MATHEMATIK
Grundlagen I
Rechenregeln
Aufgabensammlung mit vollständigen Lösungen
neo LERNHILFEN
10
licensed to:
DI Edgar Neuherz
Arbeitsblätter Mathematics
(17th January 2014 11:10) school year
2012 / 13
Responsible for content Dipl.-Ing. Edgar Neuherz
Graz, 2014
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ISBN
NEO Website: mathematik.neo-lernhilfen.at E-Mail an neo.verlag@me.com
Contents
1 Problems 1
1.1 Integer rules . . . 3
1.1.1 Positive and Negative . . . 3
1.1.2 Positive and Negative . . . 3
1.2 Factoring prime numbers . . . 4
1.2.1 Product of Prime Factors . . . 4
1.2.2 Greatest common divisor (gcd) . . . 4
1.2.3 Least common multiple (lcm) . . . 4
1.3 Fractions . . . 5
1.3.1 Addition & Subtraction (of fractions) . . . 5
1.3.2 Multiplication & Division (of fractions . . . 5
1.4 Exponents (Powers) . . . 6
1.4.1 Integer Exponents . . . 6
1.4.2 Variable Exponents . . . 6
2 Solutions 7 2.1 Integer rules . . . 9
2.1.1 Positive and Negative . . . 9
2.1.2 Positive and Negative . . . 10
2.2 Factoring prime numbers . . . 11
2.2.1 Product of Prime Factors . . . 11
2.2.2 Greatest common divisor (gcd) . . . 12
2.2.3 Least common multiple (lcm) . . . 13
2.3 Fractions . . . 14
2.3.1 Addition & Subtraction (of fractions) . . . 14
2.3.2 Multiplication & Division (of fractions . . . 15
2.4 Exponents (Powers) . . . 16
2.4.1 Integer Exponents . . . 16
2.4.2 Variable Exponents . . . 16
Problems 1
Mathematics NEO Lernhilfen 3
1.1 Integer rules
1.1.1 Positive and Negative
11:10
1
17th January 2014
Solve the following problems; make sure you have the correct sign:
1 (+5)−(+7)=
2 (−5)+(+3)=
3 (+2)+(−4)+(+3)=
4 (+4)−(−6)−(+4)=
5 (−3)+(+6)+(+7)−(+2)=
6 (−7)−(−3)+(−5)−(−2)=
7 (−5)+(−2)+(−6)+(+7)−(+5)=
8 (+7)+(+4)+(+3)+(+3)−(−3)=
9 (−3)−(+7)+(−3)+(+6)−(−7)−(−7)=
10 (−3)−(+4)+(−1)+(−5)−(−6)−(+6)=
11 (−4)+(−4)=
12 (+3)+(+3)=
13 (+4)+(+1)+(−2)=
14 (−7)+(−7)+(+3)=
15 (+2)−(−3)+(−2)−(−3)=
16 (−4)+(+7)−(+1)−(+3)=
17 (+7)+(−1)+(+6)−(+4)+(−1)=
18 (−1)−(−1)−(+1)−(−2)+(+7)=
19 (−5)+(−6)+(+4)−(−3)−(+7)+(+3)=
20 (+5)+(+7)+(−7)+(+2)−(+6)−(+6)=
1.1.2 Positive and Negative
11:10
2
17th January 2014
Calculate the following products making sure you follow the sign rules:
21 (−2)·(−4)=
22 (−3)·(+2)=
23 (+5)·(+1)·(−7)=
24 (−6)·(−5)·(+5)=
25 (−3)·(−6)·(+4)·(+4)=
26 (+4)·(+2)·(+7)·(−7)=
27 (+3)·(−3)·(+1)·(+2)·(+6)=
28 (+3)·(+2)·(−5)·(−4)·(+4)=
29 (−6)·(+7)·(−3)·(−1)·(−6)·(+3)=
30 (−5)·(+4)·(−3)·(−1)·(−5)·(−3)=
31 (−2)·(+5)=
32 (+5)·(+7)=
33 (−3)·(−3)·(−7)=
34 (−3)·(+7)·(+6)=
35 (−6)·(+4)·(+6)·(−3)=
36 (−2)·(−7)·(+4)·(−1)=
37 (−3)·(−7)·(−2)·(+5)·(+4)=
38 (+5)·(−1)·(+1)·(+2)·(−1)=
39 (+1)·(+6)·(+5)·(+1)·(−4)·(−4)=
40 (+6)·(+1)·(−7)·(+5)·(−3)·(−2)=
11:10 licensed to DI Edgar Neuherz 17th January 2014
1.2 Factoring prime numbers
1.2.1 Product of Prime Factors
11:10
