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%TCIDATA{<META NAME="Title" CONTENT="Exercises">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:10:19 1997}
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\begin{document}


\section{Exercises\label{Exercises}}

\begin{enumerate}
\item[1.]  \label{q1}Find all the primes between $100$ and $120$.\hfill 
\hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a1}

\item[2.]  \label{q2}Find two positive integers between $1000$ and $1100$
whose greatest common divisor is $23$.\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a2}

\item  \label{q3}Compare the prime factorizations of $19!$, $20!$, $21!$, $%
22!$, $23!$, $24!$, and $25!$ with the number of zeros at the end of the
evaluated factorials. Use these results to predict how many zeros will
appear at the end of $100!$. Check your prediction by direct evaluation.
Revise your method of prediction, if necessary, and predict the number of
zeros for each of $125!$, $200!$, $500!$, $625!$, and $1000!$.\hfill 
\hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a3}

\item  \label{q4}Evaluate numerically the power 
\[
\left( 1+\frac{1}{n}\right) ^{n}
\]
for $n=2,$ $4,$ $8,$ $16,$ $32,$ $64,$ $128$ and $256$. What well-known
number is starting to emerge?\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{%
0.2214in}{0.205in}{0in}}{}{}{a4}

\item  \label{q5}Experiment with numbers to test the potential identities 
\[
a\wedge \left( b\vee c\right) =\left( a\wedge b\right) \vee \left( a\wedge
c\right) 
\]
and \hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{%
a5} 
\[
a\vee \left( b\wedge c\right) =\left( a\vee b\right) \wedge \left( a\vee
c\right) 
\]

\item  \label{q6}Test the potential identity 
\[
A\cap \left( B\cup C\right) =\left( A\cap B\right) \cup \left( A\cap
C\right) 
\]
using the sets $A=\left\{ 1,3,5,7,9\right\} $, $B=\left\{ 1,4,9,16\right\} $%
, and $C=\left\{ 2,3,5,7,11\right\} $.\hfill \hyperref{\TCIIcon{%
BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{a6}

\item  \label{q7}\label{Volume, question}The mass of a block of aluminum is $%
403.2\ \unit{lb}$ and the density is $168\frac{\unit{lb}}{\unit{ft}^{3}}$.
What is its volume?\hfill \hyperref{\TCIIcon{BITMAPSETAnswer}{0.2214in}{%
0.205in}{0in}}{}{}{Volume}

\item  \label{q8}\label{Rocket, question}If a toy rocket shoots vertically
upward with an initial velocity of $80$ $\unit{m}/\unit{s}$, at $t$ seconds
after the rocket takes off, until it returns to the ground, it is at the
height $80t-16t^{2}\unit{m}$. Find the time it takes for the rocket to
return to the ground. When does it reach its highest point?\hfill \hyperref{%
\TCIIcon{BITMAPSETAnswer}{0.2214in}{0.205in}{0in}}{}{}{Rocket}
\end{enumerate}

\section{Solutions\label{Solutions}}

\begin{enumerate}
\item  \label{a1}Test the odd integers between $100$ and $120$ by factoring:

\begin{enumerate}
\item  \textsf{Factor} $101=\allowbreak 101$

\item  \textsf{Factor} $103=\allowbreak 103$

\item  \textsf{Factor} $105=\allowbreak 3\times 5\times 7$

\item  \textsf{Factor} $107=\allowbreak 107$

\item  \textsf{Factor} $109=\allowbreak 109$

\item  \textsf{Factor}\textsc{\ }$111=\allowbreak 3\times 37$

\item  \textsf{Factor} $113=\allowbreak 113$

\item  \textsf{Factor} $115=\allowbreak 5\times 23$

\item  \textsf{Factor} $117=\allowbreak 3^{2}13$

\item  \textsf{Factor} $119=\allowbreak 7\times 17$
\end{enumerate}

It follows that the primes are $101$, $103$, $107$, $109$, and $113$.\hfill 
\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}}{}{}{q1}

\item  \label{a2}\textsf{Expand} gives $\frac{1000}{23}=\allowbreak 43\frac{%
11}{23}$. Note that $44\cdot 23=\allowbreak 1012$ and $45\cdot
23=\allowbreak 1035$. Checking, we see that $\gcd (1012,1035)=\allowbreak 23$%
.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}}{%
}{}{q2}

