1. 若n 是兩位數，而20080126 是n 的倍數，求n 的最小可能值。
2. 圖中（http://yottkpcom.ckfreehost.com/images/maths.GIF），ABCD 是正方形，ABE 和CDF 是等邊三角形。AE 與DF 交於G。若 ∠EGF =
x°，求x。
3. 求 9998×9998 的最後四位數字。
4. 在首2008 個正整數中，哪一個的數字之和最大？
5. ABC 是直角三角形， 它的每隻內角以「度」表示時都是整數。若
∠A−∠B=x° ，問x 有多少個可能值？
6. 求 0− 2+ 4−6+8−10+"− 2006+ 2008 的值。
7. 某次聚會有50 名男孩和n 名女孩參加，當中某些男孩和女孩曾經互相握手。
若每名男孩和最少5 名女孩握手，而每名女孩和最多3 名男孩握手，求n 的
最小可能值。
8. 某長方形的長增加n%、闊減少n% 後，面積減少了15%。求最接近n 的整
數。
數學難題，唔該要有詳細解釋（20分）
2009-01-02 4:45 am
回答 (2)
2009-01-02 5:41 am
✔ 最佳答案
1 20080126=2*10040063=2*11*912733所以n為11
2
因為∠AEB =DEC=60﹐又EGFH是平行四邊形
∠EGF =120
3
99989998
=(10000-2)(10000-2)
=100000000-40000+4
=99960004
答案0004
4
好明顯是1999
5
因為A和B限制在1-90﹐所以可能數目為-89至89共179個
6
0+(− 2+ 4)+(−6+8)+(−10+"− 2006+ 2008
=(0+2+2+...+2)共2008/4=502個2
=1004
7
未想到
8
(1+n%)(1-n%)-1=-0.15
n^2/100^2=0.15
=>n=39
2009-01-01 21:49:47 補充：
不妨認為每個男人與女人握手5次﹐則總共握手250次﹐因每個女人頂多握手3次﹐所以至少女人數為[250/3]+1=84
2009-01-01 22:52:30 補充：
下面那位雖然用英文﹐但答案是錯的!!!
2009-01-02 6:12 am
Those are questions of F.2 Level of Pui Ching Invitational Competition 2008.
1. 20080126=2*11*912733
(912733 won't have any prime factor smaller than 13)
So n=22.
2.
∠DAB=90deg (square)
∠GAB=60deg (equilateral triangle)
So ∠DAG=30deg
Similarly <ADG=30deg
So ∠AGD=120deg (< sum of triangle)
3. 9998*9998=(10000-2)*(10000-2)
=10000^2-(2)*(10000)*(2)+4
So the last four digit is 0004.
4.
If ∠A=90°,
∠B can be 89°, 88°, ..., 1°
Then x will be 1°, 2°, ..., 89°
(89 possibilities)
If ∠B=90°,
∠A can be 89°, 88°, ..., 1°
Then x will be -1°,- 2°, ..., -89°
(89 possibilities)
If ∠C=90°,
∠A can be 89°, 88°, ..., 1°
Then ∠B will be 1°, 2°, ..., 89°
x will be 88°, 86°, ..., -88°
(89 possibilities)
So there are 89*3=267 possiblities for ∠A−∠B.
6. 0 − 2+ 4−6+8−10+"− 2006+ 2008
=(0-2)+(4-6)+...+(2004-2006)+2008
=-2-2...-2+2008 (There are (2004/4+1) "-2"s.)
=-1004+2008
=1004
7.
"每名男孩和最少5 名女孩握手，每名女孩和最多3 名男孩握手"
So we can group 3 boys and 5 girls in a group. In a group, the above sentences can be satisfied.
There are 16 groups (80 girls) and 2 boys without any group.
We need at least 5 more girls to satisfiy the sentences for the 2 boys left.
So the minimum value of n is 85.
8. Let a, b be the rectangle's original length and width respectively.
( a*(1+n%) ) * ( b*(1-n%) ) = ab * (1-15%)
ab * ( 1 - ( n/100 )^2 ) ) = ab * (85/100)
(a, b are not equalto zero.)
