Newline fixes

Advanced/Pidgeonhole Problems/main.tex

@@ -24,7 +24,7 @@
 
 
 	\begin{solution}
-		In order to completely cover an equilateral triangle, the two smaller triangles must cover all three vertices. Since the longest length of an equilateral triangle is one of its sides, a smaller triangle cannot cover more than one vertex. Therefore, we cannot completely cover the triangle with two smaller copies. \\
+		In order to completely cover an equilateral triangle, the two smaller triangles must cover all three vertices. Since the longest length of an equilateral triangle is one of its sides, a smaller triangle cannot cover more than one vertex. Therefore, we cannot completely cover the triangle with two smaller copies. \par
 
 		\textcolor{gray}{\textit{Bonus question:}} Can you cover a square with three smaller squares?
 	\end{solution}
@@ -36,15 +36,15 @@
 
 	\problem{}
 	\note{Difficulty: Hard}
-	You are given $n + 1$ integers. \\
+	You are given $n + 1$ integers. \par
 	Prove that at least two of them have a difference divisible by $n$.
 
 	\begin{solution}
-		$n~|~(a-b) \iff a \equiv b \pmod{n}$ \\
+		$n~|~(a-b) \iff a \equiv b \pmod{n}$ \par
 
-		Let $i_0 ... i_{n+1}$ be our set of integers. If we pick $i_0 ... i_{n+1}$ so that no two have a difference divisible by $n$, we must have $i_0 \not\equiv i_k \pmod{n}$ for all $1 \leq k \leq n+1$. There are $n$ such $i_k$, and there are $n$ equivalence classes mod $n$. \\
+		Let $i_0 ... i_{n+1}$ be our set of integers. If we pick $i_0 ... i_{n+1}$ so that no two have a difference divisible by $n$, we must have $i_0 \not\equiv i_k \pmod{n}$ for all $1 \leq k \leq n+1$. There are $n$ such $i_k$, and there are $n$ equivalence classes mod $n$. \par
 
-		Therefore, either, $i_1 ... i_{n+1}$ must cover all equivalence classes mod $n$ (implying that $i_0 \equiv i_k \pmod{n}$ for some k), or there exist two elements in $i_1 ... i_{n+1}$ that are equivalent mod $n$. \\
+		Therefore, either, $i_1 ... i_{n+1}$ must cover all equivalence classes mod $n$ (implying that $i_0 \equiv i_k \pmod{n}$ for some k), or there exist two elements in $i_1 ... i_{n+1}$ that are equivalent mod $n$. \par
 
 		In either case, we can find $a, b$ so that $a \equiv b \pmod{n}$, which implies that $n$ divides $a-b$.
 	\end{solution}
@@ -60,9 +60,9 @@
 	You have an $8 \times 8$ chess board with two opposing corner squares cut off. You also have a set of dominoes, each of which is the size of two squares. Is it possible to completely cover the the board with dominos, so that none overlap nor stick out?
 
 	\begin{solution}
-		A domino covers two adjacent squares. Adjacent squares have different colors. \\
+		A domino covers two adjacent squares. Adjacent squares have different colors. \par
 
-		If you remove two opposing corners of a chessboard, you remove two squares of the same color, and you're left with $32$ of one and $30$ of the other. \\
+		If you remove two opposing corners of a chessboard, you remove two squares of the same color, and you're left with $32$ of one and $30$ of the other. \par
 
 		Since each domino must cover two colors, you cannot cover the modified board.
 	\end{solution}
@@ -76,7 +76,7 @@
 	The ocean covers more than a half of the Earth's surface. Prove that the ocean has at least one pair of antipodal points.
 
 	\begin{solution}
-	Let $W$ be the set of wet points, and $W^c$ the set of points antipodal to those in $W$. $W$ and $W^c$ each contain more than half of the points on the earth. The set of dry points, $D$, contains less than half of the points on the earth. Therefore, $W^c \not\subseteq D$. \\
+	Let $W$ be the set of wet points, and $W^c$ the set of points antipodal to those in $W$. $W$ and $W^c$ each contain more than half of the points on the earth. The set of dry points, $D$, contains less than half of the points on the earth. Therefore, $W^c \not\subseteq D$. \par
 
 	\textcolor{gray}{\textit{Note:}} This solution isn't very convincing. However, it is unlikely that the students know enough to provide a fully rigorous proof.
 	\end{solution}
@@ -90,9 +90,9 @@
 	There are $n > 1$ people at a party. Prove that among them there are at least two people who have the same number of acquaintances at the gathering. (We assume that if A knows B, then B also knows A)
 
 	\begin{solution}
-		Assume that every attendee knows a different number of people. There is only one way this may happen: the most popular person knows $n-1$ people (that is, everyone but himself), the second-most popular knows $n-2$, etc. The least-popular person must then know $0$ people. \\
+		Assume that every attendee knows a different number of people. There is only one way this may happen: the most popular person knows $n-1$ people (that is, everyone but himself), the second-most popular knows $n-2$, etc. The least-popular person must then know $0$ people. \par
 
-		This is impossible, since we know that someone must know $n-1$. \\
+		This is impossible, since we know that someone must know $n-1$. \par
 		(Remember, ``knowing'' must be mutual.)
 	\end{solution}
 
@@ -106,13 +106,13 @@
 	Pick five points in $\mathbb{R}^2$ with integral coordinates. Show that two of these form a line segment that has an integral midpoint.
 
