Solutions

  1. Test the odd integers between 100 and 120 by factoring:

    1. Factor 101 = 101

    2. Factor 103 = 103

    3. Factor 105 = 3×5×7

    4. Factor 107 = 107

    5. Factor 109 = 109

    6. Factor 111 = 3×37

    7. Factor 113 = 113

    8. Factor 115 = 5×23

    9. Factor 117 = 3213

    10. Factor 119 = 7×17

    It follows that the primes are 101, 103, 107, 109, and 113. BITMAPSETProbSolvHint0.2145in0.2603in0inq1

  2. Expand gives ${\frac{{1000}}{{23}}}$ = 43${\frac{{%
11}}{{23}}}$. Note that 44⋅23 = 1012 and 45⋅23 = 1035. Checking, we see that gcd(1012, 1035) = 23.BITMAPSETProbSolvHint0.2145in0.2603in0inq2

  3. Note that Factor and Evaluate yield
    19! = 21638537211×13×17×19 = 121 645 100 408 832 000  
    20! = 21838547211×13×17×19 = 2432 902 008 176 640 000  
    21! = 21839547311×13×17×19 = 51 090 942 171 709 440 000  
    22! = 21939547311213×17×19 = 1124 000 727 777 607 680 000  
    23! = 21939547311213×17×19×23 = 25 852 016 738 884 976 640 000  
    24! = 222310547311213×17×19×23 = 620 448 401 733 239 439 360 000  
    25! = 222310567311213×17×19×23 = 15 511 210 043 330 985 984 000 000  

    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 ${\frac{{100}}{{5}}}$ + ${\frac{{100}}{{25}}}$ = 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

    $\displaystyle \left\lfloor\vphantom{ \frac{n}{5}}\right.$$\displaystyle {\frac{{n}}{{5}}}$$\displaystyle \left.\vphantom{ \frac{n}{5}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{n}{5^{2}}%
}\right.$$\displaystyle {\frac{{n}}{{5^{2}}%
}}$$\displaystyle \left.\vphantom{ \frac{n}{5^{2}}%
}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{n}{5^{3}}}\right.$$\displaystyle {\frac{{n}}{{5^{3}}}}$$\displaystyle \left.\vphantom{ \frac{n}{5^{3}}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{%
n}{5^{4}}}\right.$$\displaystyle {\frac{{%
n}}{{5^{4}}}}$$\displaystyle \left.\vphantom{ \frac{%
n}{5^{4}}}\right\rfloor$ + ...

