2019 CAT Senior澳大利亚数学信息数学竞赛真题

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2019年RMM数学竞赛试题及答案

Now we show that (i) on any move of hers, Amy does not increase the square-free part of the positive integer on the board; and (ii) on any move of his, Bob always can replace a positive integer n with a non-negative integer k with S(k) < S(n). Thus, if the game starts by a positive integer N , Bob can win in at most S(N ) moves.
Part (i) is trivial, as the deﬁnition of the square-part yields S(nk) = S(n) whenever k is odd, and S(nk) = 1 ≤ S(n) whenever k is even, for any positive integer n.
Part (ii) is also easy: if, before Bob’s move, the board contains a number n = S(n) · b2, then Bob may replace it with n = n − b2 = (S(n) − 1)b2, whence S(n ) ≤ S(n) − 1.
The 11th Romanian Master of Mathematics Competition
Day 1 — Solutions
Problem 1. Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves ﬁrst. On any move of his, Bob replaces the number n on the blackboard with a number of the form n − a2, where a is a positive integer. On any move of hers, Amy replaces the number n on the blackboard with a number of the form nk, where k is a positive integer. Bob wins if the number on the board becomes zero. Can Amy prevent Bob’s win?

2019年数学竞赛试题及答案

The 11th manian Master of Mathematics Competition
Day 1 — Solutions
Problem 1. Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves ﬁrst. On any move of his, Bob replaces the number n on the blackboard with a number of the form n − a2 , where a is a positive integer. On any move of hers, Amy replaces the number n on the blackboard with a number of the form nk , where k is a positive integer. Bob wins if the number on the board becomes zero. Can Amy prevent Bob’s win? Russia, Maxim Didin Solution. The answer is in the negative. For a positive integer n, we deﬁne its square-free part S (n) to be the smallest positive integer a such that n/a is a square of an integer. In other words, S (n) is the product of all primes having odd exponents in the prime expansion of n. We also agree that S (0) = 0. Now we show that (i) on any move of hers, Amy does not increase the square-free part of the positive integer on the board; and (ii) on any move of his, Bob always can replace a positive integer n with a non-negative integer k with S (k ) < S (n). Thus, if the game starts by a positive integer N , Bob can win in at most S (N ) moves. Part (i) is trivial, as the deﬁnition of the square-part yields S (nk ) = S (n) whenever k is odd, and S (nk ) = 1 ≤ S (n) whenever k is even, for any positive integer n. Part (ii) is also easy: if, before Bob’s move, the board contains a number n = S (n) · b2 , then Bob may replace it with n = n − b2 = (S (n) − 1)b2 , whence S (n ) ≤ S (n) − 1. Remarks. (1) To make the argument more transparent, Bob may restrict himself to subtract only those numbers which are divisible by the maximal square dividing the current number. This restriction having been put, one may replace any number n appearing on the board by S (n), omitting the square factors. After this change, Amy’s moves do not increase the number, while Bob’s moves decrease it. Thus, Bob wins. (2) In fact, Bob may win even in at most 4 moves of his. For that purpose, use Lagrange’s four squares theorem in order to expand S (n) as the sum of at most four squares of positive integers: 2 2 2 2 S (n) = a2 1 + · · · + as . Then, on every move of his, Bob can replace the number (a1 + · · · + ak )b 2 2 on the board by (a2 1 + · · · + ak−1 )b . The only chance for Amy to interrupt this process is to replace a current number by its even power; but in this case Bob wins immediately. On the other hand, four is indeed the minimum number of moves in which Bob can guarantee himself to win. To show that, let Amy choose the number 7, and take just the ﬁrst power on each of her subsequent moves.

0606_m19_qp_12(IGCSE 数学考试2019年3月真题

Identities Formulae for ∆ABC
2. TRIGONOMETRY
sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A
a sin A
=
b sin B
=
c sin C
a2 = b2 + c2 – 2bc cos A
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80.
∆
=
1 2
bc
sin
A
© UCLES 2019
0606/12/F/M/19
3
1 (a) Given that
பைடு நூலகம்
= { x : 1 1 x 1 20},
A = {multiples of 3},
B = {multiples of 4},
find
(i)
n (A),
[1]
(ii)
n (A + B) .
[1]
*6715046212*
Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education

2019年3月12亚太地区数学奥林匹克试题

2019 年亚太地区数学奥林匹克 2019 年 3 月 12 日考试时间:4 小时 5 个问题每题 7 分
1 f 是定义在正整数上且取值为正整数的函数，试求所有的 f, 使得：对任意的正整数 a 和 b， a2 + f (a)f (b) 能被 f (a) + b 整除。
2 m 是正整数，数列 an 定义如下。其中，a1 是正整数：
< 2m ; 当an ⩾ 2m ;
an+1 =
a n 2
试求所有的 a1 ，使得数列 an 的所有项都是整数。
3 三条边各不相等的三角形 ABC 内接于圆 L 中。BC 的中点为点 M ，P 点（动点）在线段 AM 上，圆 L 与三角形 BP M 的外接圆交于点 D(异于点 B ), 直线 DP 与三角形 CP M 的外接圆交于点 X (异于点 P )。而且，圆 L 与三角形 CP M 的外接圆交于点 E (异于点 C )，直线 EP 与三角形 BP M 的外接圆交于点 Y （异于点 P ）。试证：不依赖于 P 的选取，三角形 AXY 的外接圆过异于点 A 的某定点 T 。
4 对于 2018 × 2019 的方格，每个格子里写着一个整数，对于每个格子，与它有公共边的别的格子称之邻居，考虑下面的操作：任意选择几个格子，计算出格子的邻居的数的平均值，然后将这些值替换进格子。能否做到：不管最初的情况如何，总能在有限次操作后，使得所有格子里的数都相等？
5 f 是定义在实数上且取值为实数的函数，求所有的 f ，使得：对任意实数 x 和 y ，满足 f (x2 + f (y )) = f (f (x)) + f (y 2 ) + 2f (xy ).
1

