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\begin{enumerate}
\item The triangles $\vartriangle ABC$ and $\vartriangle DEF$ are not
isosceles, not congruent, and have integer-length sides. If they have the
same perimeter, what is the smallest such perimeter they could share?

\qquad A. \ $10$ \qquad\ B. \ $11$ \ \qquad\ \ C. \ $12$ \qquad\ \ D. \ $13$
\qquad E. \ $14$

Solution: \ Recall the the triangle inequality: for all triangles with sides 
$a$, $b$, and $c$,%
\begin{equation*}
a+b>c\text{ \ and \ }a+c>b\text{ \ \ and \ \ \ }b+c>a
\end{equation*}%
\thinspace Let us agree to organize all triangles by listing their sides in
an increasing order, and let us systematically look for the triangles with
the lowest possible values for perimeter. \ 

First notice that the shortest side can not be $1$ unit long. \ This is
because if the shortest side is $1$ unit long and the second shortest side
is $n$, then the longest side must be shorter than $n+1$ and there isn't an
integer between $n$ and $n+1$. So the shortest side possible is $2$ units
long. \ 

If the smallest side is $2$: \ For example, $2,3,4$ is possible as $2+3>4$.
\ But if the shortest two sides are $2$ and $3$, $4$ is the only integer
possibility.

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\begin{tabular}{lll}
If the shortest side is $2$ & ~~~~~~~~ & If the shortest side is $3$ \\ 
$2,3,4~~~~\Longrightarrow ~~P=9$ &  & $3,4,5~~~~\Longrightarrow ~~P=12$ \\ 
$2,4,5~~~~\Longrightarrow ~~P=11$ &  & $3,4,6~~~~\Longrightarrow ~~P=13$ \\ 
$2,5,6~~~~\Longrightarrow ~~P=13$ &  & $3,5,7~~~~\Longrightarrow ~~P=15$ \\ 
$2,6,7~~~~\Longrightarrow ~~P=15$ &  &  \\ 
$2,7,8~~~~\Longrightarrow ~~P=17$ &  & 
\end{tabular}

The smallest repeating value is $13,$ which is choice \fbox{D}.\vspace{0.09in%
}\vspace{0.09in}\vspace{0.09in}\vspace{0.09in}

\item At the intergalactic trading station, several different currencies are
used. Today, $15$ gleeks = $11$ zorks, $7$ gleeks = $3$ zeffs, and $5$ zeffs
= $2$ gems. A certain merchant lists prices in gleeks, but only gives change
in zorks. Can someone afford to buy an item that costs 45 gleeks if s/he has
8 gems? If so, how much change will be given (to the nearest tenth)?

\qquad A. No \ \ \ \ \ B. Yes, 3.1 zorks \ \ \ \ \ C. Yes, 1.2 zorks \ \ \ \
\ \ D. Yes, 0.3 zorks \ \ \ \ \ \ E. Yes, 1.8 zorks\vspace{0.09in}\vspace{%
0.09in}

Solution: $\ $We can use several unit conversion factors to switch from gems
to gleeks. \ She has $8~$gems, which is%
\begin{equation*}
8~\text{gems}=\dfrac{8~\text{gems}}{1}\cdot \dfrac{5~\text{zeffs}}{2~\text{%
gems}}\cdot \dfrac{7~\text{gleeks}}{3~\text{zeffs}}=\dfrac{8\cdot 5\cdot 7}{6%
}~\text{gleeks}~=\dfrac{140}{3}=46\dfrac{2}{3}~\text{gleeks}
\end{equation*}%
Thus she has enough money to buy the item, and the change is $1\dfrac{2}{3}=%
\dfrac{5}{3}$ gleeks. \ Convert the change to zorks:%
\begin{equation*}
\dfrac{5}{3}~\text{gleeks}=\dfrac{5~\text{gleeks}}{3}\cdot \dfrac{11~\text{%
zork}}{15~\text{gleeks}}=\dfrac{11}{9}~\text{zork}=\allowbreak
1.\,\allowbreak 222~\text{zork}
\end{equation*}%
So the answer is \fbox{C}.

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\item Suppose $a$ and $b$ are integers such that $(a,b)$ is a solution of $%
a^{2}+b^{2}+2ab+16a+16b=36$. Let $c$ be the average of $a$ and $b$. \ Find
the sum of all possible values of $c$.

\qquad A. $2$ \ \qquad\ \ B. $-8$ \ \qquad\ \ C. $-18$ \ \qquad\ \ D. $-2$ \
\qquad\ \ E. $8$\vspace{0.09in}\vspace{0.09in}

Solution:%
\begin{eqnarray*}
a^{2}+b^{2}+2ab+16a+16b &=&36 \\
\left( a+b\right) ^{2}+16\left( a+b\right) &=&36\text{ \ \ \ \ \ \ \ \ \ \ \
\ factor out }a+b \\
\left( a+b\right) \left( a+b+16\right) &=&36
\end{eqnarray*}%
There are only finitely many ways we can express $36$ as a product of two
integers. \ Among those, we collect the ones where the difference is $16$.%
\vspace{0.09in}

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\begin{tabular}{|l|l|l|l|}
\hline
$a+b$ & $a+b+16$ & difference &  \\ \hline
$1$ & $36$ & $35$ &  \\ \hline
$2$ & $18$ & $16$ & This one works! \\ \hline
$3$ & $12$ & $9$ &  \\ \hline
$4$ & $9$ & $5$ &  \\ \hline
$6$ & $6$ & $0$ &  \\ \hline
$-36$ & $-1$ & $35$ &  \\ \hline
$-18$ & $-2$ & $16$ & This one works! \\ \hline
$-12$ & $-3$ & $9$ &  \\ \hline
$-9$ & $-4$ & $5$ &  \\ \hline
$-6$ & $-6$ & $0$ &  \\ \hline
\end{tabular}

The value of $\ c=\dfrac{a+b}{2}$ is either $\dfrac{2}{2}=\allowbreak 1$ or $%
\dfrac{-18}{2}=\allowbreak -9$. \ The sum of the possible values is \ $%
1+\left( -9\right) =\allowbreak -8$. \ The answer is \fbox{B}.\vspace{0.09in}%
\vspace{0.09in}\vspace{0.09in}

\item If cars hold $5$ passengers and charge for $\$29$ a trip to the
airport, and vans hold $7$ passengers and charge $\$41$, find the minimum
cost to transport 49 people to the airport.