    where $\left\lfloor\vphantom{ x}\right.$x$\left.\vphantom{ x}\right\rfloor$ denotes the greatest integer ≤x. In particular,
    $\displaystyle \left\lfloor\vphantom{ \frac{125}{5}}\right.$$\displaystyle {\frac{{125}}{{5}}}$$\displaystyle \left.\vphantom{ \frac{125}{5}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{125}{5^{2}}%
}\right.$$\displaystyle {\frac{{125}}{{5^{2}}%
}}$$\displaystyle \left.\vphantom{ \frac{125}{5^{2}}%
}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{125}{5^{3}}}\right.$$\displaystyle {\frac{{125}}{{5^{3}}}}$$\displaystyle \left.\vphantom{ \frac{125}{5^{3}}}\right\rfloor$ = 25 + 5 + 1 = 31  
    $\displaystyle \left\lfloor\vphantom{ \frac{200}{5}}\right.$$\displaystyle {\frac{{200}}{{5}}}$$\displaystyle \left.\vphantom{ \frac{200}{5}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{200}{5^{2}}%
}\right.$$\displaystyle {\frac{{200}}{{5^{2}}%
}}$$\displaystyle \left.\vphantom{ \frac{200}{5^{2}}%
}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{200}{5^{3}}}\right.$$\displaystyle {\frac{{200}}{{5^{3}}}}$$\displaystyle \left.\vphantom{ \frac{200}{5^{3}}}\right\rfloor$ = 40 + 8 + 1 = 49  
    $\displaystyle \left\lfloor\vphantom{ \frac{500}{5}}\right.$$\displaystyle {\frac{{500}}{{5}}}$$\displaystyle \left.\vphantom{ \frac{500}{5}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{500}{5^{2}}%
}\right.$$\displaystyle {\frac{{500}}{{5^{2}}%
}}$$\displaystyle \left.\vphantom{ \frac{500}{5^{2}}%
}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{500}{5^{3}}}\right.$$\displaystyle {\frac{{500}}{{5^{3}}}}$$\displaystyle \left.\vphantom{ \frac{500}{5^{3}}}\right\rfloor$ = 100 + 20 + 4 = 124  
    $\displaystyle \left\lfloor\vphantom{ \frac{625}{5}}\right.$$\displaystyle {\frac{{625}}{{5}}}$$\displaystyle \left.\vphantom{ \frac{625}{5}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{625}{5^{2}}%
}\right.$$\displaystyle {\frac{{625}}{{5^{2}}%
}}$$\displaystyle \left.\vphantom{ \frac{625}{5^{2}}%
}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{625}{5^{3}}}\right.$$\displaystyle {\frac{{625}}{{5^{3}}}}$$\displaystyle \left.\vphantom{ \frac{625}{5^{3}}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{
\frac{625}{5^{4}}}\right.$$\displaystyle {\frac{{625}}{{5^{4}}}}$$\displaystyle \left.\vphantom{
\frac{625}{5^{4}}}\right\rfloor$ = 125 + 25 + 5 + 1 = 156  
    $\displaystyle \left\lfloor\vphantom{ \frac{1000}{5}}\right.$$\displaystyle {\frac{{1000}}{{5}}}$$\displaystyle \left.\vphantom{ \frac{1000}{5}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{1000}{5^{2}}%
}\right.$$\displaystyle {\frac{{1000}}{{5^{2}}%
}}$$\displaystyle \left.\vphantom{ \frac{1000}{5^{2}}%
}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{ \frac{1000}{5^{3}}}\right.$$\displaystyle {\frac{{1000}}{{5^{3}}}}$$\displaystyle \left.\vphantom{ \frac{1000}{5^{3}}}\right\rfloor$ + $\displaystyle \left\lfloor\vphantom{
\frac{1000}{5^{4}}}\right.$$\displaystyle {\frac{{1000}}{{5^{4}}}}$$\displaystyle \left.\vphantom{
\frac{1000}{5^{4}}}\right\rfloor$ = 200 + 40 + 8 + 1 = 249  

    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.BITMAPSETProbSolvHint0.2145in0.2603in0inq3

  4. Note that
    $\displaystyle \left(\vphantom{ 1+\frac{1}{2}}\right.$1 + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{2}}\right)^{{2}}_{}$ = 2. 25  
    $\displaystyle \left(\vphantom{ 1+\frac{1}{4}}\right.$1 + $\displaystyle {\frac{{1}}{{4}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{4}}\right)^{{4}}_{}$ = 2. 441 406 25  
    $\displaystyle \left(\vphantom{ 1+\frac{1}{8}}\right.$1 + $\displaystyle {\frac{{1}}{{8}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{8}}\right)^{{8}}_{}$ = 2. 565 784 513 950 347 900 4  
    $\displaystyle \left(\vphantom{ 1+\frac{1}{16}}\right.$1 + $\displaystyle {\frac{{1}}{{16}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{16}}\right)^{{16}}_{}$ = 2. 637 928 497 366 599 858 8  
    $\displaystyle \left(\vphantom{ 1+\frac{1}{32}}\right.$1 + $\displaystyle {\frac{{1}}{{32}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{32}}\right)^{{32}}_{}$ = 2. 676 990 129 378 182 684 6  
    $\displaystyle \left(\vphantom{ 1+\frac{1}{64}}\right.$1 + $\displaystyle {\frac{{1}}{{64}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{64}}\right)^{{64}}_{}$ = 2. 697 344 952 565 098 855 6  
    $\displaystyle \left(\vphantom{ 1+\frac{1}{128}}\right.$1 + $\displaystyle {\frac{{1}}{{128}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{128}}\right)^{{128}}_{}$ = 2. 707 739 019 688 020 492  
    $\displaystyle \left(\vphantom{ 1+\frac{1}{256}}\right.$1 + $\displaystyle {\frac{{1}}{{256}}}$$\displaystyle \left.\vphantom{ 1+\frac{1}{256}}\right)^{{256}}_{}$ = 2. 712 991 624 253 434 334 3  