澳大利亚数学竞赛题5-6年级10年题

小学高年级（5—6）（2007年）1.下列哪一个数是由1个一百、4个十、3个一所凑成的？（A）413（B）143（C）341（D）1043（E）1342.在下列等式中，□内的数字是什么？2□6＋497＝703（A）0（B）1（C）3（D）7（E）93.上星期二，某个班级上数学、音乐、英语及美术课的上课时数比例，如下图的饼图所示：下列哪一个叙述为真？（A）上音乐课的时间比上美术课的时间长。

（B）上音乐课的时间比上英语课的时间长。

（C）上音乐课和英语课的时间超过全部上课时间的一半。

（D）上数学课和美术课的时间超过全部上课时间的一半。

（E）上数学课的时间与上美术课的时间一样长。

4.算式14－8÷2＋2×3之值等于（A）16（B）15（C）36（D）9（E）45.下列哪一个数最大？（A）百分之三十（B）百分之十三（C）十分之三（D）百分之三十一（E）百分之三6.有一个角锥共有12条棱边。

请问它的底面是什么形状？（A）三角形（B）四边形（C）五边形（D）六边形（E）八边形7.小安的布袋内装有编号为1～20的二十颗球。

她随意从袋子中抓出一个球。

以下哪一种事件的发生有较大的可能性？（A）她抓出的球的编号是1号。

（B）她抓出的球的编号是奇数。

（C）她抓出的球的编号中有数字2。

（D）她抓出的球的编号是9或10号。

（E）她抓出的球的编号是20号。

8.凯伦驾车由市区开往湖边，她经过如下右图的路标：大约过了一个小时，她看到另一个路标显示再过5km即可抵达湖边。

请问从市区算起她已经行驶了多少路程？（A）50km（B）80km（C）125km（D）30km（E）65km9.某一班学生中有60％是女生。

这一班不可能是下列哪一种组合？（A）6位男孩9位女孩。

（B）10位男孩15位女孩（C）15位男孩10位女孩。

（D）12位男孩18位女孩。

（E）12位女孩8位男孩。

10.现在时刻是3：00pm。

在一个24小时计时的时钟上，100小时后将显示的时刻是什么？（A）7：00am （B）3：00am （C）7：00pm （D）3：00pm （E）11：00pm11.下列哪个图的阴影部分占全部面积的八分之三？12.小珍有一片巧克力，它划分成许多小正方形块，它的长有6个小正方形块，宽有8个小正方形块（如下图）。

2019年夏季数学大联盟4-5年级个人赛(Iindividual Part1)真题与解析【决赛】

Individual Questions – Part 1 of 2Each question is worth 10 points. Calculators are PROHIBITED.#1.(Time Limit: 7 minutes ) If six circles are tangent as shown, and each has an area of π, what is the area of the rectangle that just surrounds them, as shown?题目翻译: 如果6个面积为π的圆排列成图中的形状，那么图中的长方形的面积是多少？Solution:Since the area of each circle is π, each circle has a radius of 1 and a diameter of 2. The length across the rectangle consists of 3 diameters, and the width across the rectangle consists of 2 diameters. The dimensions of the rectangle are 6 by 4 and its area is 24.#2.(Time Limit: 7 minutes ) What is the least possible integer greater than 2019 that has remainder 11 when divided by 20 and 19?题目翻译: 请问大于2019的整数中满足除以20 和 19 都余11 的最小的数是多少？Solution:The least common multiple of 20 and 19 is 380, so we need an integer that is 11 more than amultiple of 380 and that is greater than 2019. Since 5 < 2019¸380 < 6, the least acceptable multiple is = 2280 and the number 11 more than this is 2291.#3. (Time Limit: 7 minutes ) I wrote each of the first 2019 positive integers in increasing order. Eachodd integer was written three consecutive times and each even integer was written two consecutive times. The first five integers I wrote were 1, 1, 1, 2, 2. What was the 2019th integer that I wrote?题目翻译：我把前2019个正整数按照从小到大的顺序写了出来。

2019年夏季数学大联盟4-5年级个人赛(Iindividual Part2)真题与解析【决赛】

Individual Questions – Part 2 of 2Each question is worth 10 points. Calculators are PROHIBITED.#2-1. (Time Limit: 7 minutes) Different letters in the word “MathLeague” represent different digits.What is the greatest possible value of the sum of the value of all 10 letters?题目翻译: MathLeague这个单词中的每一个不同的字母对应一个不同的数字(0...9这十个数字中的一个数字)，请问这十个字母所对应的数字的和的最大值是多少?Solution:Since the letters “a” and “e” appear twice, let one of them have a value of 9 and the other onea value of 8. Each of the six other letters appears only once, so their values should be 7, 6, 5,4, 3, and 2. Assigning these values to the letters, the greatest possible sum is (18 + 16 + 7 + 6 +5 + 4 + 3 + 2) =61.#2-2. (Time Limit: 7 minutes) The number N = 2666666 . . . 666666 consists of a “2” followed by 2019 “6”s. What is the remainder when N is divided by 11?题目翻译: N = 266666…666666666，数字2后面有2019个6，N除以11的余数是多少？Solution 1:66 is a multiple of 11, so is any even number of 6 in a row. Let N1 = N–66…66 (2018 “6”s) =2600…00 (2018 “0”s). The remainder when N is divided by 11 is equal to the remainderwhen N1 is divided by 11.Let N2 = N1 – 2200...00 (2018 “0”s) = 400…00 (2018 “0”s). The remainder when N1 isdivided by 11 is equal to the remainder when N2 is divided by 11.Let N3 = N2 – 4 * 99…99 (2018 “9”s) = 4. Since 99…99 (2018 “9”s) is a multiple of 11, theremainder when N2 is divided by 11 is equal to the remainder when N3 is divided by 11. So the answer is4.Solution 2:There are 2020 digits in N. As we divide, we see that the quotient is 242424. . ., with the partial remainders after each step of the division alternating between 4 and 2. Since there are 2019 6s, the last remainder will be a4.#2-3. (Time Limit: 7 minutes) If 2 watermelons can be exchanged for 7 apples and 3 apples can be exchanged for 4 bananas, for how many bananas can 12 watermelons be exchanged?题目翻译: 如果2个西瓜可以换7个苹果，3个苹果可以换4个香蕉，那么12个西瓜可以换多少个香蕉？Solution:If 2 watermelons can be exchanged for 7 apples, 12 watermelons can be exchanged for 42 apples. Similarly, if 3 apples can be exchanged for 4 bananas, then 42 apples can be exchanged for56bananas.#2-4. (Time Limit: 7 minutes) Six years ago, Steve was six times as old as Rui. Six years from now, Steve will be 10 years older than Rui. How old is Rui?题目翻译: 6年前，Steve的年纪是Rui的年纪的6倍。

