(t \neq 1\), we must have that \(f(x)=0\). Thus, we have proved lemma 2. Combining these lemmas finishes the problem.

Answer: \[ f(x) = 0 \quad \text{or} \quad f(x) = x^2 - \frac{1}{x} \]


Mathematics -> Geometry -> Solid Geometry -> 3D Shapes

A convex polyhedron has $n$ faces that are all congruent triangles with angles $36^{\circ}, 72^{\circ}$, and $72^{\circ}$. Determine, with proof, the maximum possible value of $n$.

Solution: Answer: 36 Solution: Consider such a polyhedron with $V$ vertices, $E$ edges, and $F=n$ faces. By Euler's formula we have $V+F=E+2$. Next, note that the number of pairs of incident faces and edges is both $2E$ and $3F$, so $2E=3F$. Now, since our polyhedron is convex, the sum of the degree measures at each vertex is strictly less than $360=36 \cdot 10$. As all angle measures of the faces of our polyhedron are divisible by 36, the maximum degree measure at a given vertex is $36 \cdot 9=324$. On the other hand, the total degree measure at all vertices is the total degree measure over all faces, which is $180F$. Thus we have $180F \leq 324V$, or $10F \leq 18V$. Putting our three conditions together, we have $$ 10F \leq 18V=18(E+2-F)=9(2E)+36-18F=9(3F)+36-18F=9F+36 $$ Thus $F \leq 36$. $F=36$ is attainable by taking a 9-gon antiprism with a 9-gon pyramid attached on the top and the bottom. Thus the answer is 36.

Answer: \[ 36 \]


Mathematics -> Algebra -> Abstract Algebra -> Field Theory

Does there exist a field such that its multiplicative group is isomorphic to its additive group?

Solution: There exist no such field. Suppose that \(F\) is such a field and \(g: F^{*} \rightarrow F^{+}\) is a group isomorphism. Then \(g(1)=0\). Let \(a=g(-1)\). Then \(2 a=2 \cdot g(-1)=g\left((-1)^{2}\right)=g(1)=0\); so either \(a=0\) or char \(F=2\). If \(a=0\) then \(-1=g^{-1}(a)=g^{-1}(0)=1\); we have char \(F=2\) in any case. For every \(x \in F\), we have \(g\left(x^{2}\right)=2 g(x)=0=g(1)\), so \(x^{2}=1\). But this equation has only one or two solutions. Hence \(F\) is the 2-element field; but its additive and multiplicative groups have different numbers of elements and are not isomorphic.

Answer: There exist no such field.


Mathematics -> Discrete Mathematics -> Graph Theory

There are $n$ line segments on the plane, no three intersecting at a point, and each pair intersecting once in their respective interiors. Tony and his $2 n-1$ friends each stand at a distinct endpoint of a line segment. Tony wishes to send Christmas presents to each of his friends as follows: First, he chooses an endpoint of each segment as a "sink". Then he places the present at the endpoint of the segment he is at. The present moves as follows: - If it is on a line segment, it moves towards the sink. - When it reaches an intersection of two segments, it changes the line segment it travels on and starts moving towards the new sink. If the present reaches an endpoint, the friend on that endpoint can receive their present. Prove Tony can send presents to exactly $n$ of his $2 n-1$ friends.

Solution: Draw a circle that encloses all the intersection points between line segments and extend all line segments until they meet the circle, and then move Tony and all his friends to the circle. Number the intersection points with the circle from 1 to $2 n$ anticlockwise, starting from Tony (Tony has number 1). We will prove that the friends eligible to receive presents are the ones on even-numbered intersection points. First part: at most $n$ friends can receive a present. The solution relies on a well-known result: the $n$ lines determine regions inside the circle; then it is possible to paint the regions with two colors such that no regions with a common (line) boundary have the same color. The proof is an induction on $n$ : the fact immediately holds for $n=0$, and the induction step consists on taking away one line $\ell$, painting the regions obtained with $n-1$ lines, drawing $\ell$ again and flipping all colors on exactly one half plane determined by $\ell$. Now consider the line starting on point 1. Color the regions in red and blue such that neighboring regions have different colors, and such that the two regions that have point 1 as a vertex are red on the right and blue on the left, from Tony's point of view. Finally, assign to each red region the clockwise direction and to each blue region the anticlockwise direction. Because of the coloring, every boundary will have two directions assigned, but the directions are the same since every boundary divides regions of different colors. Then the present will follow the directions assigned to the regions: it certainly does for both regions in the beginning, and when the present reaches an intersection it will keep bordering one of the two regions it was dividing. To finish this part of the problem, consider the regions that share a boundary with the circle. The directions alternate between outcoming and incoming, starting from 1 (outcoming), so all even-numbered vertices are directed as incoming and are the only ones able to receive presents. Second part: all even-numbered vertices can receive a present. First notice that, since every two chords intersect, every chord separates the endpoints of each of the other $n-1$ chords. Therefore, there are $n-1$ vertices on each side of every chord, and each chord connects vertices $k$ and $k+n, 1 \leq k \leq n$. We prove a stronger result by induction in $n$ : let $k$ be an integer, $1 \leq k \leq n$. Direct each chord from $i$ to $i+n$ if $1 \leq i \leq k$ and from $i+n$ to $i$ otherwise; in other words, the sinks are $k+1, k+2, \ldots, k+n$. Now suppose that each chord sends a present, starting from the vertex opposite to each sink, and all presents move with the same rules. Then $k-i$ sends a present to $k+i+1, i=0,1, \ldots, n-1$ (indices taken modulo $2 n$ ). In particular, for $i=k-1$, Tony, in vertex 1 , send a present to vertex $2 k$. Also, the $n$ paths the presents make do not cross (but they may touch.) More formally, for all $i, 1 \leq i \leq n$, if one path takes a present from $k-i$ to $k+i+1$, separating the circle into two regions, all paths taking a present from $k-j$ to $k+j+1, j<i$, are completely contained in one region, and all paths taking a present from $k-j$ to $k+j+1, j>i$, are completely contained in the other region. The result is true for $n=1$. Let $n>1$ and assume the result is true for less chords. Consider the chord that takes $k$ to $k+n$ and remove it. Apply the induction hypothesis to the remaining $n-1$ lines: after relabeling, presents would go from $k-i$ to $k+i+2,1 \leq i \leq n-1$ if the chord were not there. Reintroduce the chord that takes $k$ to $k+n$. From the induction hypothesis, the chord intersects the paths of the presents in the following order: the $i$-th path the chord intersects is the the one that takes $k-i$ to $k+i, i=1,2, \ldots, n-1$. Then the presents cover the following new paths: the present from $k$ will leave its chord and take the path towards $k+1$; then, for $i=1,2, \ldots, n-1$, the present from $k-i$ will meet the chord from $k$ to $k+n$, move towards the intersection with the path towards $k+i+1$ and go to $k+i+1$, as desired. Notice that the paths still do not cross. The induction (and the solution) is now complete.

