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2017澳洲AMC中學高級組參考解法

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1-10 3 ,1. 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) ,2. a = 20 b = 17 17a + 20b (A) 680 (B) 689 (C) 1720 (D) 2017 (E) 3737 17× 20 + 20 × 17 = 340 + 340 = 680 : (A)

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,3. 1 1000% (A) 0.1 (B) 1 (C) 10 (D) 100 (E) 1000 1 100% 1 1000% 10 10 : (C) ,4. 42 + 33+ 24 (A) 29 (B) 33 (C) 43 (D) 59 (E) 73 42+ 33 + 24 = 16 + 27 + 16 = 59 : (D) ,5. 中 中 (A) 30 (B) 40 (C) 45 (D) 50 (E) 60 中 360 9 = 40 180 40 90 50 50 : (D)

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,6.

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) ,7. 200− 199 + 198 − 197 + 196 − · · · + 2 − 1 (A) 1 (B) 99 (C) 100 (D) 101 (E) 200 100 (200− 199) (198 − 197) · · · (2 − 1) 1 200− 199 + 198 − 197 + 196 − · · · + 2 − 1 = 100 : (C) ,8. 360 50 8 (A) 54 (B) 55 (C) 56 (D) 57 (E) 58 8 360× 8 = 2880 2880 50 57 30 58 : (E)

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,9. 25 (A) 100 (B) 200 (C) 225 (D) 250 (E) 256 × 1 1 2 2 3 3 4 4 5 5 i (1 + 2 + 3 + 4 + 5)i = 15i 中 15× 1 + 15 × 2 + 15 × 3 + 15 × 4 + 15 × 5 = 15 × (1 + 2 + 3 + 4 + 5) = 15 × 15 = 225 : (C) ,10. 2×√4 2×√800×√4 8 (A) 2 (B) 4 (C) 8 (D) 80 (E) 800 1 2×√42×√800×√4 8 =1600×√416 = 40× 2 = 80 : (D)

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2 x =√2×√4 2×√800×√4 8 x4 = 22 × 2 × 8002× 8 = 84× 104 = 804 x = 80 : (D) 11-20 4 ,11. 中 30 小 (A) 3 (B) 15 (C) 5 (D) 13 (E) 4 2 8 12 ? 2 8 12 x a b c d e 中 中 8 + c + 12 = 30 c = 10b + 12 + x = 30 b = 18− xa + 10 + x = 30 a = 20− x 中 (20− x) + 2 + (18 − x) = 30 40− 2x = 30 x = 5 a = 15 b = 13 d = 7 e = 18 30 : (C)

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,12. 中 (A) 6 (B) 4 (C) 10 (D) 3 (E) 5 1 3 1 3 1 3 1 16 3 2 16− 4 × 3 2 = 10 5 : (E) 2 12+ 32 =10 √ (1210)2+ (1 2 10)2 =5 (5)2 = 5 : (E)

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,13. 240 m S M 60 km/h 高 80 km/h 80 km/h (A) 180 m (B) 240 m (C) 360 m (D) 300 m (E) 320 m 240 m 240 m 4 3 320 m 320 m : (E)

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,14.P Q = SQ = SR = QR ∠ST R = 90◦ T R : P S (A) 1 :3 (B) 1 : 23 (C) 1 : 2 (D) 1 :2 (E) 1 : (33)/2 P Q T R S T R △QRS T QRQR = QS = RS = 2 P Q = 2 2 1 1 2 2 12060 P Q T R S △P QS ∠P QS = 120◦ ∠P SQ = ∠SP Q = 30 ∠P SR = 90◦ P S =P R2− RS2 =12 = 23 : (B)

Note: There are several other solutions that make use of the trigonometric ratios

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,15. 2017x2− 20172017 = 0 (A) x =±2017 (B) x =±20172017 (C) x =±20171008 (D) x =±20171009 (E) x =±20172017 2017x2 = 20172017 x2 = 2017 2017 2017 = 2017 2016 x = ±(20172016)12 = ±20171008 hence (C).

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,16. (A) 221 2 (B) 30 (C) 45 (D) 60 (E) 671 2 2x x x x 2x x x 2x θθ 30/60/90 θ = 30◦ sin θ = x 2x = 1 2 θ = 30 : (B) ,17. $4000 $25 $26 (A) 153 (B) 157 (C) 155 (D) 159 (E) 161 1 160 $25 $4000 $26 160 160 $25 中 26 $25 25 $26 159 $4000 : (D)

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2 $25 $26 m n 中 m n m + n 25m + 26n = 4000 m +26 25 n = 160 m + n + 1 25 n = 160 m + n = 160− 1 25n m + n < 160 n = 25 m + n = 159 159 25 $26 134 $25 : (D) ,18. 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)

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,19. 12 cm 14π cm cm2 (A) 42π + 18 (B) 42π + 9√3 (C) 72π− 18 (D) 72π− 9√3 (E) 24π√3 1 πd = 12π 7 12× 360◦ = 210 210° 210° 150° 150° 30° 30° π 7 6 6 6 6 2× 7 12× π × 6 2 + 6× 6 × sin 30 = 42π + 18 : (A) 2 r θ 12r2− sin θ) 1 2r 2θ 1 2ab| sin θ| θ > π 0 < θ < 2π

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中 1 r = 6 θ = 7 6π A = r2(θ− sin θ) = 36 (7 6π− sin ( 6 ) ) = 42π + 36 sin(π 6 ) = 42π + 18 hence (A).

