2017澳洲AMC中學中級組參考解法

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1-10 3 ,1. 2017 5 (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 2015 5 2 : (C) ,2. A B C D E F G H 小 1 8 F G H 1 (A) 3 8 (B) 1 2 (C) 3 5 (D) 5 3 (E) 1 3 1 2 3 4 5 6 7 8 小中 1 F G H 1 3 8 : (A)

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,3. 中 (A) 14 (B) 15 (C) 16 (D) 17 (E) 18 2 3 4 3× 4 = 12 1 2 × 4 × 2 = 4 12 + 4 = 16 : (C) ,4. 1 1000% (A) 0.1 (B) 1 (C) 10 (D) 100 (E) 1000 1 100% 1 1000% 10 10 : (C)

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,5. 中中 30◦ x◦ x (A) 70 (B) 60 (C) 90 (D) 120 (E) 100 30◦ 30◦ 30◦ x◦ x◦ _x◦ 360◦ 3x + 90 = 360 3x = 270 x = 90 : (C) ,6. 中 (A) 1 2 (B) 13 42 (C) 21 43 (D) 4 123 (E) 14 23 中 1 2 小 14 23 1 2 : (E)

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,7. 小

WORDS SWORD

(A) 3 (B) 4 (C) 6 (D) 7 (E) 8

1

4

words_{→ worsd → wosrd → wsord → sword} 3 3 3 WORDS SWORD 4 1 4 中 2 2 WORDS 中 SWORD 中 3 : (B) 2 中 4 小 4 初 −1 1 0 中 X −5 → −2 → 0 → 2 → 5 X 4 X 4 初初

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X 3 ₋₅ 5 4 4 4 : (B) ,8. 3× 1 小 (A) 34 (B) 28 (C) 56 (D) 40 (E) 10 小中小 5× 4 = 20 小中小 7× 2 = 14 20 + 14 = 34 : (A) ,9. 3a = 4 9b = 7 18(a + b) (A) 38 (B) 75 (C) 198 (D) 132 (E) 22 3 18(a + b) = 6(3a) + 2(9b) = 6× 4 + 2 × 7 = 38 : (A)

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,10. 小

20

(A) 125 (B) 139 (C) 157 (D) 161 (E) 175

20 小 23 + 29 + 31 + 37 + 41 = 161

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11-20 4 ,11. 中 P Q = SQ = SR = QR || || || || P Q R S ∠P SR : ∠P QS (A) 1:1 (B) 1:2 (C) 1:3 (D) 2:3 (E) 3:4 △QRS ∠SQR = ∠QSR = 60◦ _{∠P QS = 120}◦ || || || || P Q R S 60◦ 60◦ 120◦ △P QS ∠QP S = ∠QSP ∠QP S + ∠QSP + ∠P QS = 180◦ ∠QP S = ∠QSP = 30◦ ∠P SR = 90◦ _{∠P SR : ∠P QS = 90 : 120 = 3 : 4} : (E)

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,12. 50 m 20 m 50 m 2 m 1 10 中中 20% (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 50 m 20% 10 m 60 m 5 m 50 m 5 m 中 5 m 2 m 21 2 3 8 : (C)

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,13. 25 (A) 625 (B) 269 (C) 425 (D) 225 (E) 169 13 中 13 13× 13 = 169 : (E)

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,14. 小 5 4 cm 12 cm 小 40 cm (A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm (E) 100 cm 4 cm 12 cm 12 cm 3 cm _√ 4 cm 3 cm 高 42_{+ 3}2 _{= 5 cm} ₈ ₂ 8× 5 + 2 × 4 + 40 = 88 cm 90 cm : (D)

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,15. _√ 20− √ 14 + √ 5−√1 (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 √ 5−√1 =√4 = 2 √ 14 + √ 5−√1 =√14 + 2 = 4 √ 20− √ 14 + √ 5−√1 =√20− 4 = 4 : (D)

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,16. 40 cm× 40 cm P QRS P RS 中 M QR (A) Q (C) Q 10 cm (B) Q 5 cm (D) Q 20 cm (E) P S Q R M 20 20 40 40 1 X P X M X P X = M X P Q P M X P Q QR P X2 _{= 40}2_{+ x}2 _{= x}2_{+ 1600} M X2 _{= 20}2_{+ (40}− x)2 _{= x}2− 80x + 2000 P X2− MX2 = 80x− 400 = 0 x = 400 80 = 5 40 P S Q R M X 20 40− x x X Q 5 cm : (B) 2 XY P M 中 5 cm P M 2 : 1 Y X 1 : 2 P S Q R M X Y Z