3
17th January 2014
Break down each of these numbers into their prime factors:
41 91
42 12
43 390
44 8918
45 59290
46 26
47 343
48 3185
49 264
50 4950
51 21
52 245
1.2.2 Greatest common divisor (gcd)
11:10
4
17th January 2014
Break each set into its prime factors; and determine their greatest common divisor (gcd):
53 gcd(12,325)
54 gcd(275,182)
55 gcd(325,338)
56 gcd(2197,275)
57 gcd(24843,8085)
58 gcd(3465,2475)
59 gcd(468,15015)
60 gcd(6006,1144)
61 gcd(34606,12870)
62 gcd(588,674817)
63 gcd(11375,507507)
64 gcd(1716,195195)
1.2.3 Least common multiple (lcm)
11:10
5
17th January 2014
Determine the least common multiple for each set of numbers (lcm):
65 lcm(18,110)
66 lcm(385,3003)
67 lcm(1694,3675)
68 lcm(1092,15246,156)
69 lcm(241670,2205,132)
70 lcm(10010,15125)
71 lcm(48334,11025)
72 lcm(7644,11830)
73 lcm(66,605,44)
74 lcm(10010,264,5070)
75 lcm(28561,4719,3993)
76 lcm(588,264)
Mathematics NEO Lernhilfen 5
1.3 Fractions
1.3.1 Addition & Subtraction (of fractions)
11:10
6
17th January 2014
Complete these fraction problems in three stages; first determine the lowest common de- nominator (lcd), then calculate the sum or difference, finally simplify the answer whenever possible
77 1 42+ 1
110 =
78 1 343+ 1
847 =
79 1 28+ 1
42 =
80 1 605+ 1
275 =
81 1 242− 1
66=
82 1 231− 1
30=
83 1 125+ 1
363 =
84 1 66− 1
18 =
85 1 105− 1
231 =
86 1 539+ 1
18=
87 1 539− 1
20− 1 50=
88 1 165− 1
275+ 1 105 =
89 1 30+ 1
30+ 1 20 =
90 1 27− 1
50+ 1 105 =
91 1 605− 1
42+ 1 175 =
92 1 42+ 1
275+ 1 42=
93 1
1331+ 1 44 + 1
847 =
94 1 147− 1
70− 1 66=
95 1 12− 1
539+ 1 66=
96 1 165+ 1
42− 1 105− 1
70=
1.3.2 Multiplication & Division (of fractions
11:10
7
17th January 2014
Complete these fraction problems in stages; first factorize the numerators and denominators;
then simplify as far as possible; finally note the simplest answer:
97 18 20·385
110 =
98 70 66·99
44=
99 275 125·245
242 =
100 110 66 ·231
231 =
101 63 50·98
42 =
102 42 70·363
154 ·165 20 =
103 70 385·154
12 ·242 847 =
104 1331 70 ·110
18 · 99 275 =
105 165 275·45
75·20 28=
106 42 343·20
66·165 50 =
107 44 363·110
75 · 12 385 ·363
98 =
108 147 105·75
27·42 42·847
539 =
109 70 66·175
231·154 242·165
175 =
110 539 105· 30
847 · 75 539 ·42
63=
111 105 154·1331
165 ·154 66 · 42
242 =
112 66 12 : 363
18 =
113 50 66 : 231
98 =
114 231 12 : 231
66 =
115 30 165 : 12
45 =
116 12 45 : 847
605 =
11:10 licensed to DI Edgar Neuherz 17th January 2014
1.4 Exponents (Powers)
1.4.1 Integer Exponents
11:10
8
17th January 2014
Calculate the new exponent following the rules of powers:
117 623·68
118 76·72
119 429·425
120 222 219
121 426 41
122 1319 139
123 28 22
124 (50)15
125 (49)8
126 (110)1
127 535·515
128 72·70
129 1329·1317
130 428 411
131
523 512
132 1037 1024
133 1214 124
134 (56)31
135 (106)17
136 (110)33