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,20. 10 cm 16 cm 高 小 (A) 39 cm (B) 42 cm (C) 45 cm (D) 48 cm (E) 50 cm 中 高 小 5 5 6 h 5 5 中 h = 102− 62 = 8cm 高 8 cm 高 小 10 + 4× 8 = 42cm : (B)

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21-25 5 ,21. a b c (A) 1 : 1 (B) c : (a + b) (C) ab : c2 (D) (a + b)2 : 2c2 (E) c2 : 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 + b2− 2ab 2ab = a 2+ b2 2ab = c 2 2ab, x b− x x a−x : (E)

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,22. 小 小 小 小 小 小 小 (A) 1 6 (B) 1 4 (C) 2 5 (D) 1 3 (E) 1 2 1 小 小 小 小 小 小 小 小 小 : (B)

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2x 小 1 4x 1 16x 1 64x · · · 小 1 2 = x + 1 4x + 1 16x + 1 64x +· · · = x 1 14 = 4x 3 x = 1 2 × 3 4 = 3 8 小 2× 3 8 = 3 4 小 1 4 : (B)

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,23. (A) 15 (B) 20 (C) 21 (D) 25 (E) 30 1 n 中 (n 2 ) 10n 10n≤ ( n 2 ) 10n≤ n(n− 1) 2 n ≥ 21. 21 : (C) 2 n 10n 10 21 1 21 : (C)

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,24. c s (A) c s 5% (B) c s 2% (C) c s (D) c s 小 2% (E) c s 小 5% 1 B Q C a M d a A b P D A B C D AB C D C D P Q AD BC ABM P Q//AB P QCDCP a = |AM| b = |AP | d = |CP | AM AD a2 = 2a× b b = 1 2a P D = 3 2a d 2 = (2a)2 +(3 2a )2 d2 = 16 + 9 4 a 2 d = 5 2a s = 8a c = π d c s = 16 π 小 22 7 c/s 110 112 c s 小 2% : (D)

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2 ABCD M AB△MCD M C M D CDD C A M B ABCD 8× 8 5 s = 32 c = 2π× 5 = 10π ≈ 31.4 c ss− c s 0.6 30 = 6 300 = 2 100 c s 小 2% s = 32 97% 98% 99% 31.04 31.36 31.68 c = 10π 31.36 c s 小 2% : (D)

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,25. 中 中 (A) 1 8 (B) 2 5 (C) 1 2 (D) 64 125 (E) 8 15 1 5 中 4 4 5 A1 A2 B C A1 A2 A1 B A1 C A2 B A2 C B C B C A2 C A2 B A1 C A1 B A1 A2 6 中 4 2 3 4 5 × 2 3 = 8 15 : (E) 2 6! = 720 3 23 720÷8 = 90 8 720 = 1 90 90 中 3! = 6

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2 6× 2 × 2 × 2 = 48 48 90 = 8 15 : (E)

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26-30 000-999 26 6 27 7 28 8 29 9 30 10 ,26. 小 28 m 中 小 36 m 小 小 m CIx > 0P I x IC I W P x 28− x 36 3x (28− x)2+ 362 = (3x)2 x2+ 7x− 260 = 0 x = 13 x =−20x = 13 : (013)

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,27.N N 47 N 74 小 74 4 100 4 N 4 4 中 小 n 0 1 2 3 4 5 6 7 8 9 n(n + 1) 0 2 6 2 0 0 2 6 2 0 4 小 N = n(n + 1)(n + 2) 中 n N 4 n n + 1 4 n ≤ 73 n(n + 1)(n + 2) < 753 = 421 875 < N n ≥ 80 n(n + 1)(n + 2) > 803 = 512 000 > N 74≤ n ≤ 79 n n + 1 4 n + 1 = 78 n = 77 77× 78 × 79 = 474 474 小 77 78 79 77 + 78 + 79 = 234 : (234)

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,28. n ≥ 3 nnn 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 nm 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 = 1m = 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 840 8 p1 p3 p5 p7 p0 p2 p4 p6 0 0 1 1 2 2 3 3 4 4 中 p0 · · · p7 p0 = 0 p7 = 1 n pn pn−1 pn+1 pn = P ( ) + P ( ) = 1 2pn+1+ 1 2pn−1 p0 · · · p7 pn= n 7 p3 = 3 7 840 840× 3 7 = 360 : (360)

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,30. 8× 8 小 中 2× 2 小 3× 3 3× 3 2× 2 小 8× 8 1 2× 2 小 中 n× n An BnCn n = 2 A2 = 2 B2 = 4 C2 = 0 6 A2 = 2 B2 = 2 C2 = 0 R N R R 中 – 小 N 中 小 – N

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N 中 小 N 中 小 R R 中 – – 小 N 中 小 R 中 – 小 N 中 小 N R N n× n (n + 1)× (n + 1) • • • 小 小 中 C2 = 0 n Cn= 0 An+1= An Bn+1 = An+ 2Bn n An = 2 Bn+1 = 4+2Bn (Bn+1+4) = 2(Bn+4) Bn+ 4 = 2n−2(B2+ 4) = 2n−2(4 + 4) = 2n+1 Bn = 2n+1− 4 An+ Bn= 2n+1− 2 n = 8 29− 2 = 510 : (510)

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2 nm 小 2m 中 2 2m− 2 2 2n−1 (2m− 2) + 2 × 2n−1 = 2m+ 2n− 2 n× m 28+ 28− 2 = 510 8× 8 : (510) 3 中 2 小 4 小 中 4 2 28 = 256 中 2 254

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90 254 2 + 254 + 254 = 510 8× 8

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