Answer: Tony can send presents to exactly \( n \) of his \( 2n-1 \) friends.


Mathematics -> Number Theory -> Prime Numbers

Call an ordered pair $(a, b)$ of positive integers fantastic if and only if $a, b \leq 10^{4}$ and $\operatorname{gcd}(a \cdot n!-1, a \cdot(n+1)!+b)>1$ for infinitely many positive integers $n$. Find the sum of $a+b$ across all fantastic pairs $(a, b)$.

Solution: We first prove the following lemma, which will be useful later. Lemma: Let $p$ be a prime and $1 \leq n \leq p-1$ be an integer. Then, $n!(p-1-n)!\equiv(-1)^{n-1}(\bmod p)$. Proof. Write $$\begin{aligned} n!(p-n-1)! & =(1 \cdot 2 \cdots n)((p-n-1) \cdots 2 \cdot 1) \\ & \equiv(-1)^{p-n-1}(1 \cdot 2 \cdots n)((n+1) \cdots(p-2)(p-1)) \quad(\bmod p) \\ & =(-1)^{n}(p-1)! \\ & \equiv(-1)^{n-1} \quad(\bmod p) \end{aligned}$$ (where we have used Wilson's theorem). This implies the result. Now, we begin the solution. Suppose that a prime $p$ divides both $a \cdot n!-1$ and $a \cdot(n+1)!+b$. Then, since $$-b \equiv a \cdot(n+1)!\equiv(n+1) \cdot(a \cdot n!) \equiv(n+1) \quad(\bmod p)$$ we get that $p \mid n+b+1$. Since we must have $n<p$ (or else $p \mid n$ !), we get that, for large enough $n$, $n=p-b-1$. However, by the lemma, $$a(-1)^{b-1} \equiv a \cdot b!(p-1-b)!=a \cdot b!n!\equiv b!\quad(\bmod p)$$ This must hold for infinitely many $p$, so $a=(-1)^{b-1} b$ !. This forces all fantastic pairs to be in form $((2 k-1)!, 2 k-1)$. Now, we prove that these pairs all work. Take $n=p-2 k$ for all large primes $p$. Then, we have $$\begin{aligned} a \cdot n! & \equiv(2 k-1)!(p-2 k)! \\ & \equiv(-1)^{2 k} \equiv 1 \quad(\bmod p) \\ a \cdot(n+1)! & \equiv(n+1) \cdot(a \cdot n!) \\ & \equiv(p-2 k+1) \cdot 1 \equiv-(2 k-1) \quad(\bmod p) \end{aligned}$$ so $p$ divides the gcd. The answer is $(1+1)+(6+3)+(120+5)+(5040+7)=5183$.

Answer: \[ 5183 \]


Mathematics -> Algebra -> Prealgebra -> Integers

Nine distinct positive integers summing to 74 are put into a $3 \times 3$ grid. Simultaneously, the number in each cell is replaced with the sum of the numbers in its adjacent cells. (Two cells are adjacent if they share an edge.) After this, exactly four of the numbers in the grid are 23. Determine, with proof, all possible numbers that could have been originally in the center of the grid.