\qquad A. $\$290$ \ \ \qquad B. $\$285$ \ \ \qquad C. $\$287$ \ \ \qquad D. $%
\$280$\vspace{0.09in} \ \ \qquad E. \ $\$282$

Solution: \ Let us start with all cars and no vans. \ \ The \ lowest answer
we could find is \fbox{B}.

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\begin{tabular}{|l|l|l|l|}
\hline
Number of vans & Number of cars & Capacity & Cost \\ \hline
$0$ & $10$ & $50$ & $\$290$ \\ \hline
$1$ & $9$ & $52$ & $\$302$ \\ \hline
$2$ & $7$ & $49$ & $\$285$ \\ \hline
$3$ & $6$ & $51$ & $\$333$ \\ \hline
$4$ & $5$ & $53$ & $\$309$ \\ \hline
$5$ & $3$ & $50$ & $\$292$ \\ \hline
$6$ & $2$ & $52$ & $\$304$ \\ \hline
$7$ & $0$ & $49$ & $\$287$ \\ \hline
\end{tabular}

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\pagebreak

\item In the grid (made up of $1\times 1$ squares) on the right, which of
the squares A, B, C, D, or E, when shaded, will allow the unshaded squares
to be covered by exactly 14 dominos ($1\times 2$ rectangles) with no
overlaps or gaps?\FRAME{dtbpF}{0.9772in}{0.9772in}{0pt}{}{}{pic3a.bmp}{%
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Solution 1: \ I found a covering that skipped D. \ 

Solution 2: \ While we are at coloring; consider the following. \ Suppose we
color the fields to be covered so that two fields with a side common have
different colors. \ Then A, B, C, and E are all blue and only E is white. \
The 29 fields are: 14 blue and 15 white. \ Since each domino can cover
exactly one white and one blue field, the covering is impossible if we skip
a blue field as we would have 14 dominos to cover 13 blue and 15 white
fields. \ Either way, the \ correct answer is \fbox{D}.

\ \ \qquad \qquad \qquad\ \ \ \ \ \ \FRAME{itbpF}{1.4356in}{1.4356in}{0in}{}{%
}{pic3b.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
1.4356in;height 1.4356in;depth 0in;original-width 2.3402in;original-height
2.3402in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'pic3b.bmp';file-properties "XNPEU";}} \qquad\ \qquad\ \FRAME{itbpF}{1.4356in%
}{1.4356in}{0in}{}{}{pic3c.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
1.4356in;height 1.4356in;depth 0in;original-width 2.3402in;original-height
2.3402in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'pic3c.bmp';file-properties "XNPEU";}}\vspace{0.09in}\vspace{0.09in}\vspace{%
0.09in}

\item The graph of $x^{2}+xy+x+3y=6$ is\vspace{0.09in}

A. an ellipse \ \ \ \ \ \ \ B. a parabola \ \ \ \ \ \ \ \ \ \ C. a hyperbola
\ \ \ \ \ \ \ \ \ D. 2 parallel lines \ \ \ \ \ \ \ \ E. \ two intersecting
lines.\vspace{0.09in}

Solution: \ We reduce one side to zero, factor, and apply the zero product
rule.%
\begin{eqnarray*}
x^{2}+xy+x+3y &=&6 \\
x^{2}+xy+x+3y-6 &=&0 \\
x^{2}+xy+x+3y-6 &=&0 \\
x^{2}+x-6+3y+xy &=&0 \\
\left( x+3\right) \left( x-2\right) +3y\left( x+3\right)  &=&0 \\
\left( x+3\right) \left( x-2+3y\right)  &=&0
\end{eqnarray*}%
\begin{equation*}
x_{1}=-3\text{ \ \ \ \ or \ \ \ \ }y_{2}=\dfrac{-x+2}{3}
\end{equation*}%
This are the equations of two non-parallel lines, so the correct answer is 
\fbox{E}.\vspace{0.09in}\vspace{0.09in}\vspace{0.09in}

\item Let a and b be positive integers such that $(a,b)$ is a solution to $%
\sqrt[3]{a+4\sqrt{b}}+\sqrt[3]{a-4\sqrt{b}}=3$. Find the smallest possible
value of $a+b$.

