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\begin{document}
\centerline{{\mymagtwobf MATHEMATICS 300 --- FALL 1998}}
\bigbreak
\centerline{{\mymagtwoit Introduction to Mathematical Reasoning}}
\medbreak
\centerline{{\large{\it Instructor: H. J. Sussmann}}} 

\medbreak
\centerline{{\Large{\bf HANDOUT NO. 15.}  November 30, 1998}}

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\medbreak\noindent
{\large{\bf THERE WILL BE ONE LAST HOMEWORK ASSIGNMENT (NO. 10)\\

\vspace{-3mm}
\noindent DUE ON WEDNESDAY, DECEMBER 9.}}

\medbreak\noindent  You will get the list of problems for this assignment 
on Wednesday, December 2. 

\sepa

\noindent Remember that: 

\bi
\item[1.] The {\bf second midterm test} is on Wednesday, December 2. 

\item[2.] The {\bf final exam} is on Wednesday, December 16, 8:00 pm to
11:00 pm. 

\item[3.] During the week of December 7-11, we have {\bf three} meetings, on
Monday, Wednesday and Thursday, because Thursday Dec. 10 runs on a Monday
schedule. 
\ei

\sepa

\begin{center}
{\Large{\bf SOLUTIONS TO SOME HOMEWORK PROBLEMS}}
\end{center}

\medbreak\noindent{\bf PROBLEM 1 OF ASSIGNMENT 9:} Find integers $a$, $b$
such that $37.a+53.b=1$. 

\medbreak\noindent{\bf SOLUTION:} First divide $53$ by $37$, using the
Euclidean algorithm. You get 

$\qquad\qquad\qquad\qquad\qquad\qquad 53 = 1.37 + 16\,$.\hfill(1)


Then divide $37$ by $16$. You get: 

$\qquad\qquad\qquad\qquad\qquad\qquad 37 = 2.16+5\,$.\hfill(2)

Next, divide $16$ by $5$. You get: 

$\qquad\qquad\qquad\qquad\qquad\qquad 16 = 3.5+1\,$.\hfill(3)

Then (3) gives 

$\qquad\qquad\qquad\qquad\qquad\qquad 1=16-3.5\,$,\hfill(4)

and (2) gives 

$\qquad\qquad\qquad\qquad\qquad\qquad 5=37-2.16\,$.\hfill(5)

If we plug the expression for $5$ given by (5) into (4), we get 

$\qquad\qquad\qquad\qquad\qquad\qquad
1=16-3.(37-2.16)=(1-3.(-2)).16+(-3).37\,$, 

so 

$\qquad\qquad\qquad\qquad\qquad\qquad 1=7.16+(-3).37\,$. \hfill(6)

Now (1) gives 

$\qquad\qquad\qquad\qquad\qquad\qquad 16=53-37\,$. \hfill(7)

If we plug the expression for $16$ given by (7) into (6), we get 

$\qquad\qquad\qquad\qquad\qquad\qquad
1=7.(53-37)+(-3).37=7.53+((-7)+(-3)).37\,$,

so 

$\qquad\qquad\qquad\qquad\qquad\qquad 7.53-10.37=1\,.$

So we can take \fbox{$a=7$, $b=-10$.}

\bigbreak\noindent{\bf PROBLEM 3 OF ASSIGNMENT 9:} Prove or disprove: there
exist integers $a$, $b$ such that $103.a+107.b=1$. If the numbers exists,
find them.

\medbreak\noindent{\bf SOLUTION:} First divide $107$ by $103$, using the
Euclidean algorithm. You get 

$\qquad\qquad\qquad\qquad\qquad\qquad 107 = 1.103 + 4\,$.\hfill(1)


Then divide $103$ by $4$. You get: 

$\qquad\qquad\qquad\qquad\qquad\qquad 103 = 25.4+3\,$.\hfill(2)

Next, divide $4$ by $3$. You get: 

$\qquad\qquad\qquad\qquad\qquad\qquad 4 = 1.3+1\,$.\hfill(3)

Then (3) gives 

$\qquad\qquad\qquad\qquad\qquad\qquad 1=4-3\,$,\hfill(4)

and (2) gives 

$\qquad\qquad\qquad\qquad\qquad\qquad 3=103-25.4\,$.\hfill(5)

If we plug the expression for $3$ given by (5) into (4), we get 

$\qquad\qquad\qquad\qquad\qquad\qquad
1=4-3=4-(103-25.4)\,$, 

so 

$\qquad\qquad\qquad\qquad\qquad\qquad 1=26.4-103\,$. \hfill(6)

Now (1) gives 

$\qquad\qquad\qquad\qquad\qquad\qquad 4=107-103\,$. \hfill(7)

If we plug the expression for $4$ given by (7) into (6), we get 

$\qquad\qquad\qquad\qquad\qquad\qquad
1=26.(107-103)-103\,$, 

so 

$\qquad\qquad\qquad\qquad\qquad\qquad 26.107-27.103=1\,.$
2
So we can take \fbox{$a=26$, $b=-27$.}

\end{document}