Solution: Suppose the initial grid is of the format shown below: $$\left[\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]$$ After the transformation, we end with $$\left[\begin{array}{lll} a_{n} & b_{n} & c_{n} \\ d_{n} & e_{n} & f_{n} \\ g_{n} & h_{n} & i_{n} \end{array}\right]=\left[\begin{array}{ccc} b+d & a+c+e & b+f \\ a+e+g & b+d+f+h & c+e+i \\ d+h & g+e+i & f+h \end{array}\right]$$ Since $d \neq f, a_{n}=b+d \neq b+f=c_{n}$. By symmetry, no two corners on the same side of the grid may both be 23 after the transformation. Since $c \neq g, b_{n}=a+c+e \neq a+e+g=d_{n}$. By symmetry, no two central-edge squares sharing a corner may both be 23 after the transformation. Assume for the sake of contradiction that $e_{n}=23$. Because $a_{n}, c_{n}, g_{n}, i_{n}<e_{n}$, none of $a_{n}, c_{n}, g_{n}, i_{n}$ can be equal to 23 . Thus, 3 of $b_{n}, d_{n}, f_{n}, h_{n}$ must be 23 . WLOG assume $b_{n}=d_{n}=f_{n}=23$. Thus is a contradiction however, as $b_{n} \neq d_{n}$. Thus, $e_{n} \neq 23$. This leaves the case with two corners diametrically opposite and two central edge squares diametrically opposite being 23. WLOG assume $a_{n}=b_{n}=h_{n}=i_{n}=23$. Thus, $92=4 \cdot 23=a_{n}+b_{n}+h_{n}+i_{n}=(b+d)+(a+c+e)+(e+g+i)+(f+h)=(a+b+c+d+e+f+g+h+i)+e$. Since $a+b+c+d+e+f+g+h+i=74$, this means that $e=92-74=18$. One possible example of 18 working is $\left[\begin{array}{lll}4 & 16 & 2 \\ 6 & 18 & 7 \\ 1 & 17 & 3\end{array}\right]$. Thus the only possible value for the center is 18.

Answer: The only possible value for the center is \( 18 \).


Mathematics -> Algebra -> Algebra -> Polynomial Operations

Mathematics -> Number Theory -> Factorization

A polynomial $f \in \mathbb{Z}[x]$ is called splitty if and only if for every prime $p$, there exist polynomials $g_{p}, h_{p} \in \mathbb{Z}[x]$ with $\operatorname{deg} g_{p}, \operatorname{deg} h_{p}<\operatorname{deg} f$ and all coefficients of $f-g_{p} h_{p}$ are divisible by $p$. Compute the sum of all positive integers $n \leq 100$ such that the polynomial $x^{4}+16 x^{2}+n$ is splitty.

Solution: We claim that $x^{4}+a x^{2}+b$ is splitty if and only if either $b$ or $a^{2}-4 b$ is a perfect square. (The latter means that the polynomial splits into $(x^{2}-r)(x^{2}-s)$ ). Assuming the characterization, one can easily extract the answer. For $a=16$ and $b=n$, one of $n$ and $64-n$ has to be a perfect square. The solutions to this that are at most 64 form 8 pairs that sum to 64 (if we include 0), and then we additionally have 81 and 100. This means the sum is $64 \cdot 8+81+100=693$. Now, we move on to prove the characterization. Necessity. Take a prime $p$ such that neither $a^{2}-4 b$ nor $b$ is a quadratic residue modulo $p$ (exists by Dirichlet + CRT + QR). Work in $\mathbb{F}_{p}$. Now, suppose that $$x^{4}+a x^{2}+b=(x^{2}+m x+n)(x^{2}+s x+t)$$ Then, looking at the $x^{3}$-coefficient gives $m+s=0$ or $s=-m$. Looking at the $x$-coefficient gives $m(n-t)=0$. - If $m=0$, then $s=0$, so $x^{4}+a x^{2}+b=(x^{2}+n)(x^{2}+t)$, which means $a^{2}-4 b=(n+t)^{2}-4 n t=(n-t)^{2}$, a quadratic residue modulo $p$, contradiction. - If $n=t$, then $b=n t$ is a square modulo $p$, a contradiction. (The major surprise of this problem is that this suffices, which will be shown below.) Sufficiency. Clearly, the polynomial splits in $p=2$ because in $\mathbb{F}_{2}[x]$, we have $x^{4}+a x^{2}+b=(x^{2}+a x+b)^{2}$. Now, assume $p$ is odd. If $a^{2}-4 b$ is a perfect square, then $x^{4}+a x^{2}+b$ splits into $(x^{2}-r)(x^{2}-s)$ even in $\mathbb{Z}[x]$. If $b$ is a perfect square, then let $b=k^{2}$. We then note that - $x^{4}+a x^{2}+b$ splits in form $(x^{2}-r)(x^{2}-s)$ if $\left(\frac{a^{2}-4 k^{2}}{p}\right)=1$. - $x^{4}+a x^{2}+b$ splits in form $(x^{2}+r x+k)(x^{2}-r x+k)$ if $a=2 k-r^{2}$, or $\left(\frac{2 k-a}{p}\right)=1$. - $x^{4}+a x^{2}+b$ splits in form $(x^{2}+r x-k)(x^{2}-r x-k)$ if $a=-2 k-r^{2}$, or $\left(\frac{-2 k-a}{p}\right)=1$. Since $(2 k-a)(-2 k-a)=a^{2}-4 k^{2}$, it follows that at least one of these must happen.