A. 14 \ \ \qquad \qquad B. 18 \ \ \qquad \qquad C. 22 \ \ \qquad \qquad D.
26 \ \ \qquad \qquad E. 30%
\begin{equation*}
\sqrt[3]{a+4\sqrt{b}}+\sqrt[3]{a-4\sqrt{b}}=3\text{ \ \ \ \ \ \ \ \ \ \ \ \
\ raise to third power}
\end{equation*}%
\begin{eqnarray*}
\left( \sqrt[3]{a+4\sqrt{b}}\right) ^{3}+3\left( \sqrt[3]{a+4\sqrt{b}}%
\right) ^{2}\left( \sqrt[3]{a-4\sqrt{b}}\right) +3\left( \sqrt[3]{a+4\sqrt{b}%
}\right) \left( \sqrt[3]{a-4\sqrt{b}}\right) ^{2}+\left( \sqrt[3]{a-4\sqrt{b}%
}\right) ^{3} &=&27 \\
a+4\sqrt{b}+a-4\sqrt{b}+3\left( \sqrt[3]{a+4\sqrt{b}}\right) \left( \sqrt[3]{%
a-4\sqrt{b}}\right) \left( \underset{=3}{\underbrace{\sqrt[3]{a+4\sqrt{b}}+%
\sqrt[3]{a-4\sqrt{b}}}}\right)  &=&27
\end{eqnarray*}%
\begin{eqnarray*}
2a+3\sqrt[3]{a^{2}-16b}\cdot 3 &=&27 \\
2a+9\sqrt[3]{a^{2}-16b} &=&27 \\
2a &=&27-9\sqrt[3]{a^{2}-16b} \\
2a &=&9\left( 3-\sqrt[3]{a^{2}-16b}\right) 
\end{eqnarray*}%
Thus $a$ is divisible by $9$. \ We are looking for the smallest value of $a+b
$, so let us try the smallest positive value, $a=9$.%
\begin{eqnarray*}
2\cdot 9 &=&9\left( 3-\sqrt[3]{81-16b}\right)  \\
2 &=&3-\sqrt[3]{81-16b} \\
\sqrt[3]{81-16b} &=&1 \\
81-16b &=&1 \\
80 &=&16b \\
5 &=&b
\end{eqnarray*}%
Thus $a+b=9+5=14$. \ We didn't prove yet that this is the smallest value for 
$a+b$, but it is the smallest value offered. \ So, if the problem is well
designed, the answer is \fbox{A}.\vspace{0.09in}\vspace{0.09in}

\item The matrix $A=\left[ 
\begin{array}{cc}
a & 8 \\ 
-3 & b%
\end{array}%
\right] $ is its own inverse (that is, $A$ times $A$ equals the identity
matrix). \ Find $\left\vert a-b\right\vert $.

A. 4 \ \ \qquad \qquad B. 6 \ \ \ \qquad \qquad C. 8 \ \ \qquad \qquad D. 10
\ \ \qquad \qquad E. 12

Solution: \ 
\begin{eqnarray*}
A^{2} &=&I \\
\left[ 
\begin{array}{cc}
a & 8 \\ 
-3 & b%
\end{array}%
\right] \left[ 
\begin{array}{cc}
a & 8 \\ 
-3 & b%
\end{array}%
\right]  &=&\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & 1%
\end{array}%
\right]  \\
\left[ 
\begin{array}{cc}
a^{2}-24 & 8a+8b \\ 
-3a-3b & -24-b^{2}%
\end{array}%
\right]  &=&\left[ 
\begin{array}{cc}
1 & 0 \\ 
0 & 1%
\end{array}%
\right] 
\end{eqnarray*}%
So \ $a^{2}-24=1\Longrightarrow a^{2}=25\Longrightarrow a=\pm 5$ \ \ \ and $%
a+b=0$ so $b=-\left( \pm 5\right) $. \ Either way, $\left\vert
a-b\right\vert =\left\vert \left( 5-\left( -5\right) \right) \right\vert =10$
or $\left\vert -5-5\right\vert =10$. \ The correct answer is \fbox{D}.

\item Let $f(x)=x^{2}+bx+c$. If $f\left( 4\right) =f\left( 2\right) +11$,
find $f(4)-f(0)$.

\qquad A. $-6$ \ \ \qquad \qquad B. \ $-8$ \ \qquad \qquad\ C. \ $8$ \
\qquad \qquad\ D. \ $10$ \qquad \qquad\ \ E. $14$

Solution: \ 
\begin{eqnarray*}
f\left( 4\right) &=&f\left( 2\right) +11 \\
4^{2}+b\cdot 4+c &=&2^{2}+b\cdot 2+c+11 \\
16+4b+c &=&4+2b+c+11 \\
16+4b &=&2b+15 \\
2b &=&-1 \\
b &=&-\dfrac{1}{2}
\end{eqnarray*}%
So $f(x)=x^{2}-\dfrac{1}{2}x+c$ \ Thus%
\begin{equation*}
f(4)-f(0)=\left( 4^{2}-\dfrac{1}{2}\cdot 4+c\right) -\left( 0^{2}-\dfrac{1}{2%
}\cdot 0+c\right) =16-2=14
\end{equation*}%
The answer is \fbox{E}.

\pagebreak 

\item Three people $(X,Y,Z)$ are in a room with you. One is a Knight
(Knights always tell the truth), one is a Knave (Knaves always lie), and the
other is a Spy (Spies may either lie or tell the truth), but you don't know
who is which. Each person makes exactly one statement. Which of the
following sets of three statements is NOT possible?\vspace{0.09in}

\begin{tabular}{cccccccccc}
A &  & B &  & C &  & D &  & E &  \\ 
\multicolumn{1}{l}{X: I am a Knight} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{X: I am not a Spy} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{X: I am a Spy} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{X: I am a Knight} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{X: I am not a Knave} & \multicolumn{1}{l}{} \\ 
\multicolumn{1}{l}{Y: I am a Knave} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Y: I am not a Spy} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Y: I am a Spy} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Y: I am a Knave} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Y: I am not a Knave} & \multicolumn{1}{l}{} \\ 
\multicolumn{1}{l}{Z: X is a Spy} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Z: X is not a Knight} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Z: I am a Knight} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Z: X is a Knight} & \multicolumn{1}{l}{} & 
\multicolumn{1}{l}{Z: I am not a Knave} & \multicolumn{1}{l}{}%
\end{tabular}%
\vspace{0.09in}

Solution: \ The only things impossible here are 3 true or 3 false
statements. \ 1 true because of the Knight, one false because of the Knave
and the Spy can go either way.

Consider A. \ X could be telling the truth --\TEXTsymbol{>} X is knight. \ Y
could be the spy and lying that he is a knave. \ Then Z is the knave and X
being a Spy is false. \ So, A is possible.

