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{\Large {\bf Using Place Value Cards}} 
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==============
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\noindent
{\bf Problem:}  Use place value cards to evaluate $456_{seven}+44_{seven}$.

\smallskip

\noindent
{\bf Solution:}
We begin with two place value cards.  The first card represents
$456_{seven}$ and the second represents $44_{seven}$. 

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\begin{center}
\epsfxsize 3in
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\epsffile{pictures/pcplus1.eps}
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\begin{center}
\epsfxsize 3in
\epsfysize 1.5in
\epsffile{pictures/pcplus2.eps}
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Since we are adding, we will need to combine the poker chips onto one place
value card.  This card is shown below.

\begin{center}
\epsfxsize 3in
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\epsffile{pictures/pcplus3.eps}
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Since we are working in base seven, no place can have more than six poker
chips.  Working from right to left, we see that there is a group of seven
poker chips in the one's place.  We exchange seven of these chips for one
chip in the seven's place.  The result is shown below.

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\epsfxsize 3in
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\epsffile{pictures/pcplus4.eps}
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Next we will need to exchange seven of the chips in the sevens place for
one chip in the forty-nine's place.  The result is shown below.

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\epsfxsize 3in
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\epsffile{pictures/pcplus5.eps}
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Since no place has more than six chips, we have arrived at the final
answer: $456_{seven}+44_{seven} = 533_{seven}$.
 
({\em NOTE: The students should make exchanges from right to left so that
they can be certain the algorithm will terminate after one pass.  If
students suggest that they make exchanges from left to right or in random
order, ask them to add something similar to $99+99+99+99+99+99+99+99+99+99+99$ 
(base ten) by making exchanges from left to right.  Then ask them to compare 
the different algorithms.})


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==============
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\noindent
{\bf Problem:} Use place value cards to evaluate $12_{four} \times
321_{four}$.

\smallskip

\noindent
{\bf Solution:}  Here we will need to begin with six place value cards
(because six$=12_{four}$).  Each of the six cards will need to represent
$321_{four}$. 

\begin{center}
\epsfxsize 3in
\epsfysize 1.5in
\mbox{\epsffile{pictures/pcmult1.eps}} \epsffile{pictures/pcmult1.eps} \\
\epsfxsize 3in
\epsfysize 1.5in
\mbox{\epsffile{pictures/pcmult1.eps}} \epsffile{pictures/pcmult1.eps} \\
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcmult1.eps}} \epsffile{pictures/pcmult1.eps} 
\end{center}

Since multiplication is repeated addition, and addition is the union of
disjoint sets, we will need to combine all of the poker chips on one card.

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcmult2.eps}} 
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Because we are doing multiplication in base four, no place may have more
than three poker chips.  Working from right to left, we begin by exchanging
four chips in the ones place for one chip in the fours place.

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\mbox{\epsffile{pictures/pcmult3.eps}} 
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Next we exchange twelve chips in the fours place for three 
chips in the siexteens place.

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\epsfxsize 3in
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\mbox{\epsffile{pictures/pcmult4.eps}} 
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Then we exchange twenty chips in the sixteens place for five chips in the
sixty-fours place.

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\mbox{\epsffile{pictures/pcmult5.eps}} 
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At this point, we see that we need another place since the sixty-fours
place has more than three chips.  So we add the the two hundred fifty-sixes
place and exchange four chips in the sixty-fours place for one chip in the
two-hundred fifty-sixes place.  

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\epsfxsize 4in
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\mbox{\epsffile{pictures/pcmult6.eps}} 
\end{center}

We now see that we have made all possible exchanges.  Therefore $21_{four}
\times 321_{four} = 11112_{four}$.



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==============
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\noindent
{\bf Problem:}  Use place value cards to evaluate $541_{seven}-63_{seven}$.

\smallskip

\noindent
{\bf Solution:} You only need one place value card for subtraction.  Begin
with the place value card that represents $541_{seven}$

\begin{center}
\epsfxsize 3in
\epsfysize 1.5in
\mbox{\epsffile{pictures/pcsub1.eps}} 
\end{center}

We need to remove three chips from the ones place.  Since we do not have
three chips in the ones place, we will need to exchange one chip in the
sevens place for seven chips in the ones place. 

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcsub2.eps}} 
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We can now remove three chips from the ones place.

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcsub3.eps}} 
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Next we will need to remove six chips from the sevens place.  Since there
are not six chips in the sevens place, we will need to exchange one chip in
the forty-nines place for seven chips in the sevens place.

\begin{center}
\epsfxsize 3in
\epsfysize 1.5in
\mbox{\epsffile{pictures/pcsub4.eps}} 
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We can now remove six chips from the sevens place.

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcsub5.eps}} 
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Now that we have removed three chips from the ones place and six chips from
the sevens place, we see theat $541_{seven}-63_{seven} = 455_{seven}$.

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==============
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\noindent
{\bf Problem:}  
Use place value cards to evaluate $2231_{four} \div 3_{four}$.  

\smallskip

\noindent
{\bf Solution:}  We begin with a place value card that represents
$2231_{four}$ and three blank cards since we are dividing by three.   

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcdiv1a.eps}} 
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\epsffile{pictures/pcdiv1b.eps} & 
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\epsffile{pictures/pcdiv1b.eps} \\ 
\multicolumn{2}{c}{
\epsfysize 1in
\hspace*{1in}
\epsffile{pictures/pcdiv1b.eps} 
\hspace*{1in}}
\end{tabular}
}
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We begin in the sixty-fours column.  We want to evenly distribute the chips
in the sixty-fours column of the first place value card to the sixty-fours
columns of the blank place value cards.  We cannot do this because there
are only two chips and we have three cards.  Therefore we will need to
exchange the two chips in the sixtyfours place for eight chips in the
sixteens place.


