解難小趣 - yottkp.edu.hk · 1 解難小趣 新界年小界學數學小賽界專賽特小...

12
1 ྋᙱ າҘጯᇴጯᔈᗟপΏ 2012ѐ317 Яᇴ在數學競賽中,經常會問到有關最大公因數(H.C.F.)和最小公倍數(L.C.M.)的問題。Ӣ ᏢಞኧᏢѸฅௗόӕޑ(numbers)Ǵ೭ᜪᚒϸᏢғჹኧޑΑԖӭుڅǶ ٿ܈аޑኧӅԖޑӢη(factor)ύǴനεޑᆀനεϦӢኧ(Highest Common Factor) ǹԶٿ܈аޑኧӅԖޑ७ኧύǴനλޑᆀനλϦ७ኧ(Least Common Multiple)ǶٯӵǺ 12ޑӢηǺ1ǵ2ǵ3ǵ4ǵ6ǵ12 18ޑӢηǺ1ǵ2ǵ3ǵ6ǵ9ǵ18 ٿኧӅԖޑӢη1ǵ2ǵ3ǵ6ǴՠനεϦӢኧ6Ƕ 4ޑ७ኧǺ4ǵ8ǵ12ǵ16ǵ20ǵ24ǵ6ޑ७ኧǺ6ǵ12ǵ18ǵ24ǵ30ǵ36ǵ12ک24ٿӅԖޑ७ኧǴՠനλϦ७ኧ12Ƕ ਥᏵॊۓޑကǴऩٿ܈аޑޑനεϦӢኧ1 Ǵሶ೭٤ኧᆀϕ፦ (relatively prime)Ƕ ځჴǴ 12 ё ߄Ң 3 2 2 × × ǴԶ 18 ё ߄Ң 3 3 2 × × Ǵ೭ ߄Ң ݤᆀ፦Ӣኧೱ४ᑈ (Product of Prime Factors)Ƕ೭߄ҢޑݤനεӳೀǴ൩ҺՖኧ߄Ңԋ፦Ӣኧ४ᑈǴኧ Ԅ(unique)ޑǶவٿኧԄǴόᜤว 6 3 2 = × നӭޑӅԖӢηǴΨ൩നεϦӢ ኧǶ٩೭ኬࡘޑၡǴ 36 3 3 2 2 = × × × 12ک18ޑനλϦ७ኧǶሶኬஒٿ܈аޑ ኧᙯԋ፦Ӣኧೱ४ᑈګǻ ٯӵǴТߏ48ᮯԯǴቨ42ᮯԯߎޑឦТǴӧόޑݩΠǴਭрӭϿТय़ ޑനεБߎឦТǻा،೭ᚒǴ൩ाޕၰ48ک42ډڀԖ٤܄፦ǶᡉӦǴ 48ک42ёа2ନǴ܌аǴਭᜐߏ2ᮯԯޑБฅᚒǶՠᗋёаӆεᗺ ༏ǻӧኧॶၨλਔǴךॺёаѐ၂ǶόၸǴኧॶၨεǴޣ܈ा၂ޑኧၨӭਔǴךॺሡ ׳Ԗਏޑ方法ˇˇڱ2 48 42 3 24 21 8 7 аനλޑ፦ኧѐନ48ک42ǴӃ2җۈǴޔԿ٤ኧό2ନǴ൩၂Π፦ኧǴӵ ٤ᜪǴޔԿ܌ԖᎩኧϕ፦ЗǶ೭ਔǴନ48ک42ޑ፦Ӣηޑ४ᑈ൩നεϦӢ ኧǴջ 6 3 2 = × ǴԶӢηೱኧޑ४ᑈ൩48ک42ޑനλϦ७ኧǴջ 336 8 7 3 2 = × × × Ƕ ӣډॊᔈҔᚒǴךޕၰǴߎឦТёਭрޑλБޑനεᜐߏ6ᮯԯǹӅё р56೭ኬޑБǶ