Answer: \[ 693 \]


Mathematics -> Number Theory -> Congruences

Mathematics -> Algebra -> Prealgebra -> Other

Determine whether there exists a positive integer \(n\) for which \(g(n)>n^{0.999 n}\), where \(f(n), g(n)\) are the minimal positive integers such that \(1+\frac{1}{1!}+\frac{1}{2!}+\ldots+\frac{1}{n!}=\frac{f(n)}{g(n)}\).

Solution: We show that there does exist such a number \(n\). Let \(\varepsilon=10^{-10}\). Call a prime \(p\) special, if for certain \(k \in\{1,2, \ldots, p-1\}\) there exist at least \(\varepsilon \cdot k\) positive integers \(j \leq k\) for which \(p\) divides \(f(j)\). Lemma. There exist only finitely many special primes. Proof. Let \(p\) be a special prime number, and \(p\) divides \(f(j)\) for at least \(\varepsilon \cdot k\) values of \(j \in\{1,2, \ldots, k\}\). Note that if \(p\) divides \(f(j)\) and \(f(j+r)\), then \(p\) divides \((j+r)!\left(\frac{f(j+r)}{g(j+r)}-\frac{f(j)}{g(j)}\right)=1+(j+r)+(j+r)(j+r-1)+\ldots+(j+r) \ldots(j+2)\) that is a polynomial of degree \(r-1\) with respect to \(j\). Thus, for fixed \(j\) it equals to 0 modulo \(p\) for at most \(r-1\) values of \(j\). Look at our \(\geq \varepsilon \cdot k\) values of \(j \in\{1,2, \ldots, k\}\) and consider the gaps between consecutive \(j\) 's. The number of such gaps which are greater than \(2 / \varepsilon\) does not exceed \(\varepsilon \cdot k / 2\) (since the total sum of gaps is less than \(k\) ). Therefore, at least \(\varepsilon \cdot k / 2-1\) gaps are at most \(2 / \varepsilon\). But the number of such small gaps is bounded from above by a constant (not depending on \(k\) ) by the above observation. Therefore, \(k\) is bounded, and, since \(p\) divides \(f(1) f(2) \ldots f(k), p\) is bounded too. Now we want to bound the product \(g(1) g(2) \ldots g(n)\) (for a large integer \(n\) ) from below. Let \(p \leq n\) be a non-special prime. Our nearest goal is to prove that \(\nu_{p}(g(1) g(2) \ldots g(n)) \geq(1-\varepsilon) \nu_{p}(1!\cdot 2!\cdot \ldots \cdot n!)\). Partition the numbers \(p, p+1, \ldots, n\) onto the intervals of length \(p\) (except possibly the last interval which may be shorter): \(\{p, p+1, \ldots, 2 p-1\}, \ldots,\{p\lfloor n / p\rfloor, \ldots, n\}\). Note that in every interval \(\Delta=[a \cdot p, a \cdot p+k]\), all factorials \(x\) ! with \(x \in \Delta\) have the same \(p\)-adic valuation, denote it \(T=\nu_{p}((a p)!)\) We claim that at least \((1-\varepsilon)(k+1)\) valuations of \(g(x), x \in \Delta\), are equal to the same number \(T\). Indeed, if \(j=0\) or \(1 \leq j \leq k\) and \(f(j)\) is not divisible by \(p\), then \(\frac{1}{(a p)!}+\frac{1}{(a p+1)!}+\ldots+\frac{1}{(a p+j)!}=\frac{1}{(a p)!} \cdot \frac{A}{B}\) where \(A \equiv f(j)(\bmod p), B \equiv g(j)(\bmod p)\), so, this sum has the same \(p\)-adic valuation as \(1 /(a p)\) !, which is strictly less than that of the sum \(\sum_{i=0}^{a p-1} 1 / i\) !, that yields \(\nu_{p}(g(a p+j))=\nu_{p}((a p)!)\). Using this for every segment \(\Delta\), we get \(\nu_{p}(g(1) g(2) \ldots g(n)) \geq(1-\varepsilon) \nu_{p}(1!\cdot 2!\cdot \ldots \cdot n!)\). Now, using this for all non-special primes, we get \(A \cdot g(1) g(2) \ldots g(n) \geq(1!\cdot 2!\cdot \ldots \cdot n!)^{1-\varepsilon}\) where \(A=\prod_{p, k} p^{\nu_{p}(g(k))}, p\) runs over non-special primes, \(k\) from 1 to \(n\). Since \(\nu_{p}(g(k)) \leq \nu_{p}(k!)=\sum_{i=1}^{\infty}\left\lfloor k / p^{i}\right\rfloor \leq k\), we get \(A \leq\left(\prod_{p} p\right)^{1+2+\ldots+n} \leq C^{n^{2}}\) for some constant \(C\). But if we had \(g(n) \leq n^{0.999 n} \leq e^{n} n!^{0.999}\) for all \(n\), then \(\log (A \cdot g(1) g(2) \ldots g(n)) \leq O\left(n^{2}\right)+0.999 \log (1!\cdot 2!\cdot \ldots \cdot n!)<(1-\varepsilon) \log (1!\cdot 2!\cdot \ldots \cdot n!)\) for large \(n\), a contradiction.