\item  \label{a3}Note that \textsf{Factor} and \textsf{Evaluate} yield 
\begin{eqnarray*}
19! &=&2^{16}3^{8}5^{3}7^{2}11\times 13\times 17\times 19\allowbreak
=121\,645\,100\,\allowbreak 408\,832\,000 \\
20! &=&2^{18}3^{8}5^{4}7^{2}11\times 13\times 17\times 19\allowbreak
=2432\,902\,008\,\allowbreak 176\,640\,000 \\
21! &=&\allowbreak 2^{18}3^{9}5^{4}7^{3}11\times 13\times 17\times
19=\allowbreak 51\,\allowbreak 090\,942\,171\,\allowbreak 709\,440\,000 \\
22! &=&2^{19}3^{9}5^{4}7^{3}11^{2}13\times 17\times 19=\allowbreak
1124\,\allowbreak 000\,727\,777\,\allowbreak 607\,680\,000 \\
23! &=&2^{19}3^{9}5^{4}7^{3}11^{2}13\times 17\times 19\times 23=\allowbreak
25\,852\,\allowbreak 016\,738\,884\,\allowbreak 976\,640\,000 \\
24! &=&2^{22}3^{10}5^{4}7^{3}11^{2}13\times 17\times 19\times 23=\allowbreak
620\,448\,\allowbreak 401\,733\,239\,\allowbreak 439\,360\,000 \\
25! &=&\allowbreak 2^{22}3^{10}5^{6}7^{3}11^{2}13\times 17\times 19\times
23=\allowbreak 15\,511\,210\,\allowbreak 043\,330\,985\,\allowbreak
984\,000\,000
\end{eqnarray*}
and hence $19!$ ends in $3$ zeros, $20!$ ends in $4$, and $25!$ ends in $6$
zeros. It appears that the exponent of $5$ counts the number of trailing
zeros. The number of fives in $100!$ is given by $\frac{100}{5}+\frac{100}{25%
}=\allowbreak 24$, and direct evaluation shows that $100!$ ends in $24$
zeros. In general, the number of trailing zeros in $n!$ is given by the sum 
\[
\left\lfloor \frac{n}{5}\right\rfloor +\left\lfloor \frac{n}{5^{2}}%
\right\rfloor +\left\lfloor \frac{n}{5^{3}}\right\rfloor +\left\lfloor \frac{%
n}{5^{4}}\right\rfloor +\cdots 
\]
where $\left\lfloor x\right\rfloor $ denotes the greatest integer $\leq x$.
In particular, 
\begin{eqnarray*}
\left\lfloor \frac{125}{5}\right\rfloor +\left\lfloor \frac{125}{5^{2}}%
\right\rfloor +\left\lfloor \frac{125}{5^{3}}\right\rfloor &=&25+5+1=31 \\
\left\lfloor \frac{200}{5}\right\rfloor +\left\lfloor \frac{200}{5^{2}}%
\right\rfloor +\left\lfloor \frac{200}{5^{3}}\right\rfloor &=&40+8+1=49 \\
\left\lfloor \frac{500}{5}\right\rfloor +\left\lfloor \frac{500}{5^{2}}%
\right\rfloor +\left\lfloor \frac{500}{5^{3}}\right\rfloor &=&100+20+4=124 \\
\left\lfloor \frac{625}{5}\right\rfloor +\left\lfloor \frac{625}{5^{2}}%
\right\rfloor +\left\lfloor \frac{625}{5^{3}}\right\rfloor +\left\lfloor 
\frac{625}{5^{4}}\right\rfloor &=&125+25+5+1=156 \\
\left\lfloor \frac{1000}{5}\right\rfloor +\left\lfloor \frac{1000}{5^{2}}%
\right\rfloor +\left\lfloor \frac{1000}{5^{3}}\right\rfloor +\left\lfloor 
\frac{1000}{5^{4}}\right\rfloor &=&200+40+8+1=249
\end{eqnarray*}
Go ahead and verify that these count the number of trailing zeros in $125!$, 
$200!$, $500!$, $625!$, and $1000!,$ respectively. It takes roughly
three-quarters of a 17-inch computer screen to display $1000!$, and the
results are not displayed here.\hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}}{}{}{q3}

\item  \label{a4}Note that 
\begin{eqnarray*}
\left( 1+\frac{1}{2}\right) ^{2} &=&\allowbreak 2.\,\allowbreak 25 \\
\left( 1+\frac{1}{4}\right) ^{4} &=&\allowbreak 2.\,\allowbreak 441\,406\,25
\\
\left( 1+\frac{1}{8}\right) ^{8} &=&\allowbreak 2.\,\allowbreak
565\,784\,513\,\allowbreak 950\,347\,900\,\allowbreak 4 \\
\left( 1+\frac{1}{16}\right) ^{16} &=&\allowbreak 2.\,\allowbreak
637\,928\,497\,\allowbreak 366\,599\,858\,\allowbreak 8 \\
\left( 1+\frac{1}{32}\right) ^{32} &=&\allowbreak 2.\,\allowbreak
676\,990\,129\,\allowbreak 378\,182\,684\,\allowbreak 6 \\
\left( 1+\frac{1}{64}\right) ^{64} &=&\allowbreak 2.\,\allowbreak
697\,344\,952\,\allowbreak 565\,098\,855\,\allowbreak 6 \\
\left( 1+\frac{1}{128}\right) ^{128} &=&\allowbreak 2.\,\allowbreak
707\,739\,019\,\allowbreak 688\,020\,492 \\
\left( 1+\frac{1}{256}\right) ^{256} &=&\allowbreak 2.\,\allowbreak
712\,991\,624\,\allowbreak 253\,434\,334\,\allowbreak 3
\end{eqnarray*}
The number $e=\allowbreak 2.\,\allowbreak 718\,281\,828\,\allowbreak
459\,045\,235\,\allowbreak 4$ is beginning to emerge.\hfill \hyperref{%
\TCIIcon{BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}}{}{}{q4}

