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# 2019n，求n 的最大可能整數值

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1.

例: 5! = 5  4  3  2  1

Given that 23! > 2019n, find the largest possible integer n.

23! denotes 23 factorial, the product of all the natural numbers from 1 to 23 (inclusive).

For example, 5! = 5  4  3  2  1 建議題解

23!  2.58  1022 2019n  2.019n  (103)nn = 7,

20197 = 2.0197  (103)7 = 2.0197  1021  27  1021  128  1021  1.28  1023n = 6,

20196  2.0196  (103)6  26  1018  64  1018  6.4  1019

∴ n 的最大值是 6。

Suggested Solutions:

23!  2.58  1022 2019n  2.019n  (103)n When n = 7,

20197 = 2.0197  (103)7 = 2.0197  1021  27  1021  128  1021  1.28  1023 When n = 6,

20196  2.0196  (103)6  26  1018  64  1018  6.4  1019

∴ The largest value of n is 6.

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2.

A、B、C、D 和 E 點分別是 5 個圓的圓心。各圓的半徑是 3 厘米。如圖 1所示，所有圓相交。求

Points A, B, C, D and E are the centers of five circles respectively. The radius of each circle is 3 cm. The circles intersect as shown in figure 1. Find the perimeter of the entire figure (marked in solid line). Give the answer in terms of .

1 / Figure 1

(3)

FAG = 360  60  60  108= 132 (同頂角)

 整個圖形的周界: 2 × 3 × ˚˚ × 5 = 11 厘米 Suggested Solutions:

Connect points A, B, C, D and E to form a regular pentagon.

The interior angle of a regular pentagon: [ (5  2) × 180]  5 = 108

( sum of polygon)

Connect points A, F, B and A, G, E to form two equilateral triangles.

The interior angle of an equilateral triangle: 180  3 = 60 ( sum of )

FAG = 360  60  60  108= 132 (s at a point)

 The perimeter of the entire figure: 2 × 3 × ˚˚ × 5 = 11 (cm)

(4)

3.

n  n 的方格內，包含了數字格、地雷格（ ）和安全格（）。數字格中的值顯示周邊有多少個

In n  n grid, there are Number grids, Mine grids ( ) and Safe grids (). The number inside the Number grid indicates the number of Mine grids surrounding it (vertically, horizontally, or diagonally). Figure 2a shows an example.

1

3 2

2

1 2 1  2a / Figure 2a

In figure 2b, the Mine grids and Safe grids are hidden in the blank space. Find them out and label them by ( ) and () respectively.

2 2 3

2

3 1

4

1

2 1

2b / Figure 2b

(5)

2 2 3

2

   3 1

4   

1

2     1

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4.

Figure 3a shows a 3D maze made by the surface of a regular tetrahedron. It is unfolded as a net (figure 3b). Draw the path from the starting point to the end point in figure 3b. (Show your path clearly and all the unwanted paths should be erased.)

3a / Figure 3a

3b / Figure 3b

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5.

Bus route number 23, from KWUN TONG FERRY to SHUN TIN, serves from morning 06:10 to midnight 00:25 in every 12 to 25 minutes. A bus left the terminal precisely on the minute. The average speed over the first 12 km was 44 km/h. Meanwhile, the driver consulted his watch and saw the hour-hand was overlapping with the minute-hand.

(a) 求巴士行駛該 12 公里路段所需的時間，以分鐘作單位，並以真分數表達。

Find the time needed for the bus travelled 12 km in minutes and express your answer in proper fraction.

(b) 請問巴士是何時離開車站? (以 24 小時報時制式表達)

At what time did the bus leave the terminal? (express your answer in 24-hour notation) 建議題解：

(a) 行車時間 = 公里

1 小時行走 = 30

1 分鐘行走

1 分鐘行走 = 6

(9)

Suggested Solutions:

(a) Time needed = / 60 min =16 min (b)

hour-hand 12 hours for 360

1 hour for = 30

1 min

minute-hand 1 hour for 360

1 min for = 6

Assume the driver saw the hour-hand was overlapping with the minute-hand at b minutes pass a , where 0

 a  11, 0  b  60 , a is an integer, b is a real number.

Hour-hand is pointing to 30(a) + (b) which is the same as the minute-hand 6×(b).

Therefore, we have

30(a) + (b) = 6×(b) 60a = 11b b =

As the bus left terminal precisely on the minute, the decimals part of b equals to the decimals part of the time needed in part (a).

i.e.

 16 is an integer 60a  180 is a multiple of 11

∴ a = 3、b =16 .

The driver saw the hour-hand was overlapping with the minute-hand at 16 minutes pass 3, the time for travelling 12 km is 16 min.

The bus left at 15:00 .

(10)

6.

Given that x < y < z , solve the following system of equations.