\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcdiv2a.eps}}
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\framebox{
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\epsffile{pictures/pcdiv1b.eps} &
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\epsffile{pictures/pcdiv1b.eps} \\
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\epsfysize 1in
\hspace*{1in}
\epsffile{pictures/pcdiv1b.eps}
\hspace*{1in}}
\end{tabular}
}
\end{center}

Now we need to evenly distribute the ten chips in the sixteens place to the
sixteens place of the three blank cards.  We can place three chips in
sixteens place of each of the three cards.  This will leave us with one
chip in the sixteens place, three chips in the fours place, and one chip
in the ones place of the original card.   

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcdiv3a.eps}}
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\framebox{
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\epsfysize 1in
\epsffile{pictures/pcdiv2b.eps} &
\epsfysize 1in
\epsffile{pictures/pcdiv2b.eps} \\
\multicolumn{2}{c}{
\epsfysize 1in
\hspace*{1in}
\epsffile{pictures/pcdiv2b.eps}
\hspace*{1in}}
\end{tabular}
}
\end{center}

We cannot evenly distribute the one chip that remains in the sixteens place 
among the three cards, so we will exchange it for four chips in the fours
place.  

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\epsfxsize 3in
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\mbox{\epsffile{pictures/pcdiv4a.eps}}
\end{center}

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\framebox{
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\epsffile{pictures/pcdiv2b.eps} &
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\epsffile{pictures/pcdiv2b.eps} \\
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\epsfysize 1in
\hspace*{1in}
\epsffile{pictures/pcdiv2b.eps}
\hspace*{1in}}
\end{tabular}
}
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We now wish to evenly distribute the seven chips in the fours column into
the fours columns of the three cards. We can place two chips in the fours
column of each of the three cards.  This will leave us with one chip in the
fours column and one chip in the ones column.

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcdiv5a.eps}}
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\framebox{
\begin{tabular}{cc}
\epsfysize 1in
\epsffile{pictures/pcdiv3b.eps} &
\epsfysize 1in
\epsffile{pictures/pcdiv3b.eps} \\
\multicolumn{2}{c}{
\epsfysize 1in
\hspace*{1in}
\epsffile{pictures/pcdiv3b.eps}
\hspace*{1in}}
\end{tabular}
}
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Once again, we cannot evenly distribute the one chip that remains in the
fours column among the three cards.  Therefore we will exchange it for four
chips in the ones column.

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcdiv6a.eps}}
\end{center}

\begin{center}
\framebox{
\begin{tabular}{cc}
\epsfysize 1in
\epsffile{pictures/pcdiv3b.eps} &
\epsfysize 1in
\epsffile{pictures/pcdiv3b.eps} \\
\multicolumn{2}{c}{
\epsfysize 1in
\hspace*{1in}
\epsffile{pictures/pcdiv3b.eps}
\hspace*{1in}}
\end{tabular}
}
\end{center}

Finally, we wish to evenly distribute the chips in the ones column into the
ones columns of the three cards.  We can place one chip in the ones column
of each of the three cards.  This will leave us with two chips in the ones
column of our original card.

\begin{center}
\epsfxsize 3in
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\mbox{\epsffile{pictures/pcdiv7a.eps}}
\end{center}

\begin{center}
\framebox{
\begin{tabular}{cc}
\epsfysize 1in
\epsffile{pictures/pcdiv4b.eps} &
\epsfysize 1in
\epsffile{pictures/pcdiv4b.eps} \\
\multicolumn{2}{c}{
\epsfysize 1in
\hspace*{1in}
\epsffile{pictures/pcdiv4b.eps}
\hspace*{1in}}
\end{tabular}
}
\end{center}

We cannot evenly distribute the two chips which remain in the ones
column among the three cards.  Moreover, we cannot make any more exchanges,
so we have reached the end of the algorithm.
Since each of the three cards represents $321_{four}$ and the original card
represents $2_{four}$, it follows that $2231_{four} \div 3_{four} =
321_{four} \text{ R } 2_{four}$. 

\smallskip

{\bf Notes on Division:}  In my experience, division was the most difficult
operation for the students to understand.  There were two main difficulties
they encountered.  First, some of them wanted to represent the divisior on a 
place value card. That is, in their first attempt to solve the previous
problem they had one card representing $2231_{four}$ and one card
that representing $3_{four}$.  At this point they became stuck, for obvious
reasons.  The divisor needs to be represented with empty cards.  

\smallskip
The second
difficulty occured when the divisor has more than one digit.   Even students
who understood problems similar to $2231_{four} \div 3_{four}$ often became
confused when trying to solve problems similar to $3131_{four} \div
12_{four}$.  Some did not know where to begin, and others wanted to have twelve
empty cards instead of six.  They were still having difficulty differentiating 
between numbers and numerals.  There were also some problems when the
remainder was a multi-digit numeral.  The remainder for $3131_{four} \div
12_{four}$ is five, which is represented $11_{four}$.  When they finally
obtained the place value card that had five chips in the ones place, they
saw that they could not evenly distribute the five chips among the six
cards.  Moreover, they could not make any more exchanges that would make it
possible to distribute more chips.  However, many of them forgot that $5$
is not a numeral in base four.  These students wrote $3131_{four} \div
12_{four} = 210_{four} \text{ R } 5$.  When the algorithm was complete,
they needed to make sure that all of the cards represented numerals in the
appropriate base.  Since the base was four in this example, no card should
have more than three chips in any place.  The correct answer is
$3131_{four} \div 12_{four} = 210_{four} \text{ R } 11_{four}$.
\end{document}