2019年高中数学竞赛试题及答案及答案

高中数学竞赛试题及答案一、选择题（本大题共6小题，每小题6分，共36分．每小题各有四个选择支，仅有一个选择支正确．请把正确选择支号填在答题卡的相应位置．）1．集合{0,4,}A a =，4{1,}B a =，若{0,1,2,4,16}A B ⋃=，则a 的值为A ．0B ．1C ．2D ．2．一个简单几何体的正视图、侧视图如图所示，则其俯视图不可能．．．是．①长方形；②正方形；③圆；④菱形. 其中正确的是 A ．①② B ．②③ C ．③④ D ．①④ 3．设0.50.320.5,log 0.4,cos3a b c π-===，则A ．c b a <<B ．c a b <<C ．a b c <<D ．b c a <<4. 平面上三条直线210,10,0x y x x ky -+=-=-=，如果这三条直线将平面划分为六部分，则实数k 的值为A . 1B . 2C . 0或2D . 0，1或2 5．函数()sin()f x A x ωϕ=+（其中0,||2A πϕ><）的图象如图所示，为了得到()cos 2g x x =的图像，则只要将()f x 的图像A ．向右平移6π个单位长度 B ．向右平移12π个单位长度 C ．向左平移6π个单位长度 D ．向左平移12π个单位长度6. 在棱长为1的正四面体1234A A A A 中，记12(,1,2,3,4,)i j i j a A A A A i j i j =⋅=≠，则i j a 不同取值的个数为A ．6B ．5C ．3D ．2二、填空题（本大题共6小题，每小题6分，共36分．请把答案填在答题卡相应题的横线上．） 7．已知)1,(-=m a ，)2,1(-=b ，若)()(b a b a -⊥+，则m = .8．如图，执行右图的程序框图，输出的T= . 9. 已知奇函数()f x 在(,0)-∞上单调递减，且(2)0f =，则不等式0)()1(<⋅-x f x 的解集为．10.求值：=+250sin 3170cos 1 ． 11．对任意实数y x ,，函数)(x f 都满足等式)(2)()(22y f x f y x f +=+，且0)1(≠f ，则（第5题图）（第8题图）3侧视图正视图2222=)2011(f .12.在坐标平面内，对任意非零实数m ，不在抛物线()()22132y mx m x m =++-+上但在直线1y x =-+ 上的点的坐标为 .答题卡一、选择题（本大题共6小题，每小题6分，共36分．）二、填空题（本大题共6小题，每小题6分，共36分．）7． 8． 9． 10． 11． 12．三、解答题（本大题共6小题，共78分.解答应写出必要的文字说明、证明过程或演算步骤.） 13.（本小题满分12分）为预防(若疫苗有效已知在全体样本中随机抽取1个，抽到B 组的概率是0.375. （1）求x 的值；（2）现用分层抽样的方法在全部测试结果中抽取360个，问应在C 组中抽取多少个？（3）已知465≥y ,25≥z ,求该疫苗不能通过测试的概率.已知函数x x x f 2sin )12(cos 2)(2++=π．（1）求)(x f 的最小正周期及单调增区间；（2）若),0(,1)(παα∈=f ，求α的值． 15．（本题满分13分）如图，在直三棱柱111C B A ABC -中，21===AA BC AC ，︒=∠90ACB ，G F E ,,分别是AB AA AC ,,1的中点．（1）求证：//11C B 平面EFG ；（2）求证：1AC FG ⊥；（3）求三棱锥EFG B -1的体积.ACBB 1A 1C 1FGE已知函数t t x x x f 32)(22+--=.当∈x ),[∞+t 时，记)(x f 的最小值为)(t q . (1)求)(t q 的表达式；(2)是否存在0<t ，使得)1()(tq t q =？若存在，求出t ；若不存在，请说明理由.已知圆22:228810M x y x y +---=和直线:90l x y +-=，点C 在圆M 上，过直线l 上一点A 作MAC ∆.（1）当点A 的横坐标为4且45=∠MAC 时，求直线AC 的方程；（2）求存在点C 使得45=∠MAC 成立的点A 的横坐标的取值范围.18．（本题满分14分）在区间D 上，若函数)(x g y =为增函数，而函数)(1x g xy =为减函数，则称函数)(x g y =为区间D 上的“弱增”函数．已知函数()1f x =-．（1）判断函数()f x 在区间(0,1]上是否为“弱增”函数，并说明理由；（2）设[)1212,0,,x x x x ∈+∞≠，证明21211()()2f x f x x x -<-；（3）当[]0,1x ∈时，不等式xax +≥-111恒成立，求实数a 的取值范围．参考答案一、选择题：C B A D D C二、填空题：7. 2± 8．29 9. ),2()1,0()2,(+∞--∞10.3 11．2201112. 31(,),(1,0),(3,4)22-- 三、解答题：13. （本题满分12分）解：（1）因为在全体样本中随机抽取1个，抽到B 组的概率0.375，所以375.0200090=+x ， ………………2分即660x =. ………………3分（2）C 组样本个数为y ＋z ＝2000－（673＋77＋660＋90）＝500， ………………4分现用分层抽样的方法在全部测试结果中抽取360个，则应在C 组中抽取个数为360500902000⨯=个. ………………7分（3）设事件“疫苗不能通过测试”为事件M.