1341 … … … … 1 63 … … … … 2

… … … … 3 建議題解：

x = 21  d , z = 21  d (4) 由 (4) 及 (1)，得

(21  d)2  212  (21  d)2 = 1341 2d2 = 18

d = 3 (捨去) or 3

∴ x = 18, y = 21, z = 24 Suggested Solutions:

By (2) and (3), we have 3y = 63 y = 21

Let d = x  21 = 21  z ,

x = 21  d , z = 21  d (4) By (4) and (1), we have

(21  d)2  212  (21  d)2 = 1341 2d2 = 18

d = 3 (rej.) or 3

∴ x = 18, y = 21, z = 24

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7.

400 至 800間的整數（包括400 和 800），共有多少個能符合以下所有條件：

(I) 不能被 5 整除 (II) 不能被 6 整除 (III) 不包含數字「5」

(IV) 不包含數字「6」

For all integers between 400 and 800 (inclusively), how many of them fulfill ALL the following conditions:

(I) NOT divisible by 5 (II) NOT divisible by 6

(III) DO NOT contain digit “5”

(IV) DO NOT contain digit “6”

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5 和 6 的 L.C.M. 是 30。

400 至 420 400

401, 402, 403, 404, 405, 406, 407, 408, 409, 410 411, 412, 413, 414, 415, 416, 417, 418, 419, 420

421, 422, 423, 424, 425, 426, 427, 428, 429, 430 431, 432, 433, 434, 435, 436, 437, 438, 439, 440 441, 442, 443, 444, 445, 446, 447, 448, 449, 450

451 至 480 (留意要刪去十位是「5」或「6」的數字)

(451), (452), (453), (454), 455, 456, (457), (458), (459), 460 (461), 462, (463), (464), 465, 466, (467), 468, (469), 470 471, 472, 473, 474, 475, 476, 477, 478, 479, 480

18 –7 – 5 = 6 個

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Suggested Solutions:

The least common multiple of 5 and 6 is 30.

Start from 400, 420 is the least multiple of 30.

We count the numbers between 400 and 420 first.

From 400 to 420 400

401, 402, 403, 404, 405, 406, 407, 408, 409, 410 411, 412, 413, 414, 415, 416, 417, 418, 419, 420

Remain: 11 numbers

From 421 to 450

421, 422, 423, 424, 425, 426, 427, 428, 429, 430 431, 432, 433, 434, 435, 436, 437, 438, 439, 440 441, 442, 443, 444, 445, 446, 447, 448, 449, 450

Remain: 18 numbers

From 451 to 480 (Notice that the digit in tens place: “5” or “6”) (451), (452), (453), (454), 455, 456, (457), (458), (459), 460 (461), 462, (463), (464), 465, 466, (467), 468, (469), 470 471, 472, 473, 474, 475, 476, 477, 478, 479, 480

Remain:

18 –7 – 5 = 6 numbers

Region Remain

From 481 to 510 18 – 6 = 12 From 511 to 690 0

From 691 to 720 18 – 7 = 11 From 721 to 750 18

From 751 to 780 18 –7 – 5 = 6 From 781 to 800 12

The total numbers: 11 + 18 + 6 + 12 + 11 + 18 + 6 + 12 = 94 .

(14)

8.

(a) 圖 4a 為一正方形，頂點分別為 A，B，C 和 D。開始時，4 隻靜止的螞蟻分別位於各頂點上。

Figure 4a shows a square with vertices A, B, C and D. In the beginning, 4 ants sit at different vertices. After a while, each ant moves to the adjacent vertex by randomly choosing and following a side. How many ways can the ants move such that they will not meet each other?

(b) 若(a)部的正方形改為正立方體 ABCDEFGH（圖 4b）及螞蟻數量由 4 隻改為 8 隻。螞蟻行走的 方法與(a)部相同。問螞蟻之間有多少種不會相遇的走法？

Suppose the square in part (a) is changed into a cube ABCDEFGH (figure 4b) and the number of ants is increased from 4 to 8. The movement of the ants follows the rules as part (a). How many ways can the ants move such that they will not meet each other?

A  B

D  C

4a / Figure 4a

A  B

F  G

E  H

4b / Figure 4b

(15)

Suggested solution:

(a) 2

The given square form a ring and we have clockwise and anticlockwise directions.

(b) 24

不會相遇走法的數量

= 環擺置方法的數量  螞蟻在第一環上走向

= 3  2  2

= 12

Number of ways

= orientation of the rings  the direction of the ants 1st ring  the direction of the ants 2nd ring

= 3  2  2

= 12 情況二 (1 個大環)

Case II (1 big ring)

不會相遇走法的數量

= 環擺置方法的數量螞蟻走向的數量

= 6  2

= 12

Number of ways

= orientation of the rings  the direction of the ants

= 6  2

= 12

不會相遇走法的總數量 / Total number of ways = 12  12 = 24

A  B

F  G

E  H

A  B

F  G

E  H

A  B

D  C

A  B

D  C

(16)

9.