1.4.2 Variable Exponents
11:10
9
17th January 2014
Calculate the new exponent following the rules of powers:
137 y55·y27
138 z40·z28
139 x34· x12
140 a27 a1
141
c27 c2
142 y54 y28
143 b10 b1
144 (y6)21
145 (c5)25
146 (c4)11
147 z49·z28
148 x24· x15
149 y46·y19
150 x40 x16
151 y22 y5
152 c37 c17
153 z9 z9
154 (z0)2
155 (a6)32
156 (c1)17
Solutions 2
Mathematics NEO Lernhilfen 9
2.1 Integer rules
2.1.1 Positive and Negative
11:10
1
17th January 2014
Solve the following problems; make sure you have the correct sign:
1 (+5)−(+7)=(+5)−7= −2
2 (−5)+(+3)=(−5)+3= −2
3 (+2)+(−4)+(+3)=(+2)−4+3= 1
4 (+4)−(−6)−(+4)=(+4)+6−4= 6
5 (−3)+(+6)+(+7)−(+2)= (−3)+6+7−2= 8
6 (−7)−(−3)+(−5)−(−2)= (−7)+3−5+2= −7
7 (−5)+(−2)+(−6)+(+7)−(+5)=(−5)−2−6+7−5= −11
8 (+7)+(+4)+(+3)+(+3)−(−3)=(+7)+4+3+3+3= 20
9 (−3)−(+7)+(−3)+(+6)−(−7)−(−7)=(−3)−7−3+6+7+7= 7
10 (−3)−(+4)+(−1)+(−5)−(−6)−(+6)=(−3)−4−1−5+6−6= −13
11 (−4)+(−4)=(−4)−4= −8
12 (+3)+(+3)=(+3)+3= 6
13 (+4)+(+1)+(−2)=(+4)+1−2= 3
14 (−7)+(−7)+(+3)=(−7)−7+3= −11
15 (+2)−(−3)+(−2)−(−3)= (+2)+3−2+3= 6
16 (−4)+(+7)−(+1)−(+3)= (−4)+7−1−3= −1
17 (+7)+(−1)+(+6)−(+4)+(−1)=(+7)−1+6−4−1= 7
18 (−1)−(−1)−(+1)−(−2)+(+7)=(−1)+1−1+2+7= 8
19 (−5)+(−6)+(+4)−(−3)−(+7)+(+3)=(−5)−6+4+3−7+3= −8
20 (+5)+(+7)+(−7)+(+2)−(+6)−(+6)=(+5)+7−7+2−6−6= −5
11:10 licensed to DI Edgar Neuherz 17th January 2014
2.1.2 Positive and Negative
11:10
2
17th January 2014
Calculate the following products making sure you follow the sign rules:
21 (−2)·(−4)= 8
22 (−3)·(+2)= −6
23 (+5)·(+1)·(−7)= −35
24 (−6)·(−5)·(+5)= 150
25 (−3)·(−6)·(+4)·(+4)= 288
26 (+4)·(+2)·(+7)·(−7)= −392
27 (+3)·(−3)·(+1)·(+2)·(+6)= −108
28 (+3)·(+2)·(−5)·(−4)·(+4)= 480
29 (−6)·(+7)·(−3)·(−1)·(−6)·(+3)= 2268
30 (−5)·(+4)·(−3)·(−1)·(−5)·(−3)= −900
31 (−2)·(+5)= −10
32 (+5)·(+7)= 35
33 (−3)·(−3)·(−7)= −63
34 (−3)·(+7)·(+6)= −126
35 (−6)·(+4)·(+6)·(−3)= 432
36 (−2)·(−7)·(+4)·(−1)= −56
37 (−3)·(−7)·(−2)·(+5)·(+4)= −840
38 (+5)·(−1)·(+1)·(+2)·(−1)= 10
39 (+1)·(+6)·(+5)·(+1)·(−4)·(−4)= 480
Mathematics NEO Lernhilfen 11
2.2 Factoring prime numbers
2.2.1 Product of Prime Factors
11:10
3
17th January 2014
Break down each of these numbers into their prime factors:
41 91= 7·13
91 7 13 13
1
42 12= 2·2·3
12 2
6 2
3 3
1
43 390= 2·3·5·13
390 2 195 3 65 5 13 13
1
44 8918= 2·7·7·7·13
8918 2 4459 7 637 7 91 7 13 13
1
45 59290= 2·5·7·7·11·11
59290 2 29645 5 5929 7 847 7 121 11
11 11 1
46 26= 2·13
26 2 13 13
1
47 343= 7·7·7
343 7 49 7
7 7
1
48 3185= 5·7·7·13
3185 5 637 7 91 7 13 13
1
49 264= 2·2·2·3·11
264 2 132 2 66 2 33 3 11 11
1
50 4950= 2·3·3·5·5·11
4950 2 2475 3 825 3 275 5 55 5 11 11
1 51 21= 3·7
21 3
7 7
1
52 245= 5·7·7
245 5 49 7
7 7
1
11:10 licensed to DI Edgar Neuherz 17th January 2014
2.2.2 Greatest common divisor (gcd)
11:10
4
17th January 2014
