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\centerline
{
\bf Solutions to Attendance Quiz \# 7 for Dr. Z.'s Number Theory
Course for Sept. 26, 2013
}
\bigskip
{\bf 1.}  Find  $q(101,17)$ and  $r(101,17)$.

{\bf Sol. to 1}: $5 \cdot 17=85<101$, but $6 \cdot 17=102 \geq 101$, so $q(101,17)=5$ and 
$r(101,17)=101-q(101,17) \cdot 17=101- 5 \cdot 17= 101- 85=16$.

{\bf Ans. to 1}: $q(101,17)=5$ and  $r(101,17)=16$.

{\bf Comment}:  Almost everyone got it right.

{\bf 2.} Use the clever way to find  $Div(210)$.


{\bf Sol. to 2}: Using the algorithm of the  {\bf Fundamental Theorem of Arithmetic}
$$
210= 2^1 \cdot 3^1 \cdot 5^1 \cdot 7^1 \quad
$$
So
$$
Div(210)= \{1,2\} \cdot  \{1,3 \} \cdot  \{1,5 \} \cdot  \{1,7 \}
$$
$$
Div(210)= (\{1,2 \} \cdot  \{1,3 \} )\cdot  (\{1,5\} \cdot  \{1,7\})
$$
$$
= \{ 1 \cdot 1 , 1 \cdot  3,  2 \cdot 1 , 2 \cdot 3 \} \cdot
\{ 1 \cdot 1 , 1 \cdot  7,  5 \cdot 1 , 5 \cdot 7 \} 
$$
$$
= \{ 1 , 2 , 3  , 6 \} \cdot \{ 1 ,  5,  7 , 35 \} 
$$
$$
=\{ 1 \cdot 1 ,  1 \cdot 5,  1 \cdot 7 , 1 \cdot 35  \quad ,
$$
$$
2 \cdot 1 ,  2 \cdot 5,  2 \cdot 7 , 2 \cdot 35  \quad ,
$$
$$
3 \cdot 1 , 3 \cdot 5,  3 \cdot 7 , 3 \cdot 35  \quad ,
$$
$$
6 \cdot 1 , 6 \cdot 5,  6 \cdot 7 , 6 \cdot 35  \} \quad
$$
$$
=\{ 1 ,  5,  7 ,  35  \quad ,
$$
$$
2  ,  10 ,  14 , 70  \quad ,
$$
$$
3 , 15,  21 , 105  \quad ,
$$
$$
6  , 30 ,  42 , 210  \} \quad .
$$
This is a {\bf correct} answer, but it is polite to list the elements in increasing order. so a

{\bf Polite Answer to 2} is:
$$
\{1,2,3,5,6,7,10,14, 15, 21, 30, 35, 42, 70, 105, 210 \} \quad.
$$

{\bf Comment}:  About \%60 got it right. It is a good idea to check by  using the
formula for the number of divisors
$$
|Div(n)]=\prod_{i=1}^k (a_i+1)  \quad .
$$

Here $k=4$ and $a_1=1,a_2=1,a_3=1,a_4=1$, so
$$
|Div(210)|=(1+1)(1+1)(1+1)(1+1)=2^4=16 \quad ,
$$
and indeed we got $16$ elements in our answer.

\end