(a) 求有多少種把九個相同的球分作三份的方法，使每一份至少有一個球，而每份中球的數量皆不 同。

Find the number of ways to divide 9 identical balls into 3 groups, such that each group has at least 1 ball and the numbers of balls in each group are different.

(b) 試利用下圖的等距方格來畫一個邊長分別為 1、2、3、4、5 和 6 的六邊形。

Using the isometric grid paper below, draw a hexagon with sides 1, 2, 3, 4, 5 and 6 respectively.

(c) 求符合(b)部而面積最大的六邊形的面積。

Find the area of hexagon which has the largest area and satisfies the conditions in part (b).

(17)

(a) 3 個分法/ ways

1, 2, 6; 1, 3, 5; 2, 3, 4

(b) (Accept any other reasonable answers) (接受其他合理答案)

(c) 面積/Area = 1 2 3 sq. unit.

(18)

10.

數名同學到食肆一起吃小籠包，當中一些「肚餓」的同學，他們每人會吃六隻或七隻小籠包，其他 同學則每人吃一隻或兩隻。每一籠小籠包有六隻。若他們點三籠，則不能夠滿足所有同學；若他們 點四籠，就會吃不完所有小籠包。

Several students go to a restaurant to eat steamed pork buns in bamboo steamer. Each “hungry” student eats either 6 or 7 pork buns. Everyone else eats only 1 or 2 pork buns. Each bamboo steamer has 6 steamed pork buns. Three bamboo steamers are not sufficient to serve all students while students cannot finish all the pork buns if four bamboo steamers are ordered.

How many students went to the restaurant? How many of them were "hungry"?

18 < 6x + y 及 7x + 2y < 24

，得出 x + y ≤ 4

x = 4，7x = 28 ≮ 24，捨去

x = 3，7x = 21, y = 0, 或 1，得出 7x + 2y < 24 6x = 18, 只有當 y = 1 能夠符合 19 ≤ 6x + y 若 x = 2， y 的最大值 = 2， 6x + y 的最大值= 14 ≯ 18.

若 x = 1， y 的最大值= 3， 6x + y 的最大值= 9 ≯ 18.

x = 0， y 的最大值= 4， 6x + y 的最大值= 4 ≯ 18.

∴ x = 3，y = 1

「肚餓」同學的人數 = 3 Suggested solution:

Let number of “hungry” pupils be x and number of “non-hungry” pupils be y.

The “hungry” pupils would eat 6 pork buns but would eat 7 if they were available.

18 < 6x + y and 7x + 2y < 24 i.e. 19 ≤ 6x + y and 7x + 2y ≤ 23 Taking the difference, we have x + y ≤ 4 If x = 4, 7x = 28 ≮ 24, rejected.

If x = 3, 7x = 21, y = 0, or 1, we have 7x + 2y < 24 6x = 18, only y = 1 can make 19 ≤ 6x + y

If x = 2, maximum value of y = 2, maximum value of 6x + y = 14 ≯ 18.

If x = 1, maximum value of y = 3, maximum value of 6x + y = 9 ≯ 18.

If x = 0, maximum value of y = 4, maximum value of 6x + y = 4 ≯ 18.

∴ x = 3, y = 1.

Number of pupils went to the restaurant = 4 Number of "hungry" pupils = 3

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11.

Figure 5 shows a Heron pentagon in which the sides, the diagonals and the area are natural numbers.

AB = AE = 65, AC = AD = 156 and BCDE is a rectangle. Find the length of BD.

5 / Figure 5

(20)

(x + y)2 + z2 = 1562 ~~~~~

x2 + z2 = 652 ~~~~~

   : 2xy + y2 = (156 + 65)(156  65) y(2x + y) = 221×91

=13×17×13×7

Since all x, y and z are integers, we have 由於x、 y 和 z 都是整數，因此得出:

y 2x + y x z

17 13×13×7 (13×13×7  17) > 65 rejected 捨去

13 13×17×7 (13×17×7  13) > 65 rejected 捨去

7 13×13×17 (13×13×17  7) > 65 rejected 捨去

17×13 13×7 0 > (13×7  17×13) rejected 捨去

13×7 13×17 (13×17  13×7) = 65 0 rejected 捨去

7×17 = 119 13×13 = 169 (13×13  7×17) = 25 60

13×17×7 13 ×(13  13×17×7) < 0 rejected 捨去

13×13×7 17 ×(17  13×13×7) < 0 rejected 捨去

13×13×17 7 ×(7  13×13×17) < 0 rejected 捨去

͘∴ x = 25, y =119, z = 60 BD2 = 1192 + 1202

BD = 169

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12.

