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\centerline{MONT 106Q -- Mathematical Thinking}
\centerline{Solutions for Final Exam Review Problems}
\centerline{December 7, 2016}
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{\it Some Practice Mathematical Questions}
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I. 
\item{A)}  Express the base-10 number $231$ in base-2.  
\item{} {\it Answer:}  $11100111$.
\item{B)}  (Note:  we did not discuss this, but you should be
able to see how it works if you think carefully about how we
add in our base-10 number system.)
How do you add in base-2?  Describe it in words first.  Then, for instance, what is 
$$101011101 + 100000111$$
(Express the sum first in base-2, then check your work by converting everything
to base-10 and adding that way.)
\item{} {\it Solution:}  The idea is that you align the numbers on the right
and add digit by digit just as in base-10. 
If the corresponding base-2 digits in the numbers you are adding 
are $0,0$, then that digit in the sum is $0$;
if the corresponding base-2 digits are $1,0$ or $0,1$, then that digit in 
the sum is $1$.  The main difference is if the corresponding base-2 digits
in the two numbers are $1,1$, then the sum has a $0$ in that digit and 
you carry a $1$ into the next digit.  Just as in base-10 addition, you think of any 
digits that get carried as added to the digit in the first number, and then you
follow the rules as above.  This means, for instance, that if the two numbers
both have a 1 in some digit, and you are carrying a 1 into that digit, then you 
get a 1 in that digit in the sum, {\it and} you carry a 1 into the next digit.  
For instance:
$$101011101 + 100000111 =  101100100$$
Check: $101011101$ is $1 + 4 + 8 + 16 + 64 + 256 = 349$
and $100000111$ is $1 + 2 + 4 + 256 = 263$, so in base-10, 
$349 + 263 = 612$.  Now $612 = 512 + 64 + 32 + 4$.   
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II. Compute ``the Egyptian way''
\item{A)} $67 \times 103$
\item{} {\it Solution:}  
$$\eqalign{(*) 1 \times 103 &= 103\cr
           (*) 2 \times 103 &= 206\cr
           4 \times 103 &= 412\cr
           8 \times 103 &= 824\cr
          16 \times 103 &= 1648\cr
          32 \times 103 &= 3296\cr
         (*) 64 \times 103 &= 6592\cr}$$
Then using the starred entries, $67 \times 103 = 103 + 206 + 6592 = 6901$.

\item{B)} $83 \div 14$  (For this one, don't worry about trying to follow
a systematic procedure when you get to the fractional part;
just get a sum of a whole number and a legal
sum of unit fractions any way you can!)  You can refer to a $2/n$ table
like the one from the Rhind papyrus on page 95 of {\it The Crest of the Peacock}
or the handout we discussed in class.  I would give you a copy of that
table for a question of this sort.
\item{} {\it Solution:}  We compute with 14 to get 83:
$$\eqalign{1 \times 14 &= 14\cr
           2 \times 14 &= 28\cr
           4 \times 14 &= 56\cr}$$
We can stop there since doubling again gives $112 > 83$.  Now
$83 - 56 = 27$, so ${83 \over 14} = 5 + {13\over 14}$.  There are any number of
ways to proceed from here, yielding different sums of distinct unit fractions that 
add up to ${13\over 14}$.  One way:
$${13\over 14} = {12\over 14} + {1\over 14} = {6\over 7} + {1\over 14}$$
Now we can think ${6\over 7} = {2\over 7} + {2\over 7} + {2\over 7}$.
From the $2/n$ table, this gives 
$${1\over 4} + {1\over 28} + {1\over 4} + {1\over 28} + {1\over 4} + {1\over 28}$$
Adding two of the ${1\over 4}$'s gives ${1\over 2}$.  Adding two of the ${1\over 28}$'s
gives ${1\over 14}$.  That ${1\over 14}$ plus the ${1\over 14}$ from the ${13\over 14}$
gives ${1\over 7}$.  So the final answer (this way) is
$$5 + {1 \over 2} + {1\over 4} + {1\over 7} + {1\over 28}$$
This satisfies the Egyptian requirements because it's a sum of distinct unit fractions.
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III. 
\item{A)}  Why is the Maya number system called a mixed-base system? Explain.
\item{} {\it Solution:}  It's mixed-base because it represents numbers using
$1, 20, 18\times 20, 18\times 20^2, \ldots$.  It's mainly base 20, but
with the ``wrinkle'' of the 18 starting in the third digit.  
\item{B)}  How would a Maya scribe represent the number $561$ (write it in
left-to-right format)?
\item{} {\it Solution:}  $561 = 1 \times 360 + 10\times 20 + 1$, so the 
number would be represented left-to-right like this:
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IV.