Consider B. \ X could be Spy and lying. \ Then Y must be telling the truth --%
\TEXTsymbol{>} \ he is a Knight. \ Then Z must be a knave but then he is
lying and so X is a knight. \ So if X is the knight and he is telling the
truth, then Y could be a Spy and be lying. \ then Z must be a knave and X
not a knight is then a lie. So, B is possible.

Consider C. \ X could be a spy and telling the truth. \ Then Y lies and so
he must be a knave. \ Then Z must be the knight. \ So, C is possible.

Consider D. \ The statement "I am knave" can not be true. \ Thus, Y is lying
and he is thus a Spy. \ So X and Z are the knight and knave but they
literally agree! \ So their statements aare both true or both false. \ They
are both impossible. \ So, D is impossible.

Consider E. \ X might be a knave and lying,Y might be a knight and telling
the truth and Z a spy and telling the truth. \ So, this is quite possible.

So the correct answer is \fbox{D}.\vspace{0.09in}\vspace{0.09in}

\item For a positive integer $n$, let $S\left( n\right) $ be the sum of the
first $n$ positive integers (for example, $S(5)=15$). \ For how many
positive integers, $n$, less than $2017$, will all digits of $S(n)$ be 1s?

\qquad A. 0 \ \ \qquad \qquad\ \ B. 1 \ \qquad \qquad\ \ C. 2 \ \qquad
\qquad\ \ D. 3 \ \ \qquad \qquad\ \ E. 4

Solution: \ $S\left( 1\right) =1$ \ \ \ \ \ \ \ \ \ \ $S\left( 2\right) =3$
\ \ \ \ \ \ \ \ \ \ $S\left( n\right) =\dfrac{n\left( n+1\right) }{2}$%
\begin{eqnarray*}
S\left( n\right)  &=&\dfrac{n\left( n+1\right) }{2} \\
1+2+...+n &=&\underset{k\text{ times}}{\underbrace{11....1}} \\
\dfrac{n\left( n+1\right) }{2} &=&\underset{k\text{ times}}{\underbrace{%
11....1}} \\
n\left( n+1\right)  &=&\underset{k\text{ times}}{\underbrace{22....2}} \\
n^{2}+n-\underset{k\text{ times}}{\underbrace{22....2}} &=&0 \\
n_{1,2} &=&\dfrac{-1\pm \sqrt{1+4\left( 22....2\right) }}{2}
\end{eqnarray*}%
The smaller solution is clearly negative, so we focus on the larger,
positive one.

If $k=1$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 2\right) }}{2}=1$ \ \ \ \ \
Indeed, $S\left( 1\right) =1$

If $k=2$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 22\right) }}{2}=\dfrac{\sqrt{89%
}-1}{2}$ \ \ \ not an integer

If $k=3$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 222\right) }}{2}=\dfrac{\sqrt{%
889}-1}{2}$ \ \ \ not an integer

If $k=4$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 2222\right) }}{2}=\dfrac{\sqrt{%
8889}-1}{2}$ \ \ \ not an integer

If $k=5$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 22222\right) }}{2}=\dfrac{%
\sqrt{88889}-1}{2}\approx 148.\,\allowbreak 5713$ \ \ \ not an integer

If $k=6$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 222222\right) }}{2}=\dfrac{%
\sqrt{888889}-1}{2}\approx 470.\,\allowbreak 905$ \ \ \ not an integer

If $k=7$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 2222222\right) }}{2}=\dfrac{%
\sqrt{8888889}-1}{2}\approx 1490.\,\allowbreak 212$ \ \ \ not an integer

If $k=8$ \ \ \ \ $n=\dfrac{-1+\sqrt{1+4\left( 22222222\right) }}{2}=\dfrac{%
\sqrt{88888889}-1}{2}\approx 4713.\,\allowbreak 55$ \ \ \ not an integer. \
Now we are beyond $2017$ so we can stop looking. \ $S\left( 1\right) =1$ is
the ony solution so the answer is \fbox{B}.\vspace{0.09in}\vspace{0.09in}

\item Ed filled $\dfrac{2}{3}$ of his radiator with antifreeze and then
added $4$ more quarts (a gallon) of antifreeze. After draining half the
antifreeze, he needed $11$ quarts of antifreeze to fill the radiator to
capacity. How many gallons of antifreeze can the radiator hold?\vspace{0.09in%
}

\qquad A. 4.65 \ \ \qquad \qquad\ \ B. 4.875 \ \ \qquad \qquad\ \ C. 18.6 \
\ \qquad \qquad\ \ D. 19.5 \ \ \qquad \qquad\ \ E. 78

Solution: \ Let $x$ be the capacity of the radiator.%
\begin{eqnarray*}
\dfrac{\dfrac{2}{3}x+4}{2}+11 &=&x \\
\dfrac{2}{3}x+4+22 &=&2x \\
\dfrac{2}{3}x+26 &=&2x \\
26 &=&\dfrac{4}{3}x \\
\dfrac{3\cdot 26}{4} &=&x \\
x &=&\dfrac{39}{2}=19.5
\end{eqnarray*}%
So the answer is $19.5$ quarts, but we need to present the answer in
gallons: $\dfrac{19.5\text{ quarts}}{4}=\allowbreak 4.\,\allowbreak 875$
gallons so the correct answer is \fbox{B}.

\item On a game show, the final contestant each day can win $\$1,000,000$ by
correctly guessing an integer between $1$ and $100$ inclusive (which is
chosen randomly each day). Before guessing the contestant can ask one yes/no
question of his or her choice. \ Monday's contestant asked \textquotedblleft
Is the number $57$?\textquotedblright\ and Tuesday's contestant asked
\textquotedblleft Is the number greater than $50$?\textquotedblright . Let $%
P\left( M\right) $ be the probability of Monday's contestant winning and let 
$P(T)$ be the probability of Tuesday's contestants winning (assume each
contestant properly uses the information gained from the question). Which of
the following is true?