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X Q 小 Q 5 cm : (B) ,17. 18 p 13 q 15 r 7 11 p q r (A) 7 (B) 8 (C) 9 (D) 11 (E) 12 7× 11 = 77 p + q + r = 77− (18 + 13 + 15 + 7) = 24 p q r 24÷ 3 = 8 : (B) ,18. a a(a9_{− a}8_{) + a}9 _{= a}x _x (A) 0 (B) 1 (C) 8 (D) 9 (E) 10 ax _{= a(a}9− a8_{) + a}9 _{= a}10− a9 _{+ a}9 _{= a}10 _{x = 10} : (E)

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,19. 0 9 小 (A) 1 (B) 9 (C) 99 (D) 247 (E) 315 小 1 小中中小小 9876 0123 小 50123 49876 50123− 49876 = 247 : (D) ,20. X 中 1 2X (A) 1 4X (B) 1 2X (C) 2 3X (D) 3 4X (E) 5 6X 1 1 X = 6 1 2X = 3 h 1− h 2× (12) + 4× (h × 1) = 2 + 4h = 3 h = 1 4 1− h = 3 4 2× (1 × 1) + 4 × (1 × 3 4) = 5 5 6 : (E)

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2 8 8 6X 8 6X− 1 2X = 5 6X : (E) 21-25 5 ,21. 25 cm 37 cm 中 cm (A) 3 (B) 5 (C) 6 (D) 7 (E) 12 1 L M N AD BD AC 中 A B C D L M N 25 > 37 > △ADC △ALN 2 : 1 LN // DC LN = 1 2DC = 37 2 △DAB △DLM LM // AB LM = 1 2AB = 25 2 LN // DC // LM // AB L M N M N = LN−LM = 37− 25 2 = 6 cm : (C)

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2 中 M N 中 Z x = CZ y = DZ A B C D M N Z 25 > 37 > x y △ABZ △CDZ AZ = 25 37x AC = AZ + CZ = 25 37x + x = 62 37x N C = 1 2AC = 31 37x N Z = CZ − NC = 6 37 x M Z = 6 37y △ZMN △ZDC M N = CD CZ N Z = 37 x 6x 37 = 6 cm : (C)

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,22. 1 2 3 4 5 6 7 8 9 中中 X + Y + Z (A) 9 (B) 10 (C) 11 (D) 12 (E) 13 7 Y X 5 8 Z 9 1 + 2 +· · · + 9 = 45 15 7 Y X 5 8 Z A B C 15 15 15 15 A B C 2 8 2 中 5 8 Z = 2 A = 6 X = 4 Y = 3 1 9 B = 1 C = 9 B = 9 C = 1 X + Y + Z = 4 + 3 + 2 = 9 : (A)

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,23.

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

1

abc 100a + 10b + c = 13(a + b + c) 100a + 10b + c = 13a + 13b + 13c 87a = 3b + 12c 29a = b + 4c. b + 4c 45 a = 1 29 = 1 + 4× 7 = 5 + 4 × 6 = 9 + 4 × 5 117 156 195 3 : (D) 2 n 13 n = 13k 中 k ≥ 8 k n k ≤ 9 + 9 + 9 = 27 n ≤ 13 × 27 = 351 k < 3 + 9 + 9 = 21 k = 8 9 · · · 20 k n = 13k n S k 8 9 10 11 12 13 14 15 16 17 18 19 20 n 104 117 130 143 156 169 182 195 208 221 234 247 260 S 5 9 4 8 12 16 11 15 10 5 9 13 8 n = 117 156 195 3 n 13 : (D)

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,24. 卷中年卷中中 (A) 1 8 (B) 1 6 (C) 1 2 (D) 1 4 (E) 5 16 1 2 3 1 3 x 中 1 4 − x 中中 1 2 − 1 3 = 1 6 x 1 4 − x x− ( 1 4− x ) = 1 6 2x = 1 6+ 1 4 = 5 12 x = 5 24 中 5 24 ÷ 2 3 = 5 16 : (E) 2 120 中 80 40 60 60 中 30 20 25 5

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80 中 25 25 80 = 5 16 : (E)