\item{A)} Suppose a Warlpiri woman from section 1 marries a man from section 5.
To what section do their children belong?
\item{} {\it Answer:}  See the diagrams on p. 71 of {\it Ethnomathematics}.  The
child's section is determined by the to from the mother's section, so the child
is in section 4.
\item{B)} What is the {\it mother relation} $m$ as a permutation of the 8 sections?
(That is, given the section of the child, to which section did the mother belong)?
\item{}  {\it Answer:}  The child's section determines the mother's section like
this:  
$$\eqalign{{\rm child} &\to {\rm mother}\cr
                1  &\to 3\cr
                2  &\to 4\cr
                3  &\to 2\cr
                4 &\to 1\cr
                5 &\to 8\cr
                6 &\to 7\cr
                7 &\to 5\cr
                8 &\to 6\cr}$$
\item{C)}  Same as B, but for the father relation $f$.
\item{} {\it Answer:}  The father relation is:
$$\eqalign{{\rm child} &\to {\rm father}\cr
                1  &\to 7\cr
                2  &\to 8\cr
                3  &\to 6\cr
                4 &\to 5\cr
                5 &\to 4\cr
                6 &\to 3\cr
                7 &\to 1\cr
                8 &\to 2\cr}$$
\item{D)}  Is $mf$ the same as $fm$ in the Warlpiri kinship system?  (That is,
is the mother of one's father ($mf$) in the same section as the father of one's
mother ($fm$)?)
\item{} {\it Answer:}  No they are not the same.  For instance (from the 
answers in parts B and C, if you are
in section 1, your mother was in section 3 and her father was in section 6.
But your father was in section 7 and his mother was in section 5.  That
one difference is enough to say $mf \ne fm$.  
\item{E)}  What is the name of the mathematical group that describes
the ``algebra'' of the $m$ and $f$ relations for the Warlpiri?  
\item{} {\it Answer:}  This is called {\it dihedral group of order 8}.
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V. Let $K_n$ be the graph with $n$ vertices and one edge joining every 
pair of distinct vertices (this is called the {\it complete} graph on 
$n$ vertices in graph theory).  
\item{A)}  Draw $K_2, K_3, K_4, K_5$ and $K_6$.  (The simplest forms
of these will place the vertices equally spaced 
\item{} {\it Answer:}
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\item{B)}  Which of the graphs from your part A have Eulerian paths?  
\item{} {\it Answer:}  $K_2$ does if you start at one vertex and end at the other, 
$K_3$ does since all vertices have degree 2 which is even, $K_4$ does not since
all the vertices have odd degree and there are 4 of them, $K_5$ does, and $K_6$
does not. 
\item{C)}  Now in general, for which $n$ does $K_n$ have an Eulerian path?  (Your 
answer should say exactly which $n$'s give a graph with that property.)
\item{} {\it Answer:}  The ones that do are $K_2$, and $K_n$ for all $n$ odd.
The reason is the that when $n$ is odd, from each vertex there are edges to
all of the {\it other} $n-1$ vertices in the graph, and if $n$ is odd, then 
$n - 1$ is even.
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VI. Suppose you are playing {\it mancala} and it is your turn.  
Your 6 pits contain numbers of stones left to right 
as follows:  0,5,3,0,0,0.  Is it possible to
get all 8 stones to your store before the other player gets to play?  If so,
show how to do it, step by step.  If not, explain why not.
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{\it Answer:}  No it is not possible. Playing from the rightmost ``harvestable''
pit each time (underlined), this is what happens:
$$\eqalign{
&0,\underline{5},3,0,0,0\cr
&0,0,4,1,1,\underline{1}\cr
&0,0,\underline{4},1,1,0\cr
&0,0,0,2,2,\underline{1}\cr
&0,0,0,2,\underline{2},0\cr
&0,0,0,2,0,\underline{1}\cr
&0,0,0,2,0,0\cr}$$
and you are ``stuck.''

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VII.  For each of the following symmetric strip patterns:
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\item{A)}  Say in words all the different kinds of symmetries
are present (i.e. does it have translation, glide reflection, vertical axis
reflection, or horizontal axis reflection symmetry?).
\item{} {\it Answer:}  Pattern (a) has all of them.  Pattern (b) has translation
and glide reflection symmetry, plus vertical axis reflection symmetry.  
Pattern (j) has only translation and glide reflection symmetry.  Pattern (k)
has translation, glide reflection, and horizontal axis reflection symmetry.
\item{B)}  Consult the ``decision tree'' from the course homepage and
say what is the type of the symmetry group.  
\item{} {\it Answer:}  (a) is $pmm2$, (b) is $pma2$, (j) is $p1a1$, (k) is $p1m1$ 
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