A. $\dfrac{P\left( M\right) }{P\left( T\right) }\leq .1$ \ \ \ B. $\ 0.1<%
\dfrac{P\left( M\right) }{P\left( T\right) }<0.9$ \ \ \ C. $\ 0.9\leq \dfrac{%
P\left( M\right) }{P\left( T\right) }\leq 1.1$ \ \ \ D. \ $1.1<\dfrac{%
P\left( M\right) }{P\left( T\right) }<2$ E. \ $\dfrac{P\left( M\right) }{%
P\left( T\right) }\geq 2$

Solution: \ $P\left( M\right) =\dfrac{1}{100}\left( 1\right) +\dfrac{99}{100}%
\left( \dfrac{1}{99}\right) =\dfrac{2}{100}=\dfrac{4}{200}=0.02$

$P\left( T\right) =\dfrac{1}{2}\left( \dfrac{50}{100}\right) +\dfrac{1}{2}%
\left( \dfrac{49}{100}\right) =\dfrac{99}{200}=0.495\,\ \ \ \ \ \ \ \ \ \ \
\ \ \ \dfrac{P\left( M\right) }{P\left( T\right) }=\dfrac{~\dfrac{4}{200}~}{%
\dfrac{99}{200}}=\dfrac{4}{99}\approx 0.040404$ \ \ so the answer is \fbox{A}%
. \ \ (Caution! \ This is in conflict with the official AMATYC answer....)

\item Let $P(x)$ be a degree $5$ polynomial with rational coefficients and $%
P(0)=-53\,040$. Suppose $x=12$, $x=3+5i$, and $x=4-7i$ are zeros of $P(x)$.
In which interval does the coefficient of $x^{3}$ lie?

\qquad A. $(-\infty ,-500]$ \ \qquad B. $(-500,-100]$\ \qquad\ C. $(-100,100)
$ \ \qquad\ D. $\ [100,500)$ \ \qquad\ E. $[500,\infty )$

Solution: \ $x=12$ is a zero $\Rightarrow \left( x-12\right) $ is a linear
factor by the remainder theorem.

$x=3+5i$ is a zero and $P$ has rational (thus real) coefficients $%
\Rightarrow \left( x-\left( 3+5i\right) \right) \left( x-\left( 3-5i\right)
\right) $ is a factor

$x=4-7i$ is a zero and $P$ has rational coefficients $\Rightarrow \left(
x-\left( 4-7i\right) \right) \left( x-\left( 4+7i\right) \right) $ is a
factor%
\begin{eqnarray*}
P\left( x\right)  &=&A\left( x-12\right) \left( \left( x-\left( 3+5i\right)
\right) \left( x-\left( 3-5i\right) \right) \right) \left( x-\left(
4-7i\right) \right) \left( x-\left( 4+7i\right) \right)  \\
&=&A\left( x-12\right) \left( x^{2}-6x+34\right) \left( x^{2}-8x+65\right) 
\end{eqnarray*}%
The degree is correct. \ We will find the leading coefficient $A$ using $%
P\left( 0\right) $. 
\begin{eqnarray*}
P\left( 0\right)  &=&A\left( -12\right) \left( 34\right) \left( 65\right)
=-53\,040 \\
-26\,520A &=&-53\,040 \\
A &=&2
\end{eqnarray*}%
So $P\left( x\right) =2\left( x-12\right) \left( x^{2}-6x+34\right) \left(
x^{2}-8x+65\right) $. \ Now for the cubic term: 
\begin{eqnarray*}
a_{3}x^{3} &=&2\left( x\left( x^{2}\right) 65+x\left( -6x\right) \left(
-8x\right) +x\left( 34\right) x^{2}-12x^{2}\left( -8x\right) -12\left(
-6x\right) x^{2}\right)  \\
&=&2x^{3}\left( 65+48+34+96+72\right) =630x^{3}
\end{eqnarray*}%
Since $630\in \left[ 500,\infty \right) ,$ the answer is \fbox{E}.

\item 
%TCIMACRO{\TeXButton{\begin{minipage}{5in}}{\begin{minipage}{5in}}}%
%BeginExpansion
\begin{minipage}{5in}%
%EndExpansion
A company designed a new logo by constructing semicircles inside of a unit
square (side length = 1) as shown on the right. \ Which of the following is
closest to the area of the shaded region?

A. $0.4$ \ \qquad\ \ \ B. $0.45$ \ \qquad\ C. $0.5$ \qquad\ \ \ \ D. $0.55$
\ \ \qquad\ E. $0.6$

%TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}%
%BeginExpansion
\end{minipage}%
%EndExpansion
%TCIMACRO{\TeXButton{\begin{minipage}{2.4in}}{\begin{minipage}{2.4in}}}%
%BeginExpansion
\begin{minipage}{2.4in}%
%EndExpansion
\FRAME{dtbpF}{0.9323in}{0.9461in}{0pt}{}{}{pic1a.bmp}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"USEDEF";valid_file "F";width 0.9323in;height 0.9461in;depth
0pt;original-width 0.9003in;original-height 0.9132in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'pic1a.bmp';file-properties
"XNPEU";}}%
%TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}%
%BeginExpansion
\end{minipage}%
%EndExpansion

Solution: \ Consider one-eighth of the shaded region as shown. \ Clearly it
is the difference between a square of sides $\dfrac{1}{2}$ unit and a
quarter of a circle with radius $\dfrac{1}{2}$.\FRAME{dtbpF}{3.8951in}{%
1.2073in}{0pt}{}{}{pic1b.bmp}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width
3.8951in;height 1.2073in;depth 0pt;original-width 3.8467in;original-height
1.1736in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'pic1b.bmp';file-properties "XNPEU";}}So, the area we must find is 
\begin{equation*}
A=8\left( \left( \dfrac{1}{2}\right) ^{2}-\dfrac{1}{4}\pi \left( \dfrac{1}{2}%
\right) ^{2}\right) =8\left( \dfrac{1}{4}-\dfrac{1}{16}\pi \right) =2-\dfrac{%
\pi }{2}\approx 0.4292037
\end{equation*}%
To find the correct answer, we need to find which number given is closest to
our answer. \ 