Answer: There exists a positive integer \( n \) for which \( g(n) > n^{0.999n} \).


Let $N$ be the smallest positive integer for which $$x^{2}+x+1 \quad \text { divides } \quad 166-\sum_{d \mid N, d>0} x^{d}$$ Find the remainder when $N$ is divided by 1000.

Solution: Let $\omega=e^{2 \pi i / 3}$. The condition is equivalent to $$166=\sum_{d \mid N, d>0} \omega^{d}$$ Let's write $N=3^{d} n$ where $n$ is not divisible by 3. If all primes dividing $n$ are $1 \bmod 3$, then $N$ has a positive number of factors that are $1 \bmod 3$ and none that are $2 \bmod 3$, so $\sum_{d \mid N, d>0} \omega^{d}$ has nonzero imaginary part. Therefore $n$ is divisible by some prime that is $2 \bmod 3$. In this case, the divisors of $n$ are equally likely to be 1 or $2 \bmod 3$, so the sum is $$-\frac{1}{2} \tau(n)+d \tau(n)=\frac{2 d-1}{2} \tau(n)$$ Now, $2 \cdot 166=2^{2} \cdot 83$ and 83 is prime, so we must either have $d=42$, which forces $\tau(n)=4$, or $d=1$, which forces $\tau(n)=332$. The first cases yields a lower value of $N$, namely $3^{42} 2^{3}$. Now let's try to compute this mod 1000. This is clearly divisible by 8. Modulo $125,3^{5}=243 \equiv-7$, so $3^{20} \equiv 2401 \equiv 26$ and $3^{40} \equiv 676 \equiv 51$. Therefore $3^{42} 2^{3} \equiv 72 \cdot 51=3672 \bmod 125$. Since 672 is divisible by 8, this is our answer.

Answer: \[ N \equiv 672 \pmod{1000} \]


Mathematics -> Algebra -> Linear Algebra -> Matrices

An $n \times n$ complex matrix $A$ is called $t$-normal if $A A^{t}=A^{t} A$ where $A^{t}$ is the transpose of $A$. For each $n$, determine the maximum dimension of a linear space of complex $n \times n$ matrices consisting of t-normal matrices.