An engineer is passing a tunnel on foot for structure inspection. Inside the tunnel, 4 distance marks are evenly distributed between every two safety exits. When he is walking on the way between safety exits G and H, control center informs him that a truck is coming toward him head-on. At that moment, he is at the third mark on the way from exit G to H. He realizes that he has just enough time to run toward the truck and get into the safety exit H or to run away from the truck and get into another safety exit G. If he can run 300 meters per minute, how fast is the truck going in km per hour?

*---1---1---1---1---*

Exit G Exit H 圖6 / Figure 6

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*---1---1---1---1---*

Exit G Exit H

Exit G Exit H

*---1---1---1---1---*

*---1---1---1---1---*

*---1---1---1---1---*

*---1---1---1---1---*

Speed of truck = 5×speed of engineer = 5× 300 m/min = 1500 m/min = 1.5 km/min = 90 km/hour 工程車的速度 = 5×工程師的號步速度 = 5× 300 米/分鐘 = 1500 米/分鐘= 1.5 公里/分鐘

= 90 公里/小時

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13.

7 / Figure 7

EAC = KBA = FBH = LHJ = 。

Given that EA = AB = BF and KB = BH = HL such that AB : BH = 2 : (1+√5).

EAC = KBA = FBH = LHJ = .

(a) KE 延線和 JC 延線相交於 P 點，LF 延線和 JC 延線相交於 Q 點，求 PQ 的長度。

Let P be the intersection of the extends of KE and JC, Q be the intersection of the extends of LF and JC. Find the length of PQ.

(b) 求 之值，準確至三位有效數字。

Find the value of , correct to 3 significant figures.

(24)

(a)

PA 為 x, AB 為 2h, KB 為 1 √5 h. PAE ~ PBK.

2

2 1 √5 (√5 1) x = 4h

x =

2 2

2 1 √5

1 √5 (√5 1) y = 4h

y =

∵ x = y

∴ PQ 0 (b)

∵ 5  1 = 4 (√5)2  12 = 22

(25)

(√5  1) (√5 + 1) = 2×2

∴ 及 KGE = FBH KGE ~ FBH

=0.618

(26)

Suggested solution:

(a)

Let PA be x, AB be 2h, KB be 1 √5 h. PAE ~ PBK.

2

2 1 √5 (√5 1) x = 4h

x =

Let QA be y, BF be 2h, HL be 1 √5 h. PAE ~ PBK.

2 2

2 1 √5

1 √5 (√5 1) y = 4h

y =

∵ x = y

∴ PQ 0 (b)

Let EG // CJ

∵ 5  1 = 4 (√5)2  12 = 22 (√5  1) (√5 + 1) = 2×2

(27)

∴ and also KGE = FBH KGE ~ FBH

We have =

=0.618

(28)

14.

A、B、C 和 D 為其中四個頂點。請根據以下的指示畫出八邊形:

The length of each side of an octagon is integer. All the vertices are located on the intersections of the grid.

A, B, C and D are four of the vertices. Draw the octagons with the following instructions:

(a) 面積最小的八邊形 The octagon with smallest area

(b) 周界最長的八邊形 The octagon with longest perimeter

(29)

(a) 面積最小的八邊形 The octagon with smallest area Area = 25 sq. units

(b) 周界最長的八邊形 The octagon with longest perimeter Perimeter = 60

(30)

15.

(a) 圖 8a 有一單面顏色正方形紙。它可摺成另一正方形 ABCD，使得它有四個全等長方形和一正 方形在中央（圖8b）。

Figure 8a shows a square paper with one side colored. It is folded to form another square ABCD which contains four congruent rectangles and a small square at the center as shown in figure 8b.

Without using ruler, tearing/cutting of paper or drawing any line, use the square paper provided (Appendix A) to fold the figure with the same conditions as figure 8b, but the area of the square at the center is of the area of original paper.

8a/ Figure 8a 8b/ Figure 8b



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(b) 圖 8c 有一單面顏色正方形紙。它可摺成另一正方形 EFHG，使得它有四個全等直角三角形和

Figure 8c shows another square paper with one side colored. It is folded to form another square EFHG which contains four congruent right-angled triangles and a small square at the center as shown in figure 8d.

Without using ruler, tearing/cutting of paper or drawing any line, use the square paper provided (Appendix A) to fold the figure with the same conditions as figure 8d, but the area of the square at the center is of the area of original paper.

(＊把摺好的圖形紙放入文件夾內。) (＊Put all the folded papers into the folder.)

8c/ Figure 8c 8d/ Figure 8d



E G

H F

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(a)



(b)



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