A: \ $0.4292037-0.4\approx 0.03$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C: $%
0.5-0.4292037\approx 0.071$\ \ \ \ \ \ \ \ \ E: $\ 0.6-0.4292037\approx 0.17$

B: \ $0.45-0.4292037\approx 0.021$ \ \ \ \ \ \ \ \ \ \ \ D: $%
0.55-0.4292037\approx 0.12$

So the correct answer is \fbox{B}.

\item How many positive integers less than $1000$ are divisible by exactly
one of $7$ or $11$?

\qquad A. $196$ \ \ \qquad\ \qquad\ \ B. $208$ \qquad\ \ \qquad\ \ C. $220$
\ \ \qquad \qquad D. $232$ \ \qquad \qquad\ E. $244$

Solution: \ Recall that if $\left\vert S\right\vert $ denotes the size or
cardinality of a set $S,$ then for all sets $A$, $B$, \ \ 
\begin{equation*}
\left\vert A\cup B\right\vert =\left\vert A\right\vert +\left\vert
B\right\vert -\left\vert A\cap B\right\vert 
\end{equation*}%
Define $E=\left\{ n:n\text{ is a positive integer, }n<1000\text{, }n\text{
is divisible by }11\right\} $ and \newline
$S=\left\{ n:n\text{ is a positive integer, }n<1000\text{, }n\text{ is
divisible by }7\right\} $.

Then $E\cap S=\left\{ n:n\text{ is a positive integer, \ }n<1000\text{, }n%
\text{ is divisible by }77\right\} $. \ In terms of these sets, what we are
looking for is 
\begin{equation*}
X=\left\vert E\right\vert +\left\vert S\right\vert -2\left\vert E\cap
S\right\vert 
\end{equation*}%
Let us compute the size of these sets. \ $1000\div 7=\allowbreak 142$ R $6$
\ \ \ \ $142\cdot 7=994$%
\begin{equation*}
S=\left\{ \underset{%
\begin{array}{c}
\downarrow  \\ 
1\cdot 7%
\end{array}%
}{7},\underset{%
\begin{array}{c}
\downarrow  \\ 
2\cdot 7%
\end{array}%
}{14},\underset{%
\begin{array}{c}
\downarrow  \\ 
3\cdot 7%
\end{array}%
}{21},...,\underset{%
\begin{array}{c}
\downarrow  \\ 
142\cdot 7%
\end{array}%
}{994}\right\} \text{~~}\Longrightarrow ~~\left\vert S\right\vert =142
\end{equation*}%
Similarly, $1000\div 11=90$ R $10$ \ \ \ \ \ $90\cdot 11=990$%
\begin{equation*}
E=\left\{ \underset{%
\begin{array}{c}
\downarrow  \\ 
1\cdot 11%
\end{array}%
}{11},\underset{%
\begin{array}{c}
\downarrow  \\ 
2\cdot 11%
\end{array}%
}{22},\underset{%
\begin{array}{c}
\downarrow  \\ 
3\cdot 11%
\end{array}%
}{33},...,\underset{%
\begin{array}{c}
\downarrow  \\ 
90\cdot 11%
\end{array}%
}{990}\right\} \text{~~}\Longrightarrow ~~\left\vert E\right\vert =90
\end{equation*}%
Now for $\left\vert E\cap S\right\vert $: \ \ \ \ $1000\div 77=\allowbreak 12
$ R $15$ \ \ \ \ $12\cdot 77=924$ R $76$\ \ 
\begin{equation*}
E\cap S=\left\{ \underset{%
\begin{array}{c}
\downarrow  \\ 
1\cdot 77%
\end{array}%
}{77},\underset{%
\begin{array}{c}
\downarrow  \\ 
2\cdot 77%
\end{array}%
}{154},\underset{%
\begin{array}{c}
\downarrow  \\ 
3\cdot 77%
\end{array}%
}{231},...,\underset{%
\begin{array}{c}
\downarrow  \\ 
12\cdot 77%
\end{array}%
}{924}\right\} \text{~~}\Longrightarrow ~~\left\vert E\cap S\right\vert =12
\end{equation*}%
Now our solution is 
\begin{equation*}
X=\left\vert E\right\vert +\left\vert S\right\vert -2\left\vert E\cap
S\right\vert =142+90-2\cdot 12=208
\end{equation*}%
which is choice \fbox{B}.\vspace{0.09in}\vspace{0.09in}

\pagebreak 

\item A neon light is failing. When the switch is flipped, it lights for a
second, then goes off for a second; lights for a second, then goes off for $2
$ seconds; lights for a second, then goes off for $3$ seconds, etc. Exactly
two minutes after the switch is flipped, how long (in seconds) will it stay
off before it goes on again?\vspace{0.09in}

A. 12 \ \ \qquad\ \ \ B. 13 \ \ \qquad\ \ \ C. 14 \qquad\ \ \ D. 15 \
\qquad\ \ \ \ E.16\vspace{0.09in}\vspace{0.09in}

Solution: \ 

\begin{tabular}{|lllllllllll|}
\hline
$1$ second on &  & \multicolumn{1}{l|}{} &  & $1$ second on &  & 
\multicolumn{1}{l|}{} &  & $1$ second on &  &  \\ 
$1$ second off & $\Longrightarrow $ & \multicolumn{1}{l|}{$k=1$} &  & $2$
seconds off & $\Longrightarrow $ & \multicolumn{1}{l|}{$k=2$} &  & $3$
seconds off & $\Longrightarrow $ & $k=3$ \\ \hline
\end{tabular}
\ 