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,25. a b c (A) 1 : 1 (B) c : (a + b) (C) ab : c2 (D) (a + b)2 _{: 2c}2 _{(E) c}2 _{: 2ab} _| | | | b a c x x a− x = b a ax = b(a− x) (a + b)x = ab x = ab a + b 1 2ab− x 2 x2 = ab 2x2 − 1 = (a + b) 2 2ab − 1 = a 2_{+ 2ab + b}2_{− 2ab} 2ab = a 2_{+ b}2 2ab = c 2 2ab, x b− x x a−x : (E)

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26-30 000-999 26 6 27 7 28 8 29 9 30 10 ,26. 1 2017 1× 2 × 3 × 4 + 1 = 25 = 52 2× 3 × 4 × 5 + 1 = 121 = 112 3× 4 × 5 × 6 + 1 = 361 = 192 ... 2017× 2018 × 2019 × 2020 + 1 = 16 600 254 584 281 = 4 074 3412 2017 5 11 19 _{· · · 4 074 341 中} 1 1 4× 5 × 6 × 7 + 1 = 841 = 292 5 11 19 29 41 55 · · · 6 {5, 1, 9, 9, 1} 中 1 2015 中 2 5× 2015 = 806 1 2017 1 807

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n(n + 1)(n + 2)(n + 3) + 1 n(n + 3)× (n + 1)(n + 2) + 1 = (n2+ 3n)(n2+ 3n + 2) + 1 = ((n2+ 3n + 1)− 1)((n2 + 3n + 1) + 1) + 1 = (n2+ 3n + 1)2− 12 + 1 = (n2+ 3n + 1)2 n (n + 1) (n + 1)2+ 3(n + 1) + 1− (n2+ 3n + 1) = 2n + 4, n = 1 6 n n(n + 3) + 1 n(n + 3) 10 1 n n + 3 5 中 1 : (807) 2 5 11 19 _{· · ·} 6 12 20 _{· · ·} 1 4 10 18 · · · 1 n(n + 1)(n + 2)(n + 3) 中 (n + 1) (n + 2) 中 n (n + 3) 中 n(n + 1)(n + 2)(n + 3) + 1 = (n(n + 3) + 1)2 (n(n + 3) + 1)2− 1 =(n(n + 3) + 1− 1)(n(n + 3) + 1 + 1) = n(n + 3)(n2+ 3n + 2) = n(n + 3)(n + 1)(n + 2) 1 2017 n n(n + 3) + 1 1 n(n + 3) 10 n 0 2 5 7 n 0 201 2 5 7 202 201 + 3× 202 = 807 : (807)

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3 m2 _{= n(n + 1)(n + 2)(n + 3) + 1} _中 _{x = n +} 3 2 m2 = (x−3 2)(x− 1 2)(x + 1 2)(x + 3 2) + 1 = (x2− 9 4)(x 2₋ 1 4) + 1 = x4− 5 2x 2 +25 16 = (x2− 5 4) 2 m = x2− 5 4 = n 2_{+ 3n + 1 = n(n + 3) + 1} m 1 n(n + 3) 10 n 0 2 5 7 n 0 201 2 5 7 202 201 + 3× 202 = 807 : (807) 4 m2− 1 = n(n + 1)(n + 2)(n + 3) 中 _(m− 1)(m + 1) n(n + 3) = n2+ 3n (n + 1)(n + 2) = n2+ 3n + 2 m = n2_{+ 3n + 1} m2 = (n2+ 3n + 1)2 = (n2+ 3n)(n2+ 3n + 2) + 1 = n(n + 3)(n + 1)(n + 2) + 1 n m2 _{= n(n + 1)(n + 2)(n + 3) + 1} m = n(n + 3) + 1 m 1 n(n + 3) 10 n 0 2 5 7 n 0 201 2 5 7 202 201 + 3× 202 = 807 hence (807).