(10000 - n^2) / (10000) = 85/100
10000 - n^2 = 8500
n^2 = 1500
n = 10 * √ (15)
n = 10 * 4 (correct to the nearest integer)
n = 40
(You can log onto this website for any enquiry: http://www.puichingcentre.edu.hk/pcimc/7th/resources/07-H2.pdf)
2009-01-01 22:23:12 補充：
Sorry. I make some important mistakesobviously.
For Q.1: it should be 11.
For Q.4: It should be Q.5. (The first answerer is correct. It is 1999 obviously.)
2009-01-01 22:23:22 補充：
For Q.5: The answer in case 3 (∠C=90°) is all mentioned in case 1 and case 2. So the final answer should be 179.
For Q.7: Please refer to the first answerer.
2009-01-01 22:23:30 補充：
For Q.8: √ (15) should not be corrected to the nearest integer, but should be corrected to 1 decimal places instead. √ (15)=3.9 (correct to 1 d.p.) So n=39
1. 20080126=2*11*912733
(912733 won't have any prime factor smaller than 13)
So n=22.
2.
∠DAB=90deg (square)
∠GAB=60deg (equilateral triangle)
So ∠DAG=30deg
Similarly <ADG=30deg
So ∠AGD=120deg (< sum of triangle)
3. 9998*9998=(10000-2)*(10000-2)
=10000^2-(2)*(10000)*(2)+4
So the last four digit is 0004.
4.
If ∠A=90°,
∠B can be 89°, 88°, ..., 1°
Then x will be 1°, 2°, ..., 89°
(89 possibilities)
If ∠B=90°,
∠A can be 89°, 88°, ..., 1°
Then x will be -1°,- 2°, ..., -89°
(89 possibilities)
If ∠C=90°,
∠A can be 89°, 88°, ..., 1°
Then ∠B will be 1°, 2°, ..., 89°
x will be 88°, 86°, ..., -88°
(89 possibilities)
So there are 89*3=267 possiblities for ∠A−∠B.
6. 0 − 2+ 4−6+8−10+"− 2006+ 2008
=(0-2)+(4-6)+...+(2004-2006)+2008
=-2-2...-2+2008 (There are (2004/4+1) "-2"s.)
=-1004+2008
=1004
7.
"每名男孩和最少5 名女孩握手，每名女孩和最多3 名男孩握手"
So we can group 3 boys and 5 girls in a group. In a group, the above sentences can be satisfied.
There are 16 groups (80 girls) and 2 boys without any group.
We need at least 5 more girls to satisfiy the sentences for the 2 boys left.
So the minimum value of n is 85.
8. Let a, b be the rectangle's original length and width respectively.
( a*(1+n%) ) * ( b*(1-n%) ) = ab * (1-15%)
ab * ( 1 - ( n/100 )^2 ) ) = ab * (85/100)
(a, b are not equalto zero.)
(10000 - n^2) / (10000) = 85/100
10000 - n^2 = 8500
n^2 = 1500
n = 10 * √ (15)
n = 10 * 4 (correct to the nearest integer)
n = 40
(You can log onto this website for any enquiry: http://www.puichingcentre.edu.hk/pcimc/7th/resources/07-H2.pdf)
2009-01-01 22:23:12 補充：
Sorry. I make some important mistakesobviously.
For Q.1: it should be 11.
For Q.4: It should be Q.5. (The first answerer is correct. It is 1999 obviously.)
2009-01-01 22:23:22 補充：
For Q.5: The answer in case 3 (∠C=90°) is all mentioned in case 1 and case 2. So the final answer should be 179.
For Q.7: Please refer to the first answerer.
2009-01-01 22:23:30 補充：
For Q.8: √ (15) should not be corrected to the nearest integer, but should be corrected to 1 decimal places instead. √ (15)=3.9 (correct to 1 d.p.) So n=39
參考: Myself
收錄日期: 2021-04-15 17:46:16
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090101000051KK01694