So for general $k$: \ \ \ 
\begin{eqnarray*}
\left( 1+1\right) +\left( 1+2\right) +\left( 1+3\right) +...+\left(
1+k\right)  &\geq &120 \\
k+\dfrac{k\left( k+1\right) }{2} &=&120 \\
2k+k^{2}+k-240 &=&0~~~~~~~~~~~~~~k_{1,2}=\dfrac{-3\pm \sqrt{9+4\cdot 240}}{2}
\end{eqnarray*}%
The positive root is about $\allowbreak 14.\,\allowbreak 064$. \ \ This we
are talking about between $\,k=14$ and $k=15$

$k=14$\ \ \ \ $\Longrightarrow $ \ \ \ $14+\dfrac{14\left( 15\right) }{2}=119
$ \ \ \ \ \ \ \ 

That is: as $k=14,$ the light goes on for a second and then is off for $14$
seconds. \ That is the 119th second. \ So, \ $1$ more second markes exactly
two minutes after. \ It just got dark a second ago. \ So, it will be dark
for another 13 seconds before it lights up again. \ This is answer \fbox{B}.%
\vspace{0.09in}\vspace{0.09in}

\item Let $N$ be the smallest positive integer such that ALL $N$-digit
numbers of the form $aa...a$ are divisible by $7$. Let $M$ be the smallest
positive integer such that $10^{M}$ does NOT have a factorization $ab$ in
which neither factor has any $0$ digits. Find $M+N$.\vspace{0.09in}

\qquad A. $18$ \ \ \qquad\ \ \ \ \ B. $17$ \ \qquad\ \ \ \ C. $16$ \ \
\qquad\ \ \ \ D. $15$ \ \ \qquad\ \ \ \ E. $14$\vspace{0.09in}

Solution: \ Clearly $10^{n}=2^{n}\cdot 5^{n}$.\ If a \thinspace 2 paired
with a 5, in the prime factorization of a factor, then it will end in a
zero. \ Thus, we need to look at factorizations of the form of $2^{n}\cdot
5^{n}$. \ Eventually, these factors will have a zero among their digits,
although not as the last digit.

\begin{tabular}{lllllll}
$10^{1}=2\cdot 5$ & ~~~~~~ & $10^{3}=8\cdot 125$ & ~~~~~~ & $10^{5}=32\cdot
3125$ & ~~~~~~ & $10^{7}=128\cdot 78\,125$ \\ 
$10^{2}=4\cdot 25$ &  & $10^{4}=16\cdot 625$ &  & $10^{6}=64\cdot 15\,625$ & 
& $10^{8}=256\cdot 390\,625~~\Longrightarrow ~~$Thus $M=8$%
\end{tabular}

Now for $N$: \ $11$ and $111$ and $1111$ are not divisible by $7$. \ $11111$
aren't either. \ 

But then $111111=7\cdot 15\,873$ and so all six-digit numbers $aaaaaa$ are
divisible by $7$ as $aaaaaa=a\left( 111111\right) =7\left( 15\,873a\right) $%
. \ So, $N=6$

$M+N=8+6=14$ \ which is choice \fbox{E}.\vspace{0.09in}\vspace{0.09in}

\pagebreak 

\item A triangle has vertices $A(0,0)$, $B(3,0)$, and $C(3,4)$. If the
triangle is rotated counterclockwise around the origin until $C$ lies on the
positive $y$-axis, find the area of the intersection of the region bounded
by the original triangle and the region bounded by the rotated triangle.

\qquad A. $\dfrac{21}{16}$ \ \qquad\ \ \ \ B. $\dfrac{25}{16}$ \ \qquad\ \ \
\ C. $\dfrac{29}{16}$ \ \qquad\ \ \ \ \ D. $\dfrac{35}{16}$ \ \qquad\ \ \ \
\ E. $\dfrac{75}{16}$

Solution: \ \FRAME{dtbpF}{2.4777in}{2.9265in}{0pt}{}{}{pic2.bmp}{\special%
{language "Scientific Word";type "GRAPHIC";display "USEDEF";valid_file
"F";width 2.4777in;height 2.9265in;depth 0pt;original-width
0.0813in;original-height 0.1116in;cropleft "0";croptop "1";cropright
"1.0671";cropbottom "0";filename 'pic2.bmp';file-properties "XNPEU";}}

Let $\alpha $ denote angle $CAB$. \ Clearly, $\alpha =\tan ^{-1}\left( 
\dfrac{4}{3}\right) $. \ Then angle $CAC^{\prime }=\beta =90^{\circ }-\alpha
=\tan ^{-1}\left( \dfrac{3}{4}\right) $.%
\begin{eqnarray*}
\theta &=&\alpha -\beta \\
\tan \theta &=&\tan \left( \alpha -\beta \right) =\dfrac{\tan \alpha -\tan
\beta }{1+\tan \alpha \tan \beta }=\dfrac{\dfrac{4}{3}-\dfrac{3}{4}}{1+%
\dfrac{4}{3}\cdot \dfrac{3}{4}}=\dfrac{\dfrac{7}{12}}{2}=\dfrac{7}{24}
\end{eqnarray*}%
The area of triangle $AB^{\prime }D$ is $\dfrac{1}{2}\left( AB^{\prime
}\right) \left( AB^{\prime }\tan \theta \right) =\dfrac{1}{2}\left( 3\right)
\left( 3\cdot \dfrac{7}{24}\right) =\dfrac{21}{16}$ which is answer \fbox{A}.