    The number e = 2. 718 281 828 459 045 235 4 is beginning to emerge.BITMAPSETProbSolvHint0.2145in0.2603in0inq4

  5. With the numbers 1, 2, and 3 we have
    1∧$\displaystyle \left(\vphantom{ 2\vee 3}\right.$2∨3$\displaystyle \left.\vphantom{ 2\vee 3}\right)$ = 1        and        $\displaystyle \left(\vphantom{
1\wedge 2}\right.$1∧2$\displaystyle \left.\vphantom{
1\wedge 2}\right)$$\displaystyle \left(\vphantom{ 1\wedge 3}\right.$1∧3$\displaystyle \left.\vphantom{ 1\wedge 3}\right)$ = 1  
    2∧$\displaystyle \left(\vphantom{ 3\vee 1}\right.$3∨1$\displaystyle \left.\vphantom{ 3\vee 1}\right)$ = 2        and        $\displaystyle \left(\vphantom{
2\wedge 3}\right.$2∧3$\displaystyle \left.\vphantom{
2\wedge 3}\right)$$\displaystyle \left(\vphantom{ 2\wedge 1}\right.$2∧1$\displaystyle \left.\vphantom{ 2\wedge 1}\right)$ = 2  
    3∧$\displaystyle \left(\vphantom{ 1\vee 2}\right.$1∨2$\displaystyle \left.\vphantom{ 1\vee 2}\right)$ = 2        and        $\displaystyle \left(\vphantom{
3\wedge 1}\right.$3∧1$\displaystyle \left.\vphantom{
3\wedge 1}\right)$$\displaystyle \left(\vphantom{ 3\wedge 2}\right.$3∧2$\displaystyle \left.\vphantom{ 3\wedge 2}\right)$ = 2  

    Similarly,
    1∨$\displaystyle \left(\vphantom{
2\wedge 3}\right.$2∧3$\displaystyle \left.\vphantom{
2\wedge 3}\right)$ = 2        and        $\displaystyle \left(\vphantom{ 1\vee 2}\right.$1∨2$\displaystyle \left.\vphantom{ 1\vee 2}\right)$$\displaystyle \left(\vphantom{ 1\vee 3}\right.$1∨3$\displaystyle \left.\vphantom{ 1\vee 3}\right)$ = 2  
    2∨$\displaystyle \left(\vphantom{ 1\wedge 3}\right.$1∧3$\displaystyle \left.\vphantom{ 1\wedge 3}\right)$ = 2        and        $\displaystyle \left(\vphantom{
2\vee 1}\right.$2∨1$\displaystyle \left.\vphantom{
2\vee 1}\right)$$\displaystyle \left(\vphantom{ 2\vee 3}\right.$2∨3$\displaystyle \left.\vphantom{ 2\vee 3}\right)$ = 2  
    3∨$\displaystyle \left(\vphantom{
1\wedge 2}\right.$1∧2$\displaystyle \left.\vphantom{
1\wedge 2}\right)$ = 3        and        $\displaystyle \left(\vphantom{ 3\vee 1}\right.$3∨1$\displaystyle \left.\vphantom{ 3\vee 1}\right)$$\displaystyle \left(\vphantom{ 3\vee 2}\right.$3∨2$\displaystyle \left.\vphantom{ 3\vee 2}\right)$ = 3  

    These provide experimental evidence that the following are identities: BITMAPSETProbSolvHint0.2145in0.2603in0inq5
    a$\displaystyle \left(\vphantom{ b\vee c}\right.$bc$\displaystyle \left.\vphantom{ b\vee c}\right)$ = $\displaystyle \left(\vphantom{ a\wedge b}\right.$ab$\displaystyle \left.\vphantom{ a\wedge b}\right)$$\displaystyle \left(\vphantom{ a\wedge
c}\right.$ac$\displaystyle \left.\vphantom{ a\wedge
c}\right)$  
    a$\displaystyle \left(\vphantom{ b\wedge c}\right.$bc$\displaystyle \left.\vphantom{ b\wedge c}\right)$ = $\displaystyle \left(\vphantom{ a\vee b}\right.$ab$\displaystyle \left.\vphantom{ a\vee b}\right)$$\displaystyle \left(\vphantom{ a\vee
c}\right.$ac$\displaystyle \left.\vphantom{ a\vee
c}\right)$  