Solution: Answer: The maximum dimension of such a space is $\frac{n(n+1)}{2}$. The number $\frac{n(n+1)}{2}$ can be achieved, for example the symmetric matrices are obviously t-normal and they form a linear space with dimension $\frac{n(n+1)}{2}$. We shall show that this is the maximal possible dimension. Let $M_{n}$ denote the space of $n \times n$ complex matrices, let $S_{n} \subset M_{n}$ be the subspace of all symmetric matrices and let $A_{n} \subset M_{n}$ be the subspace of all anti-symmetric matrices, i.e. matrices $A$ for which $A^{t}=-A$. Let $V \subset M_{n}$ be a linear subspace consisting of t-normal matrices. We have to show that $\operatorname{dim}(V) \leq$ $\operatorname{dim}\left(S_{n}\right)$. Let $\pi: V \rightarrow S_{n}$ denote the linear map $\pi(A)=A+A^{t}$. We have $$\operatorname{dim}(V)=\operatorname{dim}(\operatorname{Ker}(\pi))+\operatorname{dim}(\operatorname{Im}(\pi))$$ so we have to prove that $\operatorname{dim}(\operatorname{Ker}(\pi))+\operatorname{dim}(\operatorname{Im}(\pi)) \leq \operatorname{dim}\left(S_{n}\right)$. Notice that $\operatorname{Ker}(\pi) \subseteq A_{n}$. We claim that for every $A \in \operatorname{Ker}(\pi)$ and $B \in V, A \pi(B)=\pi(B) A$. In other words, $\operatorname{Ker}(\pi)$ and $\operatorname{Im}(\pi)$ commute. Indeed, if $A, B \in V$ and $A=-A^{t}$ then $$(A+B)(A+B)^{t}=(A+B)^{t}(A+B) \Leftrightarrow$$ $$\Leftrightarrow A A^{t}+A B^{t}+B A^{t}+B B^{t}=A^{t} A+A^{t} B+B^{t} A+B^{t} B \Leftrightarrow$$ $$\Leftrightarrow A B^{t}-B A=-A B+B^{t} A \Leftrightarrow A\left(B+B^{t}\right)=\left(B+B^{t}\right) A \Leftrightarrow$$ $$\Leftrightarrow A \pi(B)=\pi(B) A$$ Our bound on the dimension on $V$ follows from the following lemma: Lemma. Let $X \subseteq S_{n}$ and $Y \subseteq A_{n}$ be linear subspaces such that every element of $X$ commutes with every element of $Y$. Then $$\operatorname{dim}(X)+\operatorname{dim}(Y) \leq \operatorname{dim}\left(S_{n}\right)$$ Proof. Without loss of generality we may assume $X=Z_{S_{n}}(Y):=\left\{x \in S_{n}: x y=y x \quad \forall y \in Y\right\}$. Define the bilinear map $B: S_{n} \times A_{n} \rightarrow \mathbb{C}$ by $B(x, y)=\operatorname{tr}(\mathrm{d}[\mathrm{x}, \mathrm{y}])$ where $[x, y]=x y-y x$ and $d=\operatorname{diag}(1, \ldots, n)$ is the matrix with diagonal elements $1, \ldots, n$ and zeros off the diagonal. Clearly $B(X, Y)=\{0\}$. Furthermore, if $y \in Y$ satisfies that $B(x, y)=0$ for all $x \in S_{n}$ then $\left.\operatorname{tr}(\mathrm{d}[\mathrm{x}, \mathrm{y}])=-\operatorname{tr}([\mathrm{d}, \mathrm{x}], \mathrm{y}]\right)=0$ for every $x \in S_{n}$. We claim that $\left\{[d, x]: x \in S_{n}\right\}=A_{n}$. Let $E_{i}^{j}$ denote the matrix with 1 in the entry $(i, j)$ and 0 in all other entries. Then a direct computation shows that $\left[d, E_{i}^{j}\right]=(j-i) E_{i}^{j}$ and therefore $\left[d, E_{i}^{j}+E_{j}^{i}\right]=$ $(j-i)\left(E_{i}^{j}-E_{j}^{i}\right)$ and the collection $\left\{(j-i)\left(E_{i}^{j}-E_{j}^{i}\right)\right\}_{1 \leq i<j \leq n} \operatorname{span} A_{n}$ for $i \neq j$. It follows that if $B(x, y)=0$ for all $x \in S_{n}$ then $\operatorname{tr}(\mathrm{yz})=0$ for every $z \in A_{n}$. But then, taking $z=\bar{y}$, where $\bar{y}$ is the entry-wise complex conjugate of $y$, we get $0=\operatorname{tr}(\mathrm{y} \overline{\mathrm{y}})=-\operatorname{tr}\left(\mathrm{y} \overline{\mathrm{y}}^{\mathrm{t}}\right)$ which is the sum of squares of all the entries of $y$. This means that $y=0$. It follows that if $y_{1}, \ldots, y_{k} \in Y$ are linearly independent then the equations $$B\left(x, y_{1}\right)=0, \quad \ldots, \quad B\left(x, y_{k}\right)=0$$ are linearly independent as linear equations in $x$, otherwise there are $a_{1}, \ldots, a_{k}$ such that $B\left(x, a_{1} y_{1}+\ldots+\right.$ $\left.a_{k} y_{k}\right)=0$ for every $x \in S_{n}$, a contradiction to the observation above. Since the solution of $k$ linearly independent linear equations is of codimension $k$, $$\begin{gathered}\operatorname{dim}\left(\left\{x \in S_{n}:\left[x, y_{i}\right]=0, \text { for } i=1, \ldots, k\right\}\right) \leq \\ \leq \operatorname{dim}\left(x \in S_{n}: B\left(x, y_{i}\right)=0 \text { for } i=1, \ldots, k\right)=\operatorname{dim}\left(S_{n}\right)-k\end{gathered}$$ The lemma follows by taking $y_{1}, \ldots, y_{k}$ to be a basis of $Y$. Since $\operatorname{Ker}(\pi)$ and $\operatorname{Im}(\pi)$ commute, by the lemma we deduce that $$\operatorname{dim}(V)=\operatorname{dim}(\operatorname{Ker}(\pi))+\operatorname{dim}(\operatorname{Im}(\pi)) \leq \operatorname{dim}\left(S_{n}\right)=\frac{n(n+1)}{2}$$

Answer:
\[
\frac{n(n+1)}{2}
\]


In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?

Solution:
We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$ , who do not know each other.)
Base case: $n = 4$ is obvious.
Inductive step: Suppose in a party with $k$ people (with $k \ge 4$ ), at least $(k-3)$ people know everyone else. Consider a party with $(k+1)$ people. Take $k$ of the people (leaving another person, $A$ , out) and apply the inductive step to conclude that at least $(k-3)$ people know everyone else in the $k$ -person group, $G$ .
Now suppose that everyone in the group $G$ knows each other. Then take $3$ of these people and $A$ to deduce that $A$ knows a person $B \in G$ , which means $B$ knows everyone else. Then apply the inductive step on the remaining $k$ people (excluding $B$ ) to find $(k-3)$ people out of them that know everyone else (including $B$ , of course). Then these $(k-3)$ people and $B$ , which enumerate $(k-2)$ people, know everyone else.
Suppose that there exist two people $B, C \in G$ who do not know each other. Because $k-3 \ge 1$ , there exist at least one person in $G$ , person $D$ , who knows everyone else in $G$ . Now, take $A, B, C, D$ and observe that because $B, C$ do not know each other, either $A$ or $D$ knows everyone else of $A, B, C, D$ (by the problem condition), so in particular $A$ and $D$ know each other. Then apply the inductive step on the remaining $k$ people (excluding $D$ ) to find $(k-3)$ people out of them that know everyone else (including $D$ , of course). Then these $(k-3)$ people and $D$ , which enumerate $(k-2)$ people, know everyone else.
This completes the inductive step and thus the proof of this stronger result, which easily implies that at least $1982 - 3 = \boxed{1979}$ people know everyone else.