由（2）知 500y z +=，且,y z N ∈,所以C 组的测试结果中疫苗有效与无效的可能的情况有：（465，35）、（466，34）、（467，33）、……（475，25）共11个. ……………… 9分由于疫苗有效的概率小于90%时认为测试没有通过，所以疫苗不能通过测试时，必须有9.02000660673<++y， …………………10分即1800660673<++y ，解得467<y ，所以事件M 包含的基本事件有：（465，35）、（466，34）共2个. …………………11分所以112)(=M P ，故该疫苗不能通过测试的概率为211. …………………12分14. （本小题满分12分）解：x x x f 2sin )62cos(1)(+++=π…………………1分x x x 2sin 6sin2sin 6cos 2cos 1+-+=ππx x 2sin 212cos 231++= ………………… 2分 1)32sin(++=πx . …………………4分（1）)(x f 的最小正周期为ππ==22T ； …………………5分又由]22,22[32πππππ+-∈+k k x ， …………………6分得)](12,125[Z k k k x ∈+-∈ππππ， …………………7分从而)(x f 的单调增区间为)](12,125[Z k k k ∈+-ππππ． …………………8分（2）由11)32sin()(=++=πααf 得0)32sin(=+πα， …………………9分所以ππαk =+32，62ππα-=k )(Z k ∈． …………………10分又因为),0(πα∈，所以3πα=或65π． …………………12分15. （本题满分13分）解：（1）因为E G 、分别是AC AB 、的中点，所以BC GE //；……1分又BC C B //11，所以GE C B //11； …………2分又⊆GE 平面EFG ，⊄11C B 平面EFG ，所以//11C B 平面EFG ． …………3分（2）直三棱柱111C B A ABC -中，因为︒=∠90ACB ，所以⊥BC 平面C C AA 11； ……………4分又BC GE //，所以⊥GE 平面C C AA 11，即1AC GE ⊥； ……………5分又因为21==AA AC ，所以四边形11A ACC 是正方形，即11AC C A ⊥； ……………6分又F E ,分别是1,AA AC 的中点，所以C A EF 1//，从而有1AC EF ⊥， ……………7分由E GE EF =⋂，所以⊥1AC 平面EFG ，即1AC FG ⊥． ……………8分（3）因为//11C B 平面EFG ，所以111EFC G EFG C EFG B V V V ---==． ……………10分由于⊥GE 平面C C AA 11，所以GE S V EFC EFC G ⋅=∆-1131，且121==BC GE ．…………11分又由于2321114111111=---=---=∆∆∆∆ECC FC A AEF A ACC EFC S S S S S 正方形，……………12分所以21123313111=⋅⋅=⋅=∆-GE S V EFC EFC G ，即211=-EFG B V ． ……………13分16. （本题满分13分）解：（1）t t x x x f 32)(22+--=13)1(22-+--=t t x ． ……………1分①当1≥t 时，)(x f 在∈x ),[∞+t 时为增函数，所以)(x f 在∈x ),[∞+t 时的最小值为t t f t q ==)()(；……………3分②当1<t 时，13)1()(2-+-==t t f t q ； ……………5分综上所述，2(1)()31(1)t t q t t t t ≥⎧=⎨-+-<⎩． ……………6分ACBB 1A 1C 1FGE（2）由（1）知，当0<t 时，13)(2-+-=t t t q ，所以当0<t 时，131)1(2-+-=tt tq ． ……………7分由)1()(t q t q =得：1311322-+-=-+-tt t t ， ……………8分即013334=-+-t t t ， ……………9分整理得0)13)(1(22=+--t t t ， ……………11分解得：1±=t 或253±=t . ……………12分又因为0<t ，所以1-=t .即存在1-=t ，使得)1()(tq t q =成立. ……………13分17. （本题满分14分）解：（1）圆M 的方程可化为：2217(2)(2)2x y -+-=，所以圆心M （2，2），半径r=2. ……1分由于点A 的横坐标为4，所以点A 的坐标为（4，5），即AM =……………2分若直线AC 的斜率不存在，很显然直线AM 与AC 夹角不是45，不合题意，故直线AC 的斜率一定存在，可设AC 直线的斜率为k ，则AC 的直线方程为5(4)y k x -=-，即540kx y k -+-=. ……………3分由于45=∠MAC 所以M 到直线AC 的距离为226||22==AM d ，此时r d <，即这样的点C 存在. ……………4分2=，2=，解得15 5k k =-=或. ……………5分所以所求直线AC 的方程为0255=-+y x 或0215=+-y x . ……………6分 (2)当r AM 2||=时，过点A 的圆M 的两条切线成直角，从而存在圆上的点C （切点）使得45=∠MAC . ……………7分设点A 的坐标为),(y x ，则有⎪⎩⎪⎨⎧=-+=⋅=-+-09172342)2()2(22y x y x ， ……………8分解得⎩⎨⎧==63y x 或⎩⎨⎧==36y x . ……………9分记点)6,3(为P ，点)3,6(为Q ，显然当点A 在线段PQ 上时，过A 的圆的两条切线成钝角，从而必存在圆上的一点C 使得45=∠MAC ；……当点A 在线段PQ 的延长线或反向延长线上时，过A 的圆的两条切线成锐角，从而必不存在圆上的点C 使得45=∠MAC ， …………所以满足条件的点A 为线段PQ 上的点，即满足条件的点的横坐标取值范围是.……14分18．（本题满分14分）解：（1）由()1f x =-可以看出，在区间(0,1]上，()f x 为增函数. ………………1分又11()(1f x x x ===3分显然)(1x f x在区间(0,1]∴ ()f x 在区间(0,1]为“弱增”函数. ………………4分（2）21()()f x f x -===.…6分[)1212,0,,x x x x ∈+∞≠,∴111≥+x ，112≥+x ，21121>+++x x ，即2>，………………8分21()()f x f x ∴-2112x x <-. ………………9分（3）当0x =时,不等式xax +≥-111显然成立. ………………10分“当(]0,1x ∈时，不等式xax +≥-111恒成立”等价于“ 当(]0,1x ∈时，不等式)111(1xx a +-≤即)(1x f x a ≤恒成立” . ………………11分也就等价于:“ 当(]0,1x ∈时， min )](1[x f xa ≤成立” . ………………12分由(1)知1()f x x 在区间(0,1]上为减函数, 所以有221)1()](1[min -==f x f x . ……………13分 ∴221-≤a ，即221-≤a 时，不等式xax +≥-111对[]0,1x ∈恒成立. ……………14分。