Break each set into its prime factors; and determine their greatest common divisor (gcd):
53 gcd(12,325)=1= 1
12 2
6 2
3 3
1
325 5 65 5 13 13
1
54 gcd(275,182)=1= 1
275 5 55 5 11 11
1
182 2 91 7 13 13
1
55 gcd(325,338)=13= 13
325 5
65 5
13 13 1
338 2
169 13 13 13
1 56 gcd(2197,275)=1= 1
2197 13 169 13 13 13
1
275 5 55 5 11 11
1
57 gcd(24843,8085)=147= 3·7·7
24843 3 8281 7 1183 7 169 13
13 13 1
8085 3 2695 5 539 7 77 7 11 11
1
58 gcd(3465,2475)=495= 3·3·5·11
3465 3 1155 3
385 5
77 7
11 11 1
2475 3
825 3
275 5
55 5
11 11 1
gcd(468, = =
60 gcd(6006,1144)=286= 2·11·13
6006 2 3003 3 1001 7 143 11
13 13 1
1144 2
572 2
286 2
143 11 13 13
1
61 gcd(34606,12870)=286= 2·11·13
34606 2 17303 11
1573 11 143 11 13 13
1
12870 2 6435 3 2145 3
715 5
143 11 13 13
1 62 gcd(588,674817)=3= 3
588 2 294 2 147 3 49 7
7 7
1
674817 3 224939 11
20449 11 1859 11 169 13 13 13
1
63 gcd(11375,507507)=91= 7·13
11375 5 2275 5
455 5
91 7
13 13 1
507507 3 169169 7 24167 11
2197 13 169 13 13 13
1
64 gcd(1716,195195)=429= 3·11·13
1716 2
858 2
429 3
143 11
195195 3 65065 5 13013 7 1859 11
Mathematics NEO Lernhilfen 13
11:10 licensed to DI Edgar Neuherz 17th January 2014
2.2.3 Least common multiple (lcm)
11:10
5
17th January 2014
Determine the least common multiple for each set of numbers (lcm):
65 lcm(18,110)=
=990= 2·3·3·5·11 18 2
9 3
3 3
1
110 2
55 5
11 11 1
66 lcm(385,3003)=
=15015= 3·5·7·11·13
385 5
77 7
11 11 1
3003 3 1001 7 143 11
13 13 1 67 lcm(1694,3675)=
=889350= 2·3·5·5·7·7·11·11 1694 2
847 7
121 11 11 11
1
3675 3 1225 5 245 5 49 7
7 7
1 68 lcm(1092,15246,156)=
=396396= 2·2·3·3·7·11·11·13 1092 2
546 2
273 3
91 7
13 13 1
15246 2 7623 3 2541 3
847 7
121 11 11 11
1
156 2 78 2 39 3 13 13
1
69 lcm(241670,2205,132)=
=213152940= 2·2·3·3·5·7·7·11·13·13·13 241670 2
120835 5 24167 11
2205 3 735 3 245 5
132 2 66 2 33 3
71 lcm(48334,11025)=
=532882350= 2·3·3·5·5·7·7·11·13·13·13 48334 2
24167 11 2197 13 169 13 13 13
1
11025 3 3675 3 1225 5 245 5 49 7
7 7
1 72 lcm(7644,11830)=
=496860= 2·2·3·5·7·7·13·13 7644 2
3822 2 1911 3 637 7 91 7 13 13
1
11830 2 5915 5 1183 7 169 13
13 13 1
73 lcm(66,605,44)=
=7260= 2·2·3·5·11·11 66 2
33 3 11 11
1
605 5
121 11 11 11
1
44 2 22 2 11 11
1
74 lcm(10010,264,5070)=
=1561560= 2·2·2·3·5·7·11·13·13 10010 2
5005 5 1001 7 143 11
13 13 1
264 2 132 2 66 2 33 3 11 11
1
5070 2 2535 3
845 5
169 13 13 13
1
75 lcm(28561,4719,3993)=
=114044073= 3·11·11·11·13·13·13·13
28561 13 4719 3 3993 3
Mathematics NEO Lernhilfen 15
11:10 licensed to DI Edgar Neuherz 17th January 2014
2.3 Fractions
2.3.1 Addition & Subtraction (of fractions)
11:10
6
17th January 2014
Complete these fraction problems in three stages; first determine the lowest common de- nominator (lcd), then calculate the sum or difference, finally simplify the answer whenever possible
77
1 42+ 1
110 =
= 1
2·3·7 + 1
2·5·11 = . . .