Answer: \[ 1979 \]


Mathematics -> Geometry -> Plane Geometry -> Triangulations

Mathematics -> Algebra -> Intermediate Algebra -> Other

Find all integers $n \ge 3$ such that among any $n$ positive real numbers $a_1$ , $a_2$ , $\dots$ , $a_n$ with \[\max(a_1, a_2, \dots, a_n) \le n \cdot \min(a_1, a_2, \dots, a_n),\] there exist three that are the side lengths of an acute triangle.

Solution:
Without loss of generality, assume that the set $\{a\}$ is ordered from least to greatest so that the bounding condition becomes $a_n \le n \cdot a_1.$ Now set $b_i \equiv \frac{a_i}{a_1},$ and since a triangle with sidelengths from $\{a\}$ will be similar to the corresponding triangle from $\{b\},$ we simply have to show the existence of acute triangles in $\{b\}.$ Note that $b_1 = 1$ and for all $i$ , $b_i \le n.$
Now three arbitrary sidelengths $x$ , $y$ , and $z$ , with $x \le y \le z,$ will form a valid triangle if and only if $x+y>z.$ Furthermore, this triangle will be acute if and only if $x^2 + y^2 > z^2.$ However, the first inequality can actually be inferred from the second, since $x+y>z \longrightarrow x^2 + y^2 +2xy > z^2$ and $2xy$ is trivially greater than $0.$ So we just need to find all $n$ such that there is necessarily a triplet of $b$ 's for which $b_i^2 + b_j^2 > b_k^2$ (where $b_i < b_j < b_k$ ).
We now make another substitution: $c_i \equiv b_i ^2.$ So $c_1 = 1$ and for all $i$ , $c_i \le n^2.$ Now we examine the smallest possible sets $\{c\}$ for small $n$ for which the conditions of the problem are not met. Note that by smallest, we mean the set whose greatest element is as small as possible. If $n=3$ , then the smallest possible set, call it $\{s_3\},$ is trivially $\{1,1,2\}$ , since $c_1$ and $c_2$ are obviously minimized and $c_3$ follows as minimal. Using this as the base case, we see inductively that in general $\{s_n\}$ is the set of the first $n$ Fibonacci numbers. To show this note that if $\{s_n\} = \{F_0, F_1, ... F_n\}$ , then $\{s_{n+1}\} = \{F_0, F_1, ... F_n, c_{n+1}\}.$ The smallest possible value for $c_{n+1}$ is the sum of the two greatest values of $\{s_n\}$ which are $F_{n-1}$ and $F_n$ . But these sum to $F_{n+1}$ so $\{s_{n+1}\} = \{F_0, F_1, ... F_{n+1}\}$ and our induction is complete.
Now since we know that the Fibonacci set is the smallest possible set which does not satisfy the conditions of the problem, then any set $\{c\}$ whose greatest term is less than $F_{n-1}$ must satisfy the conditions. And since $\{c\}$ is bounded between $1$ and $n^2$ , then the conditions of the problem are met if and only if $F_{n-1} > n^2$ . The first $n$ for which this restriction is satisfied is $n=13$ and the exponential behavior of the Fibonacci numbers ensure that every $n$ greater than $13$ will also satisfy this restriction. So the final solution set is $\boxed{\{n \ge 13\}}$ .

Answer: \(\{n \ge 13\}\)


For any positive real numbers \(a\) and \(b\), define \(a \circ b=a+b+2 \sqrt{a b}\). Find all positive real numbers \(x\) such that \(x^{2} \circ 9x=121\).

Solution: Since \(a \circ b=(\sqrt{a}+\sqrt{b})^{2}\), we have \(x^{2} \circ 9x=(x+3\sqrt{x})^{2}\). Moreover, since \(x\) is positive, we have \(x+3\sqrt{x}=11\), and the only possible solution is that \(\sqrt{x}=\frac{-3+\sqrt{53}}{2}\), so \(x=\frac{31-3\sqrt{53}}{2}\).

Answer: \frac{31-3\sqrt{53}}{2}


Problem
Find all pairs of primes $(p,q)$ for which $p-q$ and $pq-q$ are both perfect squares.