Upload: others

Post on 29-Feb-2020

11 views

Category:

Documents


0 download

TRANSCRIPT

1

2012 3 17

在數學競賽中,經常會問到有關最大公因數(H.C.F.)和最小公倍數(L.C.M.)的問題。

(numbers)

(factor) (Highest Common

Factor) (Least Common

Multiple)

12 1 2 3 4 6 12

18 1 2 3 6 9 18

1 2 3 6 6

4 4 8 12 16 20 24 …

6 6 12 18 24 30 36 …

12 24 12

1

(relatively prime)

12 322 ×× 18 332 ××(Product of Prime Factors)

(unique) 632 =×363322 =××× 12 18

48 42

48 42

48 42 2 2

方法 。

2 48 42

3 24 21

8 7

48 42 2 2

48 42

632 =× 48 42 3368732 =×××

6

56

2

H.C.F. L.C.M. 18 24 32 H.C.F.

L.C.M.

2 18 24 32

9 12 16

9 12 16 H.C.F. 2 L.C.M. 3456161292 =×××L.C.M.

9 12 12 16

2 18 24 32

2 9 12 16

2 9 6 8

3 9 3 4

3 1 4 ←三數再無公因子

18 24 32 L.C.M. 2884133222 =××××××

32 24 18

288 932288 =÷ 1224288 =÷1618288 =÷ 172816129 =××

288

2884133222 =××××××(index notation)

422222 =××× 2 (base) 4 (index) 2

25 32288 ×=

1

5766 31

5766 2 3 312= × × 31

31 2 3 31 186× × =

2 31 62× = 3 31 93× =

3

1186 1 1661 2 2209 3

118511186 =−

165921161 =−

221232209 =+

1185 1659 2212

1185 3 5 79= × ×

1659 3 7 79= × ×

2212 2 2 7 79= × × × 79

165 15

165

165 3 5 11= × ×

= ×15 11 15 15x 15y

x + y = 11

x = 1 y = 10 15 150

x = 2 y = 9 30 135

x = 3 y = 8 45 120

x = 4 y = 7 60 105

x = 5 y = 6 75 90

x > 6

100 6666

100 100

66.661006666 =÷ 6666

Q 1011006666 ×=

99 66 1 132266 =× 100

66

14 30 33 35 39 75 143 169

4

8

14 = 2 × 7

30 = 2 × 3 × 5

33 = 3 × 11

35 = 5 × 7

39 = 3 × 13

75 = 3 × 25

143 = 11 × 13

169 = 213

↑ ↑ ↑

{14} {30, 35} 11 33 143

A B

14, 33 30, 35, 143

14, 143 30, 35, 33

B 143 13 39 B 169

A

A B

14, 33, 75, 169 30, 35, 143, 39

14, 143, 39, 75 30, 35, 33, 169

5

3 2 5 3 7 2 (solutions) 23 23 105

3 70 5 21 7 15 3

105 105

2 70 3 21 2 15 233× + × + × = 233 105 233 105 128 105 23

23 128 233 338……

3 70

5 21

7 15

3 105 3 5 7

5 2 7 3 11 4

7 3 11 4 59 7 11

77 59 77 5 2 59

59 + 77 = 136 5 1

136 + 77 = 213 5 3

213 + 77 = 290 5 0

290 + 77 = 367 5 2

367

7 3 11 4

6

2 3 4 5 6 1 61

2 3 4 5 6 60 61 60 7

61 7

61 + 60 = 121 7 2

121 + 60 = 181 7 6

181 + 60 = 241 7 3

241 + 60 = 301 7 0

301

6 7 8 9

6 7 8 9 504 504 10000

6 7 8 9 10000 504 19÷ = ......424 504

957619504 =×

504 10080

99999 504 99999 504 198÷ = ......207

504 99792198504 =×

10080 504 20

99792 504 198

10080 99792 198 – 20 + 1 = 179

2C(94)