\item  \label{a5}With the numbers $1$, $2$, and $3$ we have 
\begin{eqnarray*}
1\wedge \left( 2\vee 3\right) &=&\allowbreak 1\qquad \text{and}\qquad \left(
1\wedge 2\right) \vee \left( 1\wedge 3\right) =1 \\
2\wedge \left( 3\vee 1\right) &=&\allowbreak 2\qquad \text{and}\qquad \left(
2\wedge 3\right) \vee \left( 2\wedge 1\right) =2 \\
3\wedge \left( 1\vee 2\right) &=&\allowbreak 2\qquad \text{and}\qquad \left(
3\wedge 1\right) \vee \left( 3\wedge 2\right) =2
\end{eqnarray*}
Similarly, 
\begin{eqnarray*}
1\vee \left( 2\wedge 3\right) &=&\allowbreak 2\qquad \text{and}\qquad \left(
1\vee 2\right) \wedge \left( 1\vee 3\right) =2 \\
2\vee \left( 1\wedge 3\right) &=&\allowbreak 2\qquad \text{and}\qquad \left(
2\vee 1\right) \wedge \left( 2\vee 3\right) =2 \\
3\vee \left( 1\wedge 2\right) &=&\allowbreak 3\qquad \text{and}\qquad \left(
3\vee 1\right) \wedge \left( 3\vee 2\right) =3
\end{eqnarray*}
These provide experimental evidence that the following are identities:
\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}}{}{%
}{q5} 
\begin{eqnarray*}
a\wedge \left( b\vee c\right) &=&\left( a\wedge b\right) \vee \left( a\wedge
c\right) \\
a\vee \left( b\wedge c\right) &=&\left( a\vee b\right) \wedge \left( a\vee
c\right)
\end{eqnarray*}

\item  \label{a6}Note that 
\[
\left\{ 1,3,5,7,9\right\} \cap \left( \left\{ 1,4,9,16\right\} \cup \left\{
2,3,5,7,11\right\} \right) =\allowbreak \left\{ 1,3,5,7,9\right\} 
\]
and \hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}%
}{}{}{q6} 
\[
\left( \left\{ 1,3,5,7,9\right\} \cap \left\{ 1,4,9,16\right\} \right) \cup
\left( \left\{ 1,3,5,7,9\right\} \cap \left\{ 2,3,5,7,11\right\} \right)
=\allowbreak \left\{ 1,3,5,7,9\right\} 
\]

\item  \label{a7}\label{Volume}The volume of the block of aluminum is $\frac{%
403.2\ \unit{lb}}{168\frac{\unit{lb}}{\unit{ft}{}^{3}}}=\allowbreak .0\,6796%
\unit{m}^{3}$. The volume in cubic feet is the solution to the equation $%
.0\,6796\unit{m}^{3}=\allowbreak x\unit{ft}^{3}$. The solution is $x=2.\,4$%
.\hfill \hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}}{%
}{}{Volume, question}

\item  \label{a8}\label{Rocket}The rocket returns to the ground when its
height is $0\unit{m}$. Solving $\left( 80t-16t^{2}\right) \unit{m}=0\unit{m}$%
, gives the two solutions $t=0$ and $t=5$. The rocket thus returns to the
ground in $5\unit{s}$. The rocket reaches its highest point in half this
time, that is, in $\frac{5}{2}\unit{s}$ $=\allowbreak 2.5\unit{s}$. The
maximum height of the rocket is $80\left( 2.5\right) -16\left( 2.5\right)
^{2}=\allowbreak 100.0\unit{m}$.\hfill \hyperref{\TCIIcon{%
BITMAPSETProbSolvHint}{0.2145in}{0.2603in}{0in}}{}{}{Rocket, question}
\end{enumerate}

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Integers and Fractions}{}{}{DM2-1.tex#Addition}

\item  \hyperref{Elementary Number Theory}{}{}{DM2-2.tex#Prime factorization}

\item  \hyperref{Real Numbers}{}{}{DM2-3.tex#Basic operations}

\item  \hyperref{Functions and Relations}{}{}{DM2-4.tex#Absolute value}

\item  \hyperref{Complex Numbers}{}{}{DM2-5.tex#Complex numbers}

\item  \hyperref{Units and Measurements}{}{}{DM2-6.tex#Units}
\end{itemize}

\rule{0.67in}{0.01in}

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"XNPEU";}}}{}{}{DMIndex.tex#Math Index}\quad {\small Index entries: 
\hyperref{Complex numbers}{}{}{DMIndex.tex#Complex numbers}, \hyperref{%
Functions}{}{}{DMIndex.tex#Functions}, \hyperref{Real numbers}{}{}{%
DMIndex.tex#Real numbers}, \hyperref{Solve}{}{}{DMIndex.tex#Solve}, 
\hyperref{Units}{}{}{DMIndex.tex#Units}}

\end{document}