Solution:
We first consider the case where one of $p,q$ is even. If $p=2$ , $p-q=0$ and $pq-q=2$ which doesn't satisfy the problem restraints. If $q=2$ , we can set $p-2=x^2$ and $2p-2=y^2$ giving us $p=y^2-x^2=(y+x)(y-x)$ . This forces $y-x=1$ so $p=2x+1\rightarrow 2x+1=x^2+2 \rightarrow x=1$ giving us the solution $(p,q)=(3,2)$ .
 Now assume that $p,q$ are both odd primes. Set $p-q=x^2$ and $pq-q=y^2$ so $(pq-q)-(p-q)=y^2-x^2 \rightarrow p(q-1)$  $=(y+x)(y-x)$ . Since $y+x>y-x$ , $p | (x+y)$ . Note that $q-1$ is an even integer and since $y+x$ and $y-x$ have the same parity, they both must be even. Therefore, $x+y=pk$ for some positive even integer $k$ . On the other hand, $p>p-q=x^2 \rightarrow p>x$ and $p^2-p>pq-q=y^2 \rightarrow p>y$ . Therefore, $2p>x+y$ so $x+y=p$ , giving us a contradiction.
 Therefore, the only solution to this problem is $(p,q)=(3,2)$ .
 ~BennettHuang

Answer: \((p, q) = (3, 2)\)


Find all the polynomials of degree 1 with rational coefficients that satisfy the condition that for every rational number $r$, $f(r)$ is rational, and for every irrational number $r$, $f(r)$ is irrational.

Solution: Let $f(x)$ be a polynomial of degree $n$ such that $f(r) \in \mathbb{Q}$ for every $r \in \mathbb{Q}$. For distinct rational numbers $r_{0}, r_{1}, \ldots, r_{n}$, where $n=\operatorname{deg} f(x)$, let us define the polynomial $$g(x)= c_{0}\left(x-r_{1}\right)\left(x-r_{2}\right) \ldots\left(x-r_{n}\right)+c_{1}\left(x-r_{0}\right)\left(x-r_{2}\right) \ldots\left(x-r_{n}\right)+\ldots +c_{i}\left(x-r_{0}\right) \ldots\left(x-r_{i-1}\right)\left(x-r_{i+1}\right) \ldots\left(x-r_{n}\right)+\ldots +c_{n}\left(x-r_{0}\right)\left(x-r_{1}\right) \ldots\left(x-r_{n-1}\right)$$ where $c_{0}, c_{1}, \ldots, c_{n}$ are real numbers. Suppose that $g\left(r_{i}\right)=f\left(r_{i}\right), i=0,1, \ldots, n$. Since $g\left(r_{i}\right)=c_{i}\left(r_{i}-r_{0}\right) \ldots\left(r_{i}-r_{i-1}\right)\left(r_{i}-r_{i+1}\right) \ldots\left(r_{i}-r_{n}\right)$ then $$c_{i}=\frac{g\left(r_{i}\right)}{\left(r_{i}-r_{0}\right) \ldots\left(r_{i}-r_{i-1}\right)\left(r_{i}-r_{i+1}\right)}=\frac{f\left(r_{i}\right)}{\left(r_{i}-r_{0}\right) \ldots\left(r_{i}-r_{i-1}\right)\left(r_{i}-r_{i+1}\right)}$$ Clearly $c_{i}$ is a rational number for $i=0,1, \ldots, n$ and therefore the coefficients of $g(x)$ are rational. The polynomial $g(x)$ defined with coefficients $c_{0}, c_{1}, \ldots, c_{n}$ coincides with $f(x)$ in $n+1$ points and both polynomials $f$ and $g$ have degree $n$. It follows that for every real $x, f(x)=g(x)$. Therefore the coefficients of $f(x)$ are rational. Thus if a polynomial satisfies the conditions, all its coefficients are rational. It is easy to see that all the polynomials of first degree with rational coefficients satisfy the conditions of the problem and polynomials of degree 0 do not satisfy it. Let us prove that no other polynomials exist. Suppose that $f(x)=a_{0}+a_{1} x+\ldots+a_{n} x^{n}$ is a polynomial with rational coefficients and degree $n \geq 2$ that satisfies the conditions of the problem. We may assume that the coefficients of $f(x)$ are integers, because the sets of solutions of equations $f(x)=r$ and $a f(x)=a r$, where $a$ is an integer, coincide. Moreover let us denote $g(x)=a_{n}^{n-1} f\left(\frac{x}{a_{n}}\right) \cdot g(x)$ is a polynomial with integer coefficients whose leading coefficient is 1 . The equation $f(x)=r$ has an irrational root if and only if $g(x)=a_{n}^{n-1} r$ has an irrational root. Therefore, we may assume WLOG that $f(x)$ has integer coefficients and $a_{n}=1$. Let $r$ be a sufficiently large prime, such that $$r>\max \left\{f(1)-f(0), x_{1}, x_{2}, \ldots, x_{k}\right\}$$ where $\left\{x_{1}, x_{2}, \ldots, x_{k}\right\}$ denote the set of all real roots of $f(x)-f(0)-x=0$. Putting $q=r+f(0) \in \mathbb{Z}$, and considering the equality $$f(x)-q=f(x)-f(0)-r$$ we then have $$f(1)-q=f(1)-f(0)-r<0$$ On the other hand, by the choice of $r$, we have $$f(1)-q=f(r)-f(0)-r>0$$ It follows from the intermediate value theorem that there is at least one real root $p$ of $f(x)-q=0$ between 1 and $r$. Notice that from the criterion theorem for rational roots, the possible positive rational roots of the equation $f(x)-q=f(x)-f(0)-r=0$ are 1 and $r$. Thus $p$ must be irrational.

Answer: There are no polynomials of degree 1 with rational coefficients that satisfy the given conditions.