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5 n m2 中 n = 1 · · · 2017 n(n + 1)(n + 2)(n + 3) = m2− 1 = (m − 1)(m + 1) . 中 n(n + 1)(n + 2)(n + 3) 52_{− 1 =} ₄_{× 6} _{= (1}_{× 4) × (2 × 3)} 112− 1 = 10 × 12 = (2 × 5) × (3 × 4) 192− 1 = 18 × 20 = (3 × 6) × (4 × 5) m− 1 = n(n + 3) m + 1 = (n + 1)(n + 2) m = n2_{+ 3n + 1} m− 1 = n2+ 3n = n(n + 3) m + 1 = n2+ 3n + 2 = (n + 1)(n + 2) m2− 1 = (m − 1)(m + 1) = n(n + 3)(n + 1)(n + 2) m2 = n(n + 1)(n + 2)(n + 3) + 1 n m = n2_{+ 3n + 1} m− 1 = n(n + 3) 10 m 1 n 10 n + 3 10 n n + 3 5 n + 3 n 5 n 0 2 5 7 n 0 201 2 5 7 202 201 + 3× 202 = 807 : (807) 6 n m2 中 n = 1 · · · 2017 m2 = n(n + 1)(n + 2)(n + 3) + 1 = n4+ 6n3+ 11n2+ 6n + 1 = n2(n2+ 6n + 9) + 2n2+ 6n + 1 = n2(n + 3)2+ 2n(n + 3) + 1

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z = n(n + 3) m2 = z2+ 2z + 1 = (z + 1)2 m = z + 1 = n2_{+ 3n + 1} m = n2+ 3n + 1 n(n + 3) = m− 1 (n + 1)(n + 2) = n2+ 3n + 2 = m + 1 n(n + 1)(n + 2)(n + 3) + 1 = (m− 1)(m + 1) + 1 = m2 m = n2_{+ 3n + 1} m 1 z = n(n + 3) 10 n n + 3 n n + 3 5 n n = 5, 10, 15, . . . , 2015 (403 ) n = 2, 7, 12, . . . , 2017 (404 ) n 403 + 404 = 807 m 1 : (807)

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,27. 900 1 小 7× 6 × = 42× 12× = 6× 6× = 2× 50× = 900 = 18 18× 36 = 648 : (648) 2 小 1 2 中 P R = AB = 2 + 1 + 2 = 5 P Q = 5 2 P QS P S

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A B P Q R S 5 2 60◦ 2 1 2 P S = P Q sin 60◦ = 5 2 √ 3 2 = √5 3. 小 6× 1 2 (√5 3) 2 = 18 25, 18 25 × 900 = 648 : (648) 3 小 x 小 2x 中 P R = AB = 2x + x + 2x = 5x △P QR 30◦ 60◦ 90◦ △P′Q′R′ 5x A B P Q R 30◦ 2x x Q′ 1 R′ √ 3 P′ 2 1 √ 3 = QR P R = QR 5x QR = √5x 3 小 √ 3 5 小 ( √ 3 5 )2 = 3 25 18 25 18 25 × 900 = 648

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,28. n ≥ 3 n 中 n 中 n 1 8 22 43 · · · 1 8 2222 434343 n 中中 2017 n 小中 m k kn 1 + n + 2n +· · · + mn = 1 + nm(m + 1) 2 2017 nm(m + 1) = 2× 2016 = 26× 32× 7 n 小 m m m + 1 4032 1 3 7 9 21 63 4032 中 1 1 2 4 6 8 64 小 m = 63 m + 1 = 64 n = 1 小 m = 8 m + 1 = 9 n = 56 1 + 56 + 2× 56 + · · · + 8 × 56 = 1 + 36 × 56 = 1 + 2016 = 2017 n 小 56 : (056)

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,29. 中 100 5 5 初中 11 11 1 r c r× c r× c − 100 = (r − 5) × (c + 5) ⇒ rc− 100 = rc + 5r − 5c − 25 ⇒ 5c− 5r = 75 ⇒ c− r = 15. 初 11 11 r× c − (r − 11) × (c + 11) = rc − (rc + 11r − 11c − 121) = 11c− 11r + 121 = 11(c − r) + 121. c− r = 15 11× 15 + 121 = 286 : (286) 2 高 5 6 (i) 5 90◦ (ii) 100 5× 20 6 5 (i) 6 5 5 5× 20 (ii) 6 5 5 6 11× 26 (iii)

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(iii) 6 6 高初中 11× 26 = 286 : (286) ,30. G = 10100 _googol ₁₀G _googolplex _n _nn _{< 10}G n nn < 10G n n = 10k 10G> nn = (10k)(10k)= 10(k10k) ⇐⇒ G > k × 10k k× 10k _{< 10}100 _{k < 10}100−k k = 98 k× 10k= 98× 1098= 0.98× 10100< G nn< 10G k = 99 k× 10k_{= 99}_{× 10}99_{= 9.9}_{× 10}100 _{> G} _nn _{> 10}G n nn _{< 10}G ₁₀98 ≤ n < 1099 _n ₉₉ : (099)