2019年袋鼠数学竞赛L3

Kangaroo Aus 2019(Grade 5-6)一、三分题1、小明画一只猫，他接下来加上眼睛。

请问下图中拿一幅图展示了他最后完成这幅图的情景？2、玛丽用点和线来代表数字。

点代表1，线代表5，请问下面哪副图代表17？3、一个救援队里有14个女生和12个男生。

队里有一半的人出去散步了。

请问最少有多少女生去散步了？4、一个数字钟显示时间如图：，请问下面哪副图使用了上面4个数字，并且是在上面时间之后的第一个时间？5、下面哪个骰子的对面数字之和为7？6、哪一个几何在图形中找不到？7、有一群袋鼠，他们的年龄和是36岁。

再过2年，他们的年龄和是60岁。

请问一共有多少只袋鼠?8、劳拉想在图中给一个2×2的正方形涂色。

请问有多少种方法？二、四分题9、每一张卡片上都有一位三位数。

这三个三位数的和是826。

那么隐藏的两个数字之和是多少？10、大卫把6个最小的奇数写在骰子上。

然后，他抛了3次，这3次的和不可能为？11、吉姆有一个可以折叠的由10根相同长度的小棒连成的棒子。

以下哪个图形不能用这个棒子做成？12、哪一幅图的黑色面积最大？13、一个小院子里有30只小动物（狗、猫、老鼠）。

牧师把6只狗变为6只猫，然后再把5只猫变为5只老鼠。

这个时候，狗、猫、老鼠的只数一样多。

请问最开始有多少只猫？14、大卫用1cm×1cm×2cm的砖头搭建城堡。

如图；他搭的最后一个塔共用了28块砖。

请问最后他搭的那个塔是多高？15、吉姆按照如图方式把纸折碟，然后再按照图示的方法，沿着两条线把这个纸剪开。

请问，他最后得到了多少张纸块？16、每一张图上都画了一条线。

现在我们把他们叠成一个正方体。

请问哪一幅图在折叠成正方体后，它上面的线是一个封闭的图形？三、五分题17、骰子的每一面都写了一个数字。

如图。

已知每个对面的数字积都相等。

请问6个面的数字都加起来和最小为多少？18、如图，三个黑色的小球加上一个白色的小球放在天平左端，右端放一块30克的冰和一个黑色的球。

澳大利亚数学竞赛cat往年真题18-sen-EN

Part A:Questions1–6Each question should be answered by a single choice from A to E.Questions are worth3points each.1.DNA sequenceMolly is processing long DNA sequences such as the following:A A A T C C C C C A A G A A A A AShe encodes this sequence as A3T1C5A2G1A5,a saving of five characters. How many characters are saved if she encodes the following sequence?T T T T T G G G A C C C C C C G A A(A)6(B)7(C)8(D)9(E)102.ArtworksJermaine creates most wonderful artworks!She paints in a very particular style. She starts with a9by9grid of squares,all white except for the one in the middle; it is black.Then,she picks any white square that shares a side with one and only one other black square and colours it black,repeating this until she is happy.You attend an art show hoping to buy one of Jermaine’s works.But there are lots of imitations and fakes on display!Which one is Jermaine’s?A B C D E(A)A(B)B(C)C(D)D(E)E3.BookshelfTerri is moving several numbered books from her bookshelf to her desk.She takes a book from either end of her bookshelf,but always puts it at the left on the desk.For instance,if she had books 213on her bookshelf,and took the leftmost book(2)from the bookshelf,then the rightmost book (3),then the remaining book (1),they would be in order 132on her desk.BookshelfDeskMove book 2Move book 3Move book 1Terri has 7books on her bookshelf,in the order 2364715.She notices that the books form a number.She decides to make that number as large as possible,still following her ‘take from either end,place on left end’rule.What is the number on the fifth book from the left after she has moved all 7books?(A)1(B)2(C)3(D)4(E)54.Taking HalfAmy and Bob are playing a game.They start with a large pile of pebbles and take turns to remove one or more pebbles.On a player’s turn:•If there is only one pebble left,the player removes that pebble.•If there is more than one pebble left,the player can remove up to half of the pebbles remaining.The winner is the player who removes the last pebble.There are 24pebbles and Amy plays first.How many pebbles should she take on her first turn to ensure that she can win the game,assuming she plays optimally throughout?(A)1(B)3(C)7(D)10(E)125.Flow DiagramFlow diagrams provide a visual way of showing a process or algorithm:a box is used for an action,a diamond(shaded)for making a decision,and arrows indicate the flow of control.Each of the values23,47,119,and123456in turn is input to the flow diagram below.How many of the outputs are even?(A)0(B)1(C)2(D)3(E)46.Triangular WalkA robot can only walk horizontally to the left,up right at60°from the horizontal,or down right at60°from the horizontal.←←←The robot goes on the walk shown below.←←←←←←←←←←←←←←←←←←←←←←←How many more instructions are needed for the robot to return to its starting position?(A)4(B)5(C)6(D)7(E)8Part B:Questions7–9Each question has three parts,each of which is worth2points.Each part should be answered by a number in the range0–999.7.Maximum Plus–Minus SumA sublist of a list of numbers is a set of one or more adjacent elements from the list.For example,546and4641are both sublists of the list546412.(There are many other sublists.)The plus–minus sum of a sublist is obtained by alternately adding and subtracting the numbers in the sublist.For example,the plus–minus sum of546is+5−4+6=7,whilst that of464 1is+4−6+4−1=1.In each of the following lists of numbers,find the sublist with the largest plus–minus sum.Your answer will be this sum.A.46823472B.42687437C.86476156824758.Tweet SpotsTwo or more robots are walking from left to right along separate tracks.The tracks are made up of squares and each square of the track is either a plain or a tweet spot square.During each second:•If a robot is on a plain square,it will take a step to the right.•If a robot is on a tweet spot square,it will not move,unless all robots are on a tweet spot square.It will then tweet to the other robots(which takes no time)and then will take a step to the right.On the tracks below,the tweet spot is shaded,and each robot is represented by an R.RRStart After2secs After4secs After5secs After7secs The robots will tweet after4seconds,and both robots will be on the last square after7seconds.For each of the tracks below,how many seconds will it take for all robots to reach the last square?(7in the example above.)A.B.C.9.Evolving NumbersA number with an even number of digits can evolve into a new number by extract-ing pairs of adjacent digits and placing them at the start of a new number. Consider the number451972.It could evolve into429751as shown in the table below.Extract51514972Extract97975142Extract42429751–By choosing different pairs of adjacent digits along the way,the original number could have evolved into several different numbers.Your task is to find the largest possible evolved number.For each of the following strings of digits,what are the last three digits in the largest number that can be evolved from the string?A.37594156B.52730819C.9453678152SolutionsPart A:Questions1–61.DNA sequenceThe sequence T T T T T G G G A C C C C C C G A A would be encoded asT5G3A1C6G1A2The uncoded sequence had18characters.The encoded sequence has12characters. Molly has saved18−12=6characters.Hence(A).An alternative solution is to note that each letter in the rewritten sequence requires 2characters(if under10)and so L1means+1,L2means stays the same,L2+ means saving of excess.