Added solutions

Advanced/Pidgeonhole Problems/main.tex

@@ -19,7 +19,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Easy}
 	Is it possible to cover an equilateral triangle with two smaller equilateral triangles? Why or why not?
 
 
@@ -34,8 +33,7 @@
 
 
 
-	\problem{}
+	\problem{}<divisibledifference>
-	\note{Difficulty: Hard}
 	You are given $n + 1$ integers. \par
 	Prove that at least two of them have a difference divisible by $n$.
 
@@ -56,7 +54,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Easy}
 	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}
@@ -72,7 +69,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Easy}
 	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}
@@ -86,7 +82,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Normal}
 	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}
@@ -102,7 +97,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Normal}
 	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}
@@ -122,7 +116,6 @@
 
 
 	\problem{}<line_threecolor>
-	\note{Difficulty: Normal}
 	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}
@@ -194,7 +187,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Easy}
 	Every point on a plane is painted black or white. Show that there exist two points in the plane that have the same color and are located exactly one foot away from each other.
 
 	\begin{solution}
@@ -206,38 +198,50 @@
 
 
 
-
+	\problem{}<multipleofones>
-	\problem{}
-	\note{Difficulty: Normal}
-	Each point on a circle is colored either black or white. Prove that there exist three equally spaced points of the same color.
-
-	\begin{solution}
-		This problem is exactly the same as \ref{line_threecolor}.
-	\end{solution}
-
-	\vfill
-
-
-
-	\problem{}
-	\note{Difficulty: Hard}
 	Let n be an integer not divisible by $2$ and $5$. Show that n has a multiple consisting entirely of ones.
 
+	\begin{solution}
+		Let $a_1 = 1, a_2 = 11, a_3 = 111$, and so on.
+
+		\vspace{2mm}
+
+		Consider the sequence $a_1, ..., a_{n+1}$. \par
+		By \ref{divisibledifference}, there exist $a_i$ and $a_j$ in this list so that $a_i - a_j \equiv 0 \pmod{n}$. \par
+
+		\vspace{2mm}
+
+		Since $a_i$ and $a_j$ are both made of ones, $a_i - a_j = 11...111 \times 10^j$. \par
+		$n$ must be a factor of either $11...111$ or $10^j$. \par
+		Since $n$ isn't divisible by $2$ or $5$, $10^j$ cannot be divisible by $n$, so $11...111$ must be a factor of $n$.
+	\end{solution}
+
+
 	\vfill
 
 
 
 
 	\problem{}
-	\note{Difficulty: Brutal}
 	Prove that for any $n > 1$, there exists an integer made of only sevens and zeros that is divisible by $n$.
 
+
+	\begin{solution}
+		If $n$ is not divisible by $2$ or $5$, the solution to this problem is the same as \ref{multipleofones}: \par
+		just multiply the number of all ones by $7$.
+
+		\vspace{2mm}
+
+		If $n$ is divisible by $2$ or $5$, set $p$ to the largest power of $2$ or $5$ in $n$. \par
+		Multiply the above number by $10^p$ to get a number that satisfies the conditions above.
+	\end{solution}
+
+
 	\vfill
 
 
 
 	\problem{}
-	\note{Difficulty: Hard}
 	Choose $n + 1$ integers between $1$ and $2n$. Show that at least two of these are co-prime.
 
 	\vfill
@@ -245,7 +249,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Hard}
 	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}
@@ -267,8 +270,21 @@
 
 
 	\problem{}
-	\note{Difficulty: Hard}
+	Show that it is always possible to choose a subset of the set of integers $\{a_1, a_2, ... , a_n\}$ so that the sum of the numbers in the subset is divisible by $n$.
-	Show that it is always possible to choose a subset of the set of integers $a_1, a_2, ... , a_n$ so that the sum of the numbers in the subset is divisible by $n$.
+
+	\begin{solution}
+		Let $\{a_1^\prime, a_2^\prime, ..., a_n^\prime\}$ be this set mod $n$. \par
+		If any $a_i^\prime$ is zero, we're done: $\{a_i^\prime\}$ satisfies the problem.
+
+		\vspace{2mm}
+
+		If none are zero, consider the set $\{a_1^\prime,~ a_1^\prime + a_2^\prime,~ ...,~ a_1^\prime + a_2^\prime + ... + a_n^\prime\}$. \par
+		If any element of this set is zero, we're done.
+
+		\vspace{2mm}
+
+		If zero is not in this set, we have $n$ numbers with $n-1$ possible remainders. Therefore, at least two elements in this set must be equivalent mod $n$. If we subtract these two elements, we get a sum divisible by $n$.
+	\end{solution}
 
 	\vfill
 
@@ -276,16 +292,25 @@
 
 
 	\problem{}
-	\note{Difficulty: Hard}
 	Show that there exists a positive integer divisible by $2013$ that has $2014$ as its last four digits.
 
+	\begin{solution}
+		Let $n$ be this number. \par
+		First, note that $n - 2013$ has $0001$ as its last four digits. \par
+
+		\vspace{2mm}
+
+		So, we see that $n - 2013 = 2013k \equiv 1 \pmod{1000}$. \par
+		Of course, $k$ = $2013^{-1} \pmod{1000}$, which exists because 2013 and 1000 are coprime. \par
+		And finally, we see that $n = 2013 \times (k + 1)$.
+	\end{solution}
+
 	\vfill
 
 
 
 
 	\problem{}
-	\note{Difficulty: Normal}
 	Let $n$ be an odd number. Let $a_1, a_2, ... , a_n$ be a permutation of the numbers $1, 2, ... , n$. \par
 	Show that $(a_1 - 1) \times (a_2 - 2) \times ... \times (a_n - n)$ is even.
 
@@ -303,7 +328,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Hard}
 	A stressed-out student consumes at least one espresso every day of a particular year, drinking $500$ overall. Show the student drinks exactly $100$ espressos on some consecutive sequence of days.
 
 	\begin{solution}
@@ -366,7 +390,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Hard}
 	Show that there are either three mutual acquaintances or four mutual strangers at a party with ten or more people.
 
 	\vfill
@@ -374,7 +397,6 @@
 
 
 	\problem{}
-	\note{Difficulty: Brutal}
 	Given a table with a marked point, $O$, and with $2013$ properly working watches put down on the table, prove that there exists a moment in time when the sum of the distances from $O$ to the watches' centers is less than the sum of the distances from $O$ to the tips of the watches' minute hands.
 
 	\vfill