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\begin{document}
\noindent{\color{blue}{\Large\bf Homework 1 \hfill 21-127 Concepts of Math \hfill due 09.06.2012}\\
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\noindent{\color{blue}\large\bf Name:} % WRITE YOUR NAME HERE
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{\color{blue}\large\bf Section:} % WRITE YOUR RECITATION SECTION HERE
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{\color{blue}\large\bf Collaborators:} % LIST ANYONE YOU WORKED WITH HERE
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\begin{problem}[The First Rule Of Logic Club Is \ldots\hfill (15 pts)]
To join Logic Club, you must decide to {\it always} tell the truth or {\it always} lie. Members of Logic Club know who lies and who soothsays. I do not belong to Logic Club, but I encounter three members of Logic Club on the street. They make the following statements:
\begin{itemize}
\item {\bf Jack:} ``All three of us are liars.''
\item {\bf Tyler:} ``Exactly two of us are liars.''
\item {\bf Chuck:} ``Jack and Tyler are liars.''
\end{itemize}
Who is telling the truth and who is lying? Are there multiple possibilities?\vspace{0.3cm}\\
{\bf Explain} your answer and how you arrived at your conclusion.
\vspace{0.2cm}\\{\color{blue}\noindent\rule{\linewidth}{0.03cm}\vspace{0.3cm}}
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\end{problem}
\begin{problem}[Integrrrrrs \hfill(15 pts)]
Consider the variable equation $6x+15y=93$. We want to find some {\it integral} solutions; that is, we want to find values of $x$ and $y$ that are both {\it integers} (natural numbers, zero, and negative natural numbers) that satisfy the equation.
\begin{enumerate}[{\bf\color{blue}(a)}]
\item Find a solution where both $x$ and $y$ are positive integers. Describe {\bf how} you came up with this solution.
\item Find a solution where one of the values, $x$ or $y$, is positive and the other is negative. Again, describe {\bf how} you came up with this solution.
\item How {\bf many} solutions do you think there are? Try to write down a {\bf characterization} of all possible solutions, or {\bf describe} how you might find them all.
\end{enumerate}
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\end{problem}
\begin{problem}[Tetrominohno!\hfill(20 pts)]
Consider a standard $8\times 8$ chessboard. Remove two squares from the lower-left corner, and remove two squares from the upper-right corner.\vspace{0.3cm}\\
Can you tile the remaining squares with T-shaped tetrominoes?\vspace{0.3cm}\\
(Tetrominoes are similar to dominoes but they have {\bf four} blocks, like a Tetris piece.)\vspace{0.3cm}\\
If so, {\bf how}? If not, {\bf why} not? {\bf Explain} your answer as if you were trying to convince a friend!
\vspace{0.2cm}\\{\color{blue}\noindent\rule{\linewidth}{0.03cm}\vspace{0.3cm}}
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\end{problem}
\begin{problem}[Quadratic Spoofula\hfill(15 pts)]
Let $a,b,c$ be real numbers, with $a\neq 0$. What is wrong with the following proposed ``proof'' that $-\frac{b}{2a}$ is a solution to the equation $ax^2+bx+c=0$?
\begin{quote}
Let $x$ and $y$ be solutions to the equation.

Subtracting $ay^2+by+c=0$ from $ax^2+bx+c=0$ yields $a(x+y)(x-y)+b(x-y)=0$.

Hence, $a(x+y)+b=0$, and so $x+y=-\frac{b}{a}$.

Since $x$ and $y$ were {\it any} solutions, we may repeat this computation with $x=y$.

Thus, $2x=-\frac{b}{a}$.

Therefore, $x=-\frac{b}{2a}$ is a solution.
\end{quote}
Be as {\bf specific} as possible about the error(s). Explain {\bf why} they are problems. ({\bf Don't} just disagree with the conclusion.)\vspace{0.3cm}\\
Then, choose {\bf specific values} of $a,b,c$ and demonstrate that $x=-\frac{b}{2a}$ is {\bf not} a solution to the equation.
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\end{problem}
\begin{problem}[Sum Thing\hfill(15 pts)]
For every natural number $n$, define the expression $S_n$ to be this summation:
\[ S_n =  \sum_{k=1}^n \frac{1}{k(k+1)}\]
\begin{enumerate}[{\bf\color{blue}(a)}]
\item Find a {\bf simpler expression} for $S_n$ that does {\bf not} involve a summation.\vspace{0.3cm}\\
{\bf Hint:} $\displaystyle{\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}}$\vspace{0.3cm}\\
{\bf Show} your steps to derive your expression.
\item {\bf Show} that your expression holds true for the value $n=1$.
\item {\bf Suppose} that your expression holds true for the value $n=m$. Use this assumption to {\bf show} that your expression also holds true for the value $n=m+1$.
\end{enumerate}
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\end{problem}
\begin{problem}[The Cutest Infestation Problem Ever\hfill(20 pts)]
We have several piles of koala bears in our apartment. In an attempt to disperse them, we remove exactly one koala bear from {\bf each} pile and place all of those koalas into one {\bf new} pile.\vspace{0.3cm}\\
For example, if we started with koala piles of sizes 1, 4, and 4, and applied the operation once, we would then end up with koala piles of sizes 3, 3, and 3:
\[
\begin{tabular}{ccccccc}
&{\huge$\cancel{\cdot}$}&{\huge$\cancel{\cdot}$}&&&&\\
&{\huge$\cdot$}&{\huge$\cdot$}&&{\huge$\cdot$}&{\huge$\cdot$}&{\huge$\cdot$}\\
&{\huge$\cdot$}&{\huge$\cdot$}&$\rightarrow$&{\huge$\cdot$}&{\huge$\cdot$}&{\huge$\cdot$}\\
{\huge$\cancel{\cdot}$}&{\huge$\cdot$}&{\huge$\cdot$}&&{\huge$\cdot$}&{\huge$\cdot$}&{\huge$\cdot$}
\end{tabular}
\]
If we started with piles of size 3 and 4, we would end up with piles of size 2, 3, and 2:
\[
\begin{tabular}{cccccc}
&{\huge$\cancel{\cdot}$}&&&&\\
{\huge$\cancel{\cdot}$}&{\huge$\cdot$}&&&{\huge$\cdot$}&\\
{\huge$\cdot$}&{\huge$\cdot$}&$\rightarrow$&{\huge$\cdot$}&{\huge$\cdot$}&{\huge$\cdot$}\\
{\huge$\cdot$}&{\huge$\cdot$}&&{\huge$\cdot$}&{\huge$\cdot$}&{\huge$\cdot$}
\end{tabular}
\]\vspace{0.3cm}\\It {\bf is} possible that we do this operation {\bf exactly once} and end up with the {\bf exact same pile sizes} as we started with (where the {\bf order} of the piles is irrelevant; only the {\bf sizes} matter).\vspace{0.3cm}\\
Identify all of the starting situations that have this property. {\bf Explain} why they are the {\bf only} ones.\vspace{0.3cm}\\
{\bf Hint:} An example of a starting situation with this property is when we have just one pile of size 1. We do the operation and again obtain one pile of size 1. Bingo.\vspace{0.3cm}\\
{\bf Hint 2:} Be sure to also explain why your situations are the {\bf only} ones that work. How can we be sure you didn't miss some situations?
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\end{problem}
\end{document}