(Where L is T,G,A or C.)The sequence of repetitions in the encoded sequence is531612.This would result in a saving of3+1−1+ 4−1+0=6characters.Hence(A).2.ArtworksThere cannot be any2×2sets of black squares in Jessica’s art-works as the4th white square would have two black neighbours.C and E contain2×2sets of black squares,so they are not Jessica’s artworks. Again,there cannot be any loops Jessica’s artworks as the finalwhite square would have two black neighbours.A contains two loops of black squares,so it is not one of Jessica’s artworks. Finally,in D none of the squares in the bottom left or the top right share any sideswith those in the middle,so it is not one of Jessica’s artworks.Hence(B).3.BookshelfThe last book to be moved should be book7.Then the second last should be the larger of its neighbours,book4.Working outward in the same manner gives 7463215The fifth book is2.Hence(B).4.Taking HalfIf,after a turn late in the game,Amy leaves2pebbles,she will win,for Bob has to take1leaving the last one for Amy to take.Again,if Amy leaves5pebbles she can win,for Bob must take2or1leaving3or 4.Amy can then take1or2,leaving2,and she wins.Further,if Amy leaves11pebbles she can win,for Bob must leave between6and 10.Amy can then take sufficient to leave5,from which she can win.Finally,if Amy leaves23pebbles she can win,for Bob must leave between12and 22.Amy can then take sufficient to leave11,from which she can win.Bob’s moves 0123456789101112131415161718192021222324Amy’s movesTo leave23,Amy must take1.Hence(A).5.Flow DiagramThe number123456is far too large to trace through the flow diagram.So we have to determine what the flow diagram’s algorithm does.We start by tracing the two smaller numbers through the flow diagram.Input23to the flow diagram:A:231333no→no→yes→B:0125Output5.Input47to the flow diagram:A:4737271777no→no→no→no→yes→B:0123411Output11.We can now see that the effect of the flowchart’s algorithm is23→2+3=5and47→4+7=11.(In mathematical terms,if A is input,the output=A div10+A mod10.) Then119→11+9=20and123456→12345+6=12351.So when23,47,119,and123456are input the outputs are5,11,20and12351. One of these is even.Hence(B).6.Triangular WalkIn total the robot moves9times←,6times←,and8times←.N moves in each direction(in any order)returns the robot to its starting position. So we need another3+1=4moves to have9moves in each direction.Hence(A).Part B:Questions7–97.Maximum Plus–Minus SumConsider the list245343.In the sublist24534,the2and the4sum to–2,so we would be better off omitting them,leaving534.This sums to5−3+4=6, which is better than the sublist5.This leads to an algorithm where we write down the running sum,starting back from0if the running sum goes negative.Number245343Running sum2–25263The largest running sum is6,corresponding to the sublist534.But there is a further factor.In the above running sum,we only considered sublists starting with the1st,3rd,5th,…numbers.We also need to consider numbers starting at the2nd,4th,…numbers.Number245343Starting odd2–25263Starting even-4–13–13The best sublist starting from an even position is4,whose sum is also4.So the largest sublist is534with sum6.A.Number46823472Starting odd4–286951210Starting even-6–22–14–32The largest sum is12,corresponding to the sublist82347.B.Number42687437Starting odd4280736–1Starting even-2–481529The largest sum is9,corresponding to the sublist87437.C.Number8647615682475 Starting odd826–165104121014712Starting even-62934–16–22–272 The largest sum is14,corresponding to the sublist6156824.8.Tweet SpotsA simple way to solve this problem is to number the squares to show the times(in seconds)when the robots will be in the square.For the example this givesA more systematic approach is to tabulate the distance on each robot’s track to the next tweet spot or the last square.Then the total time will be the sum of the largest of these times.Tweet spot1Last TotalRobot123Robot241Largest437This approach will be used in the solutions below.A.Tweet spot12Last TotalRobot1253Robot2325Largest35513The last robot will reach the final square after13seconds.B.Tweet spot123Last TotalRobot11433Robot24241Robot33242Largest444315The last robot will reach the final square after15seconds.C.Tweet spot12345Last TotalRobot1423220Robot2142321Robot3251221Robot4324220Largest45432119The last robot will reach the final square after19seconds.9.Evolving NumbersBecause pairs are put on the left of the evolved number,the last pair extracted will be the most significant digits of the new number:we aim for them to be as large as possible.This will be the first pair identified.The algorithm is a backward algorithm,identifying the last pair to be extracted, the second last,…We will used the notation pq for the largest pair of numbers that can be extracted from a list.Then p is the largest digit in an odd position,and q is the largest digit in an even position to the right of p.Further,once a pair pq has been identified,a subsequent pair cannot embrace p or q,as pq will be extracted later.We extract or strike out the pq pairs as we identify them.This splits the remaining digits into up to three sublists,from which we can see their pq pairs more readily. If solving the problem by hand,we would probably just strike out the pq pairs, but making the lists explicit would be preferred in a computer program.In the solutions below,the largest pq pair is in bold and the lists are between[ and].A.p q By hand Lists Number37594156[37594156]-59→37A5A94156[37][4156]5956→37A5A941A5A6[37][41]595641→37A5A9A4A1A5A6[37]59564137→A3A7A5A9A4A1A5A6[]59564137The last3digits are137.B.p q By hand Lists Number52730819[52730819]-79→52A73081A9[52][3081]7981→52A730A8A1A9[52][30]798152→A5A2A730A8A1A9[30]79815230→A5A2A7A3A0A8A1A9[]79815230The last3digits are230.C.p q By hand Lists Number-9453678152[9453678152]-97→A94536A78152[4536][8152]9782→A94536A7A815A2[4536][15]978246→A9A453A6A7A815A2[53][15]97824653→A9A4A5A3A6A7A815A2[15]9782465315→A9A4A5A3A6A7A8A1A5A2[]9782465315 The last3digits are315.DNA sequenceT T T T T G G G A C C C C C C G A A ArtworksA B C D EBookshelf2364715Flow DiagramTriangular Walk←←←←←←←←←←←←←←←←←←←←←←←Maximum Plus–Minus Sum46823472426874378647615682475 Tweet SpotsEvolving Numbers37594156527308199453678152。