 	\begin{solution}
-		Let $e, o$ represent even and odd integers. \\
-		There are four possible classes of points: $(e, e)$, $(o, o)$, $(e, o)$, $(o, e)$. \\
+		Let $e, o$ represent even and odd integers. \par
+		There are four possible classes of points: $(e, e)$, $(o, o)$, $(e, o)$, $(o, e)$. \par
 
-		$\text{midpoint}(a, b) = (\frac{a_x + b_x}{2}, \frac{a_y + b_y}{2})$. If $a_x + b_x$ and $a_y + b_y$ are both even, the midpoint of points $a$ and $b$ will have integer coordinates. \\
+		$\text{midpoint}(a, b) = (\frac{a_x + b_x}{2}, \frac{a_y + b_y}{2})$. If $a_x + b_x$ and $a_y + b_y$ are both even, the midpoint of points $a$ and $b$ will have integer coordinates. \par
 
-		Since we pick five points from four classes, at least two must come from the same class. \\
-		$e + e = e$ and $o + o = e$, so the midpoint between two points of the same class must have integral coordinates. \\
+		Since we pick five points from four classes, at least two must come from the same class. \par
+		$e + e = e$ and $o + o = e$, so the midpoint between two points of the same class must have integral coordinates. \par
 
 	\end{solution}
 
@@ -126,7 +126,7 @@
 	Every point on a line is painted black or white. Show that there exist three points of the same color where one is the midpoint of the line segment formed by the other two.
 
 	\begin{solution}
-		This is a proof by contradiction. We will try to construct a set of points where three points have such an arrangement. \\
+		This is a proof by contradiction. We will try to construct a set of points where three points have such an arrangement. \par
 
 		We know that some two points on the line will have the same color:
 
@@ -164,7 +164,7 @@
 			\end{tikzpicture}
 		\end{center}
 
-		Our original assumption also implies that the center point is white. \\
+		Our original assumption also implies that the center point is white. \par
 		This, however, creates a line of equidistant white points:
 
 		\begin{center}
@@ -249,7 +249,7 @@
 	Choose $n + 1$ integers between $1$ and $2n$. Show that you must select two numbers $a$ and $b$ such that $a$ divides $b$.
 
 	\begin{solution}
-		Split the the set $\{1, ..., 2n\}$ into classes defined by each integer's greatest odd divisor. There will be $n$ classes since there are $\frac{k}{2}$ odd numbers between $1$ and $n$. Because we pick $n + 1$ numbers, at least two will come from the same class---they will be divisible. \\
+		Split the the set $\{1, ..., 2n\}$ into classes defined by each integer's greatest odd divisor. There will be $n$ classes since there are $\frac{k}{2}$ odd numbers between $1$ and $n$. Because we pick $n + 1$ numbers, at least two will come from the same class---they will be divisible. \par
 
 		For example, if $n = 5$, our classes are
 		\begin{itemize}
@@ -286,12 +286,12 @@
 
 	\problem{}
 	\note{Difficulty: Normal}
-	Let $n$ be an odd number. Let $a_1, a_2, ... , a_n$ be a permutation of the numbers $1, 2, ... , n$. \\
+	Let $n$ be an odd number. Let $a_1, a_2, ... , a_n$ be a permutation of the numbers $1, 2, ... , n$. \par
 	Show that $(a_1 - 1) \times (a_2 - 2) \times ... \times (a_n - n)$ is even.
 
 	\begin{solution}
-		If $n$ is odd, there will be $m$ even and $m + 1$ odd numbers between $1$ and $n$. \\
-		Therefore, if we match each $a_n$ with an integer in $[1, ..., n]$, we will have to match at least one odd number with an odd number. \\
+		If $n$ is odd, there will be $m$ even and $m + 1$ odd numbers between $1$ and $n$. \par
+		Therefore, if we match each $a_n$ with an integer in $[1, ..., n]$, we will have to match at least one odd number with an odd number. \par
 
 		The difference of two odd numbers is even, so the product above will have at least one factor of two.
 	\end{solution}
@@ -349,13 +349,13 @@
 			\end{tikzpicture}
 		\end{center}
 
-		If there exists a sequence of days where the student drinks exactly 100 espressos, we must have at least one ``block'' (in orange, above) of 100 espressos that both begins and ends on a ``clean break'' between days. \\
+		If there exists a sequence of days where the student drinks exactly 100 espressos, we must have at least one ``block'' (in orange, above) of 100 espressos that both begins and ends on a ``clean break'' between days. \par
 
-		There are $499$ ``breaks'' between $500$ espressos. \\
-		In a year, there are $364$ clean breaks. This leaves $499 - 364 = 135$ ``dirty'' breaks. \\
-		We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \\
+		There are $499$ ``breaks'' between $500$ espressos. \par
+		In a year, there are $364$ clean breaks. This leaves $499 - 364 = 135$ ``dirty'' breaks. \par
+		We therefore have $135$ places to start a block on a dirty break, and $135$ places to end a block on a dirty break. This gives us a maximum of $270$ dirty blocks. \par
 
-		However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \\
+		However, there are $401$ possible blocks, since we can start one at the $1^{\text{st}}, 2^{\text{nd}}, ..., 401^{\text{st}}$ espresso. \par
 
 		Out of $401$ blocks, a maximum of $270$ can be dirty. We are therefore guaranteed at least $131$ clean blocks. This completes the problem---each clean block represents a set of consecutive, whole days during which exactly 100 espressos were consumed.