2·3·5·7·11
| {z }
lcm(42,110)
= 1
2·3·7 ·5·11 5·11+ 1
2·5·11·3·7 3·7 =
= 1 42·55
55 + 1 110 ·21
21= 55+21 2310 = 76
2310 = 38 1155
78 1
343+ 1 847 =
= 1
7·7·7 + 1
7·11·11 = . . .
7·7·7·11·11
| {z }
lcm(343,847)
= 1
7·7·7·11·11
11·11+ 1
7·11·11·7·7 7·7 =
= 1 343·121
121+ 1 847 ·49
49 = 121+49
41503 = 170
41503 = 170 41503
79 1 28+ 1
42 =
= 1
2·2·7 + 1
2·3·7 = . . .
2·2·3·7
| {z }
lcm(28,42)
= 1 2·2·7 ·3
3+ 1 2·3·7·2
2 =
= 1 28·3
3 + 1 42·2
2 =3+2 84 = 5
84 = 5 84
80 1
605+ 1 275 =
= 1
5·11·11+ 1
5·5·11 = . . .
5·5·11·11
| {z }
lcm(605,275)
= 1 5·11·11·5
5+ 1
5·5·11·11 11 =
= 1 605·5
5 + 1 275 ·11
11= 5+11 3025 = 16
3025 = 16 3025
Mathematics NEO Lernhilfen 17
82
1 231 − 1
30 =
= 1
3·7·11− 1
2·3·5 = . . .
2·3·5·7·11
| {z }
lcm(231,30)
= 1
3·7·11·2·5 2·5 − 1
2·3·5·7·11 7·11 =
= 1 231·10
10− 1 30·77
77 =10−77 2310 = −67
2310= −67 2310
83 1
125 + 1 363 =
= 1
5·5·5+ 1
3·11·11 = . . .
3·5·5·5·11·11
| {z }
lcm(125,363)
= 1
5·5·5 ·3·11·11
3·11·11+ 1
3·11·11·5·5·5 5·5·5 =
= 1 125·363
363 + 1 363 ·125
125 = 363+125
45375 = 488
45375 = 488 45375
84 1 66 − 1
18 =
= 1
2·3·11− 1
2·3·3 = . . .
2·3·3·11
| {z }
lcm(66,18)
= 1 2·3·11·3
3 − 1 2·3·3 ·11
11=
= 1 66·3
3− 1 18·11
11 =3−11 198 = −8
198 = −4 99
85 1
105 − 1 231 =
= 1
3·5·7− 1
3·7·11 = . . .
3·5·7·11
| {z }
lcm(105,231)
= 1 3·5·7 ·11
11− 1 3·7·11·5
5 =
= 1 105·11
11− 1 231·5
5 =11−5 1155 = 6
1155 = 2 385
86 1
539 + 1 18 =
= 1
7·7·11+ 1
2·3·3 = . . .
2·3·3·7·7·11
| {z }
lcm(539,18)
= 1
7·7·11·2·3·3 2·3·3+ 1
2·3·3·7·7·11 7·7·11 =
= 1 539·18
18+ 1 18·539
539 = 18+539 9702 = 557
9702 = 557 9702
11:10 licensed to DI Edgar Neuherz 17th January 2014
87
1 539− 1
20− 1 50 =
= 1
7·7·11− 1
2·2·5− 1
2·5·5 = . . .
2·2·5·5·7·7·11
| {z }
lcm(539,20,50)
=
= 1
7·7·11·2·2·5·5 2·2·5·5− 1
2·2·5·5·7·7·11 5·7·7·11− 1
2·5·5 ·2·7·7·11 2·7·7·11 =
= 1 539·100
100− 1 20·2695
2695− 1 50·1078
1078= 100−2695−1078
53900 = −3673
53900 = −3673 53900
88 1
165− 1 275+ 1
105 =
= 1
3·5·11− 1
5·5·11+ 1
3·5·7 = . . .