【独家带详解答案】IGCSE2019年数学真题卷1(060612)_20200830123606

Candidates answer on the Question Paper. Additional Materials: Electronic calculator
0606/12 May/June 2019
2 hours
READ THESE INSTRUCTIONS FIRST
Write your centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES.
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 80.
*0773973091*
Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education
ADDITIONAL MATHEMATICS Paper 1

澳大利亚数学竞赛题3-4年级10年题

小学中年级（3—4）（2007年）1.下列哪一个数最大？（A）231（B）201（C）321（D）123（E）3022.下列哪一个数是由1个一百、4个十、3个一所凑成的？（A）413（B）143（C）341（D）1043（E）1343.在以下的方格表中，不同的位置用不同的图案标记。

例如位于B4。

请问位于哪个小方格内？（A）A3（B）C3（C）D4（D）B4（E）C24.小欧现在年龄为2岁，她的哥哥小马比她大5岁。

请问小马现在几岁？（A）2（B）3（C）5（D）7（E）105.我购买了一件＄25的衬衫和一件＄12的T恤衫。

若我支付＄50，请问应该找回多少钱？（A）＄13（B）＄23（C）＄25（D）＄33（E）＄376.一个正立方体有多少条棱边？（A）4（B）6（C）8（D）10（E）127.若小安正好排在一列队伍的正中央，请问下列哪个数是这列队伍可能的总共人数？（A）4（B）6（C）8（D）9（E）128.五只猫跑进一间刚刷过油漆的屋子。

有四只猫的两只前腿沾上了油漆，有一只猫的两只后腿沾上了油漆。

请问这些猫共有多少只腿没有沾上油漆？（A）5（B）6（C）8（D）10（E）129.小杰在4：35pm电视卡通节目开演前必须把家庭作业写完。

他写家庭作业需时17分钟，请问他最迟在什么时刻要开始写作业才能及时观看卡通？（A）4：52pm（B）4：20pm（C）4：18pm（D）4：28pm（E）4：08pm10.上星期二，某个班级上数学、音乐、英语及美术课的上课时数比例如下图的饼图所示：下列哪一个叙述为真？（A）上音乐课的时间比上美术课的时间长。

（B）上音乐课的时间比上英语课的时间长。

（C）上音乐课和英语课的时间超过全部上课时间的一半。

（D）上数学课和美术课的时间超过全部上课时间的一半。

（E）上数学课的时间与上美术课的时间一样长。

11.某个家庭中，每位小孩都至少有一位兄弟且至少有二位姊妹。

请问这个家庭中至少有几位小孩？（A）1（B）2（C）3（D）4（E）512.在章鱼大战游戏中，喷水一次可得5分、钩住一次可得3分、缠住一次可得2分。

中学7年级袋鼠数学2019竞赛题

(A) 40
(B) 5 and a half (C) 4
(D) 3
(E) 2 and a half
Question 3 A 3 × 3 × 3 cube is built from 1 × 1 × 1 cubes. The cubes which are in the middle of each face are removed, and the cube in the very center of the 3 × 3 × 3 cube is also removed as shown below. How many 1 × 1 × 1 cubes are left?
(A) 60
(B) 65
(C) 70
(D) 75
(E) 80
Section B (Correct – 4 points | Unanswered – 0 points | Wrong – deduct 1 point)
Question 11 A four-digit integer is written on each of three pieces of paper. The pieces of paper are arranged so that three of the digits are covered as shown below. The sum of the three 4-digit integers is 10126. Which of the following are the covered digits?
(A) 5, 6 and 7 (B) 4, 5 and 7 (C) 4, 6 and 7 (D) 4, 5 and 6 (E) 3, 5 and 6

