\documentclass[12pt]{article}
\usepackage{amsmath}
\title{Math Wrangle Problems}
\author{American Mathematics Competitions}
\date{\today}
\begin{document}
\maketitle
\begin{enumerate}
\item Find the sum of all positive two-digit integers that are
divisible by each of their digits.
Let $a$ represent the tens digit and $b$ the units digit of an
integer with the required property. Then $10a+b$ must be divisible
by both $a$ and $b$. It follows that $b$ must be divisible by $a$,
and that $10a$ must be divisible by $b$. The former condition
requires that $b=ka$ for some positive integer $k$, and the latter
condition implies that $k=1$ or $k=2$ or $k=5$. Thus the requested
two-digit numbers are 11, 22, 33, \ldots, 99, 12, 24, 36, 48, and
15. Their sum is $11 \cdot 45 + 12 \cdot 10 + 15 = 630$.
\item An equilateral triangle is inscribed in the ellipse whose
equation is ${x^2 + 4y^2 = 4}$. One vertex of the triangle is $(0,1)$,
one altitude is contained in the $y$-axis. Find the length of each
side of this equilateral triangle.
Let the other two vertices of the triangle be $(x,y)$ and $(-x,y)$,
with $x>0$. Then the line through $(0,1)$ and $(x,y)$ forms a
120-degree angle with the positive $x$-axis, and its slope is
$\tan(120^{\circ}) = -\sqrt{3}$. Therefore, the line's equation is
$y=-\sqrt{3}x+1$. Substituting this into the equation of the ellipse
and simplifying yields
\[
13x^2-8\sqrt{3}x=0 \qquad \text{or} \qquad x=\frac{8\sqrt{3}}{13}.
\]
The triangle has sides of length $2x=(16\sqrt{3})/13 =
\sqrt{768/169}$.
\item A fair die is rolled four times. Find the probability that each of
the final three rolls is at least as large as the roll preceding it.
Any particular outcome of the four rolls has probability $1/6^4$.
Given the values of four rolls, there is exactly one order that
satisfies the requirement. It therefore suffices to count all the
sets of values that could be produced by four rolls, allowing
duplicate values. This is equivalent to counting the number of ways
to put four balls into six boxes labeled 1 through 6. By thinking of
4 balls and 5 dividers to separate the six boxes, this can be seen
to be $\binom{9}{4}=126$. The requested probability is thus
$126/6^4=7/72$.
\vskip.1in\centerline{\textbf{OR}}\vskip.1in
Let $a_1$, $a_2$, $a_3$, and $a_4$ be the sequence of values rolled,
and consider the difference between the last and the first: If
$a_4-a_1=0$, then there is 1 possibility for $a_2$ and $a_3$, and 6
possibilities for $a_1$ and $a_4$. If $a_4-a_1=1$, then there are 3
possibilities for $a_2$ and $a_3$, and 5 possibilities for $a_1$ and
$a_4$. In general, if $a_4-a_1=k$, then there are $6-k$
possibilities for $a_1$ and $a_4$, while the number of possibilities
for $a_2$ and $a_3$ is the same as the number of sets of 2 elements,
with repetition allowed, that can be chosen from a set of $k+1$
elements. This is equal to the number of ways to put 2 balls in
$k+1$ boxes, or $\binom{k+2}{2}$. Thus there are $\sum_{k=0}^{5}
\binom{k+2}{2}(6-k)=126$ sequences of the type requested, so the
probability is $126/6^4=7/72$.
\item
Three of the vertices of a cube are $P=(7,12,10)$, $Q=(8,8,1)$,
and $R=(11,3,9)$. What is the surface area of the cube?
% Rick Parris
% rparris@exeter.edu
% o2 packet item 44
Notice that $PQR$ is an equilateral triangle, because
$PQ=QR=RP=7\sqrt2$. This implies that each edge of the cube is 7
units long. Hence the surface area of the cube is $6(7^2)=294$.
\item
Find the integer that is closest to
$\displaystyle\;{1000\sum_{n=3}^{10000} {1\over n^2-4}}$.
% Leo Schneider
% leo@jcvaxa.jcu.edu
% 02 packet item 47
Because $\displaystyle{
{1\over n^2-4}={1\over4}\left({1\over n-2}-{1\over n+2}\right)
}$, the series telescopes, and it follows that
\begin{align*}
1000\sum_{n=3}^{10000} {1\over n^2-4} &=
250\left({1\over1}+{1\over2}+{1\over3}+{1\over4}
-{1\over9999}-{1\over10000}-{1\over10001}-{1\over10002}\right)\\
&=250+125+{250\over3}+{250\over4}
-{250\over9999}-{250\over10000}-{250\over10001}-{250\over10002}\\
&=520+{5\over6}-r
\end{align*}
where the positive number $r$ is less
than $1/3$. Thus the requested integer is 521.
\item
Let $\cal S$ be the set $\{1,2,3,\ldots,10\}$. Let $n$ be the
number of sets of two non-empty disjoint subsets of $\cal S$.
({\it Disjoint sets\/} are defined as sets that have no common
elements.) What is $n$\,?
% Chris Jeuell --- Department of Computer Science, Cornell University
% chj2@cornell.edu
%02 packet item 6
Let $k$ be the number of elements of $\cal S$, and let $A$ and $B$
be two empty jars into which elements of $\cal S$ will be placed
to create two disjoint subsets. For each element $x$ in $\cal S$,
there are three possibilities: place $x$ in $A$, place $x$ in $B$,
or place $x$ in neither $A$ nor $B$. Thus the number of ordered
pairs of disjoint subsets $(A,B)$ is $3^k$. However, this counts
the pairs where $A$ or $B$ is empty. Note that for $A$ to be
empty, there are two possibilities for each element $x$ in $\cal
S$: place $x$ in $B$, or do not place $x$ in $B$. The number of
pairs for which $A$ or $B$ is empty is thus $2^k+2^k-1=2^{k+1}-1$.
Since interchanging $A$ and $B$ does not yield a different set of
subsets, there are ${1 \over 2} (3^k-2^{k+1}+1)={1 \over2}
(3^k+1)-2^k$ sets. When $k=10$, $n={3^{10}+1\over 2}
-2^{10}=28501$.
\end{enumerate}
\end{document}