\item Consider a game where a player bets $\$X$ and then flips a biased coin
where the probability of flipping heads is $0.4$. If the result is heads,
she wins $\$X$; if it is tails, she loses $\$X$. Suppose she starts with $%
\$25$ and her first bet is $\$5$. Every time she wins, she will bet double
what she won on the next flip. Whenever she loses, she will bet $\$5$ on the
following flip. If she has $\$100$ or more at any point, she will quit. What
is the probability (rounded to the nearest thousandth) that she will quit
with $\$100$ or more in 7 flips or less?

\qquad A. $0.026$ \ \qquad\ \ \ \ B. $0.035$ \ \qquad\ \ \ \ \ C. $0.038$ \
\qquad\ \ \ \ D. $0.052$ \ \qquad\ \ E. $0.070$

Solution: As we investigate the situation, it occurs to us that the stakes
are higher and higher in a winning streak and so a loss takes one down
drasticallly. \ Therefore, she can win this game with either all wins, or,
if losses occur, they should be at the early flips. \ 

Let us organize our cases of winning within 7 flips by the number of lost
games.

All wins:

\qquad\ \ 
\begin{tabular}{|l|l|}
\hline
before 1st flip & has \$25, bets \$5 \\ \hline
1st flip & win $\Longrightarrow \,$has \$30, bets \$10 \\ \hline
2nd flip & win $\Longrightarrow \,$has \$40, bets \$20 \\ \hline
3rd flip & win $\Longrightarrow \,$has \$60, bets \$40 \\ \hline
4th flip & win $\Longrightarrow \,$has \$100 \ \ \ DING! DING! \ $%
\Longrightarrow $\ \ $P=0.4^{4}$ \\ \hline
\end{tabular}

All wins and one loss: if we lose only in the first flip:

\qquad\ \ 
\begin{tabular}{|l|l|}
\hline
before 1st flip & has \$25, bets \$5 \\ \hline
1st flip & lose $\Longrightarrow \,$has \$20, bets \$5 \\ \hline
2nd flip & win $\Longrightarrow \,$has \$25, bets \$10 \\ \hline
3rd flip & win $\Longrightarrow \,$has \$35, bets \$20 \\ \hline
4th flip & win $\Longrightarrow \,$has \$55, bets \$40 \\ \hline
5th flip & win $\Longrightarrow \,$has \$95, bets \$80 \\ \hline
6th flip & win $\Longrightarrow \,$has \$175 \ \ \ DING! \ DING! \ $%
\Longrightarrow $\ \ $P=0.6\cdot 0.4^{5}$ \\ \hline
\end{tabular}

or if we lose first in the second flip:

\qquad\ \ 
\begin{tabular}{|l|l|}
\hline
before 1st flip & has \$25, bets \$5 \\ \hline
1st flip & win $\Longrightarrow \,$has \$30, bets \$10 \\ \hline
2nd flip & lose $\Longrightarrow \,$has \$20, bets \$5 \\ \hline
3rd flip & win $\Longrightarrow \,$has \$25, bets \$10 \\ \hline
4th flip & win $\Longrightarrow \,$has \$35, bets \$20 \\ \hline
5th flip & win $\Longrightarrow \,$has \$55, bets \$40 \\ \hline
6th flip & win $\Longrightarrow \,$has \$95, bets \$80 \\ \hline
7th flip & win $\Longrightarrow \,$has \$175 \ \ \ DING! \ DING! \ $%
\Longrightarrow $\ \ $P=0.6\cdot 0.4^{6}$ \\ \hline
\end{tabular}

if we lose first in the third flip, we will not make to $100$ in seven steps:

\qquad\ \ 
\begin{tabular}{|l|l|}
\hline
before 1st flip & has \$25, bets \$5 \\ \hline
1st flip & win $\Longrightarrow \,$has \$30, bets \$10 \\ \hline
2nd flip & win $\Longrightarrow \,$has \$40, bets \$20 \\ \hline
3rd flip & lose $\Longrightarrow \,$has \$20, bets \$5 \\ \hline
4th flip & win $\Longrightarrow \,$has \$25, bets \$10 \\ \hline
5th flip & win $\Longrightarrow \,$has \$35, bets \$20 \\ \hline
6th flip & win $\Longrightarrow \,$has \$55, bets \$40 \\ \hline
7th flip & win $\Longrightarrow \,$has \$95 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ almost... \\ \hline
\end{tabular}

We can imagine it is even worse if the first loss occurs in the fourth flip.

What if losing first and second is better than losing just once but in the
third flip?

\qquad\ \ 
\begin{tabular}{|l|l|}
\hline
before 1st flip & has \$25, bets \$5 \\ \hline
1st flip & lose $\Longrightarrow \,$has \$20, bets \$5 \\ \hline
2nd flip & lose $\Longrightarrow \,$has \$15, bets \$5 \\ \hline
3rd flip & win $\Longrightarrow \,$has \$20, bets \$10 \\ \hline
4th flip & win $\Longrightarrow \,$has \$30, bets \$20 \\ \hline
5th flip & win $\Longrightarrow \,$has \$50, bets \$40 \\ \hline
6th flip & win $\Longrightarrow \,$has \$90, bets \$80 \\ \hline
7th flip & win $\Longrightarrow \,$has \$170 \ \ \ DING! DING! \ $%
\Longrightarrow $\ \ $P=0.6^{2}\cdot 0.4^{5}$ \\ \hline
\end{tabular}

There are no other ways feasable to win within seven flips. \ So our
probability is%
\begin{equation*}
P\left( \text{win in }\leq \text{7 steps}\right) =0.4^{4}+0.6\cdot
0.4^{5}+0.6\cdot 0.4^{6}+0.6^{2}\cdot 0.4^{5}\approx 0.037888
\end{equation*}%
So the answer is $0.038,$ which is \fbox{C}.
\end{enumerate}

\bigskip

{\small Last revised: May 15, 2017 - Marta Hidegkuti}

\end{document}