136

73

9

2

xm

m

x

x

9

2

136

73 =−−

m

(73 – x) 2 (136 – x) 9 m

(136 – x) – (73 – x) = 9m – 2m

7m = 63 m = 9

∴ 9273 ×=− x 99136 ×=− x

x = 55

9m(73 – x) = 2m(136 – x)

m[9(73 – x) – 2(136 – x)] = 0 Q m 0

∴ 9(73 – x) – 2(136 – x) = 0

7x = 385

x = 55

7

400

10 8

10 8 40 40

= 10

400

= 40

= 8

400

= 50

= 40

400

= 10

∴ = 40 + 50 – 10

= 80

8

1. 320 240 200

2. 52 57 65 69 95 28 9 161 3. 21 22 34 39 44 45 65 76 133 153

4. 1111 1

5. 433 260 30 50

13 8 6. 115 142 72 7 2

7. 360 314 245 8. 9cm 6cm 7cm 9. 3 4 5 6 2 7 10. 5 2 7 3 9 4 11. 10 9 8 7 6 5

4 3 2 1 12. 5 6 7 1 10000

13. 6

10

14. 20

15

25

9

1. 320 240 200

52320 6 ×=

532240 4 ××=

23 52200 ×=

40523 =×

2. 4

9 = 23

28 = 22 × 7

52 = 22 × 13

57 = 3 × 19

65 = 5 × 13

69 = 3 × 23

95 = 5 × 19

161 = 7 × 23

2 2

( ) 28 52

( ) ( ) {28,69} {52,161}

( ) {28, 57, 69} {9, 52, 161}

( ) {28, 57, 65, 69} {9, 52, 95, 161}

3. 2

21 = 3 × 7

22 = 2 × 11

34 = 2 × 17

39 = 3 × 13

44 = 22 × 11

45 = 23 × 5

65 = 5 × 13

76 = 22 × 19

133 = 7 × 19

153 = 23 × 17

10

19

( ) 76 133

( ) {21, 76} {133}

( ) 21 39 3 39 133 {21, 76}

{39, 133}

( ) ( ) {21, 65, 76} {39, 45, 133}

( ) 45 23 45 153 ( )

153 17 34 153 {21, 65,

76, 153} {34, 39, 45, 133}

( ) ( ) {21, 22, 65, 76, 153} {34, 39, 44, 45, 133}

4. 1111 1111

101111111 ×=

101×a 101×b 101×c 101×d 11=+++ dcba

101

5. 433 13 260 8 420 252

7532420 2 ×××=

732252 22 ××=

847322 =×× 30

50 42732 =××

6. 115 142 72 7 2

115 – 7 = 108

144 + 2 = 144

108 144 72 = 36

7. 360 314 245

360 – 314 = 46 = 232×

314 – 245 = 69 = 233×

360 – 245 = 115 = 235×

23

8. 9cm 6cm 7cm

9 6 7 9 6

18 7 9 6 (relatively prime)

126187 =×

= )7126()6126()9126( ÷+÷+÷ = 5292

11

9. 3 4 5 6 = 605322 =×× 62

62 60 7

62 + 60 = 122 7 3

122 + 60 = 182 7 0

182

10. 5 2 7 3 17 5 7 35 17

35 9 4

17 + 35 = 52 9 7

52 + 35 = 87 9 6

87 + 35 = 122 9 5

122 + 35 = 157 9 4

157 + 35 = 192 9 3

192

11. 10 9 … 2 = 2520

5000 10 9 … 2 1

= 122520 −×

= 5039

12. 5 6 7 L.C.M. = 765 ×× = 210

5 6 7 1 = 210 + 1 = 211

130......4721010000 =÷

5 6 7 1 = 147210 +×

= 9871 < 10000

47

13. 6 10

6 10

Q 6 10 = 30

∴ = 630÷

= 5

= 1030÷

= 3

= 5 – 3

= 2

∴ 15230 =÷

12

14. 20 15 (60x)

20 15 60 x

60 2

3

5

(60x) (6x + 1)

6x + 1 = 25

x = 4

240