  6. Note that

    $\displaystyle \left\{\vphantom{ 1,3,5,7,9}\right.$1, 3, 5, 7, 9$\displaystyle \left.\vphantom{ 1,3,5,7,9}\right\}$$\displaystyle \left(\vphantom{ \left\{ 1,4,9,16\right\} \cup \left\{
2,3,5,7,11\right\} }\right.$$\displaystyle \left\{\vphantom{ 1,4,9,16}\right.$1, 4, 9, 16$\displaystyle \left.\vphantom{ 1,4,9,16}\right\}$$\displaystyle \left\{\vphantom{
2,3,5,7,11}\right.$2, 3, 5, 7, 11$\displaystyle \left.\vphantom{
2,3,5,7,11}\right\}$$\displaystyle \left.\vphantom{ \left\{ 1,4,9,16\right\} \cup \left\{
2,3,5,7,11\right\} }\right)$ = $\displaystyle \left\{\vphantom{ 1,3,5,7,9}\right.$1, 3, 5, 7, 9$\displaystyle \left.\vphantom{ 1,3,5,7,9}\right\}$

    and BITMAPSETProbSolvHint0.2145in0.2603in0inq6

    $\displaystyle \left(\vphantom{ \left\{ 1,3,5,7,9\right\} \cap \left\{ 1,4,9,16\right\} }\right.$$\displaystyle \left\{\vphantom{ 1,3,5,7,9}\right.$1, 3, 5, 7, 9$\displaystyle \left.\vphantom{ 1,3,5,7,9}\right\}$$\displaystyle \left\{\vphantom{ 1,4,9,16}\right.$1, 4, 9, 16$\displaystyle \left.\vphantom{ 1,4,9,16}\right\}$$\displaystyle \left.\vphantom{ \left\{ 1,3,5,7,9\right\} \cap \left\{ 1,4,9,16\right\} }\right)$$\displaystyle \left(\vphantom{ \left\{ 1,3,5,7,9\right\} \cap \left\{ 2,3,5,7,11\right\} }\right.$$\displaystyle \left\{\vphantom{ 1,3,5,7,9}\right.$1, 3, 5, 7, 9$\displaystyle \left.\vphantom{ 1,3,5,7,9}\right\}$$\displaystyle \left\{\vphantom{
2,3,5,7,11}\right.$2, 3, 5, 7, 11$\displaystyle \left.\vphantom{
2,3,5,7,11}\right\}$$\displaystyle \left.\vphantom{ \left\{ 1,3,5,7,9\right\} \cap \left\{ 2,3,5,7,11\right\} }\right)$ = $\displaystyle \left\{\vphantom{ 1,3,5,7,9}\right.$1, 3, 5, 7, 9$\displaystyle \left.\vphantom{ 1,3,5,7,9}\right\}$

  7. The volume of the block of aluminum is ${\frac{{%
403.2\ \unit{lb}}}{{168\frac{\unit{lb}}{\unit{ft}{}^{3}}}}}$ = .0 6796$\unit$m3. The volume in cubic feet is the solution to the equation .0 6796$\unit$m3 = x$\unit$ft3. The solution is x = 2. 4.BITMAPSETProbSolvHint0.2145in0.2603in0inVolume, question

  8. The rocket returns to the ground when its height is 0$\unit$m. Solving $\left(\vphantom{ 80t-16t^{2}}\right.$80t - 16t2$\left.\vphantom{ 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$s = 2.5$\unit$s. The maximum height of the rocket is 80$\left(\vphantom{ 2.5}\right.$2.5$\left.\vphantom{ 2.5}\right)$ -16$\left(\vphantom{ 2.5}\right.$2.5$\left.\vphantom{ 2.5}\right)^{{2}}_{}$ = 100.0$\unit$m.BITMAPSETProbSolvHint0.2145in0.2603in0inRocket, question



Subsections