%q5S.tex: 
%begin macros
\baselineskip=14pt
\parskip=10pt
\magnification=\magstephalf
\parindent=0pt
\overfullrule=0in
\def\frac#1#2{{#1 \over #2}}
\def\g{\bigtriangledown}
\def\S{\cal S}
\nopagenumbers
%end macros
 

\centerline
{
\bf  Solutions to the Quiz \# 5 for Dr. Z.'s Number Theory
Course for Oct. 24, 2013
}
\bigskip


{\bf 1.} (5 points) Using divisiblity tests, determine
for each of the following 
integers (written in the usual, base $10$, way)
whether it is divisible by (i) $7$ (ii)$13$ . Explain!

{\bf a.} $490735357707$

{\bf b.} $169130403130$

{\bf Sol. to 1a}:
We break it up into blocks of three:
$$
490|735|357|707
$$

For (i)(divisibility by $7$) we need to find
$$
490-735+357-707 \pmod{7} \quad .
$$

Most people first found
$$
490-735+357-707 = -595
$$
and then took it $\pmod{7}$ getting zero (and some people complained that they can't use a calculator!, these
are simple addition/subtraction problems!). But people who do not like to calculate could have
done $\pmod{7}$ for each individual number and realized that they are all $0$!

{\bf Ans. to 1a(i)}: $490735357707$ is divisible by $7$

For 1a(ii), it is the same way, but now we take it $\pmod{13}$.

{\bf First way}: Again we get $-495$. Since $495 \pmod{13}=10$, it is {\bf not divisible by} by $13$.

{\bf Second way:} 
$490 \pmod {13} = 9$, $735 \pmod {13} = 7$, $357 \pmod {13} = 6$, $707 \pmod {13} = 5$
So
$$
490-735+357-707 \pmod{13} = 9 -7 + 6 -5 =3 \quad .
$$
So it is {\bf not} divisible by $13$.

{\bf Sol. to b}:
We break it up into blocks of three:
$$
169|130|403|130
$$
The alternating sum is
$$
(169+403)- (130+130)=312 \quad.
$$

Since $312$ is not divisible by $7$, $169130403130$ is not divisible by $7$, but since
$312$ is divisible by $13$, $169130403130$ is  divisible by $13$.

{\bf Ans. to 1b(i)}: {\bf not} divisible by $7$. {\bf Ans. to 1b(ii)}:  divisible by $13$.

{\bf 2.} (5 points)

What day of the week is

(i) Oct. 24, 2014

(ii) Oct. 24, 2015

(iii) Oct. 24, 2016

(iv) Oct. 24, 2017

(v) Oct. 24, 2018

{\bf Sol. of 2}: Every non-leap year, it advances by $1$ day (since $365 \pmod{7}=1$).
In a leap year (after Feb. 29), it advances by $2$ days.

So, since today is Thursday, (i) Friday (ii) Sat. (iii) Monday (since 2016 is a leap year, and Oct. is after Feb. 29))
(iv) Tues. (v) Wed.


\end