3·5·5·7·11
| {z }
lcm(165,275,105)
=
= 1
3·5·11·5·7 5·7 − 1
5·5·11·3·7 3·7+ 1
3·5·7·5·11 5·11 =
= 1 165·35
35− 1 275·21
21+ 1 105·55
55 =35−21+55 5775 = 69
5775 = 23 1925
89 1 30+ 1
30+ 1 20 =
= 1
2·3·5 + 1
2·3·5 + 1
2·2·5 = . . .
2·2·3·5
| {z }
lcm(30,30,20)
=
= 1 2·3·5 ·2
2+ 1 2·3·5·2
2 + 1 2·2·5 ·3
3 =
= 1 30·2
2 + 1 30·2
2+ 1 20·3
3 = 2+2+3 60 = 7
60 = 7 60
90
1 27− 1
50+ 1 105 =
= 1
3·3·3 − 1
2·5·5 + 1
3·5·7 = . . .
2·3·3·3·5·5·7
| {z }
lcm(27,50,105)
=
= 1
3·3·3 ·2·5·5·7 2·5·5·7 − 1
2·5·5 ·3·3·3·7 3·3·3·7 + 1
3·5·7 ·2·3·3·5 2·3·3·5 =
= 1 27·350
350 − 1 50·189
189+ 1 105·90
90 =350−189+90 9450 = 251
9450 = 251 9450
1 − 1 + 1
=
Mathematics NEO Lernhilfen 19
92
1 42 + 1
275 + 1 42 =
= 1
2·3·7+ 1
5·5·11+ 1
2·3·7 = . . .
2·3·5·5·7·11
| {z }
lcm(42,275,42)
=
= 1
2·3·7·5·5·11 5·5·11+ 1
5·5·11·2·3·7 2·3·7+ 1
2·3·7·5·5·11 5·5·11 =
= 1 42·275
275+ 1 275·42
42+ 1 42·275
275 = 275+42+275
11550 = 592
11550 = 296 5775
93 1
1331 + 1 44 + 1
847 =
= 1
11·11·11 + 1
2·2·11+ 1
7·11·11 = . . .
2·2·7·11·11·11
| {z }
lcm(1331,44,847)
=
= 1
11·11·11 ·2·2·7 2·2·7 + 1
2·2·11·7·11·11
7·11·11+ 1
7·11·11·2·2·11 2·2·11 =
= 1 1331·28
28+ 1 44·847
847 + 1 847 ·44
44 =28+847+44
37268 = 919
37268 = 919 37268
94 1
147 − 1 70 − 1
66 =
= 1
3·7·7− 1
2·5·7 − 1
2·3·11 = . . .
2·3·5·7·7·11
| {z }
lcm(147,70,66)
=
= 1
3·7·7·2·5·11 2·5·11− 1
2·5·7 ·3·7·11 3·7·11− 1
2·3·11·5·7·7 5·7·7 =
= 1 147·110
110 − 1 70·231
231 − 1 66·245
245 =110−231−245
16170 = −366
16170 = −61 2695
95 1 12 − 1
539 + 1 66 =
= 1
2·2·3− 1
7·7·11+ 1
2·3·11 = . . .
2·2·3·7·7·11
| {z }
lcm(12,539,66)
=
= 1
2·2·3·7·7·11 7·7·11− 1
7·7·11·2·2·3 2·2·3+ 1
2·3·11·2·7·7 2·7·7 =
= 1 12·539
539− 1 539·12
12+ 1 66·98
98 =539−12+98 6468 = 625
6468 = 625 6468
96 1
165 + 1 42 − 1
105− 1 70 =
= 1
3·5·11+ 1
2·3·7− 1
3·5·7− 1
2·5·7 = . . .
2·3·5·7·11
| {z }
lcm(165,42,105,70)
=
= 1
3·5·11·2·7 2·7 + 1
2·3·7 ·5·11 5·11 − 1
3·5·7·2·11 2·11− 1
2·5·7 ·3·11 3·11 =
= 1 165·14
14+ 1 42·55
55− 1 105 ·22
22− 1 70·33
33 = 14+55−22−33
2310 = 14
2310 = 1 165
11:10 licensed to DI Edgar Neuherz 17th January 2014