2019年夏季数学大联盟4-5年级个人赛(Iindividual Part2)真题【决赛】

#2-5. (Time Limit: 7 minutes) The lengths of three sides of a rectangle are x – 5, 2x + 7, and 3x – 17. What is the greatest possible numerical value of the perimeter of this rectangle?
2019 Math League SuMmAmTHerLTEoAuGrUnEaPmReEnSSt
Math League Press, PO BPo.Ox 1. 7B,oTxe1n7a,flTye,nNaeflwy, JNeersweyJe0r7s6ey700-70607107-0U0S1A7
4 5 \July, 2019 – Grades &
#2-3. (Time Limit: 7 minutes) If 2 watermelons can be exchanged for 7 apples and 3 apples can be exchanged for 4 bananas, for how many bananas can 12 watermelons be exchanged?
Individual Questions – Part 2 of 2
Each question is worth 10 points. Calculators are PROHIBITED. #2-1. (Time Limit: 7 minutes) Different letters in the word “MathLeague” represent different digits.
Part 2 ANSWERS (Individual Questions)

2019年高斯数学竞赛7年级真题(Gauss7Contest)

Scoring:There is no penalty for an incorrect answer.Each unanswered question is worth 2,to a maximum of 10unanswered questions.Part A:Each correct answer is worth 5.1.Erin receives $3a day.How many days will it take Erin to receive a total of $30?(A)8(B)12(C)14(D)27(E)102.In the diagram,point F has coordinates (5,5).The pointwith coordinates (2,4)is located at(A)A(B)B (C)C (D)D (E)E (5, 5)3.In the diagram,square P QRS is divided into fouridentical squares.What percentage of square P QRS isshaded?(A)33%(B)40%(C)25%(D)50%(E)15%P Q RS 4.The value of 0.9+0.09is(A)1.08(B)0.909(C)1.8(D)0.99(E)0.185.Based on the graph shown,what is the mode forthe amount of rainfall for the week?(A)9mm(B)12mm (C)3mm (D)15mm (E)6mmDay of the WeekRainfall in Gaussburg 6.If x =3,which of the following is true?(A)2x =5(B)3x −1=8(C)x +5=3(D)7−x =2(E)6+2x =147.When two numbers are added,the result is −26.If one of the numbers is 11,what isthe other number?(A)−37(B)37(C)−15(D)15(E)−488.Joshua is reading a 396-page book.He has read the ﬁrst third of the book only.Howmany pages does he have left to read to ﬁnish the rest of the book?(A)264(B)124(C)250(D)199(E)2449.In thediagram,the value of k is(A)180(B)210(C)240(D)270(E)300˚10.The mean (average)of the numbers 20,30,40is equal to the mean of the numbers(A)28,30,31(B)24,30,38(C)22,30,39(D)23,30,37(E)25,30,34Part B:Each correct answer is worth 6.11.The value of √81is equal to(A)3(B)32(C)33(D)34(E)3512.In the diagram,what is the area of rectangleP QRS ?(A)36(B)32(C)40(D)20(E)442)P (– 4, – – 2)S (– 4, x 13.A piano has 52white keys that occur in a repeating pattern of ABCDEFG.The ﬁrstwhite key is A.What letter is associated with the 33rd white key?(A)A (B)B (C)C (D)D (E)E14.A circular spinner is divided into 8equal sections,asshown.An arrow is attached to the centre of the spinner.The arrow is spun once.What is the probability that thearrow stops in asection containing a prime number thatis odd?(A)18(B)28(C)38(D)48(E)7815.Canadian currency has coins with values $2.00,$1.00,$0.25,$0.10,and $0.05.Barryhas 12coins including at least one of each of these coins.What is the smallest total amount of money that Barry could have?(A)$3.75(B)$3.90(C)$3.70(D)$3.40(E)$3.9516.A positive integer whose digits are the same when read forwards or backwards is calleda palindrome .For example 474and 222are palindromes.How manypalindromes are there between 100and 1000?(A)10(B)90(C)100(D)900(E)100017.The twoare balanced.Ofthefollowing,hasthe samemass as(A)(B)(C)(D)(E)18.A rectangle has length x and width y .A triangle has base 16and height x .If thearea of the rectangle is equal to the area of the triangle,then the value of y is(A)16(B)4(C)8(D)12(E)3219.Each of a,b,c,and d is a positive integer and is greater than3.If1 a−2=1b+2=1c+1=1d−3then which ordering of these four numbers is correct?(A)a<b<c<d(B)c<b<a<d(C)b<a<c<d(D)d<a<c<b(E)b<c<a<d20.The positive integer n has exactly8positive divisors including1and n.Two of thesedivisors are14and21.What is the sum of all8positive divisors of n?(A)35(B)47(C)53(D)96(E)103Part C:Each correct answer is worth8.21.Kathy owns more cats than Alice and more dogs than Bruce.Alice owns more dogsthan Kathy and fewer cats than Bruce.Which of the statements must be true?(A)Bruce owns the fewest cats.(B)Bruce owns the most cats.(C)Kathy owns the most cats.(D)Alice owns the most dogs.(E)Kathy owns the fewest dogs.22.Each of the integers334and419has digits whose product is36.How many3-digitpositive integers have digits whose product is36?(A)21(B)15(C)18(D)24(E)1223.Points T,U,V,W,X,Y lie on square P QRS,as shown.IfP T=T U=UQ=QV=V W=W R=XS=SY,what fraction of the area of square P QRS is shaded?(A)518(B)13(C)29(D)14(E)16SY WV24.A dot starts at(20,19).It can move one unit vertically or horizontally to one of thepoints(21,19),(19,19),(20,20),or(20,18).From there it can move two units in either direction that is perpendicular to theﬁrst move.All moves thereafter increase in length by one unit(three units,four units,ﬁve units,etc.)and must be perpendicular to the direction of the previous move.The dot stops after ten moves.Which of the followingﬁnal locations is not possible?(A)(27,33)(B)(30,40)(C)(21,21)(D)(42,44)(E)(37,37)25.An8×8×n rectangular prism is made up from1×1×1cubes.Suppose thatA is the surface area of the prism andB is the combined surface area of the1×1×1cubes that make up the prism.What is the sum of the values of n for which BAis aninteger?(A)86(B)90(C)70(D)78(E)96。

澳大利亚数学竞赛数学试题

澳大利亚数学竞赛数学试题在澳大利亚，数学竞赛是一项广泛参与的活动，为学生提供了锻炼解题能力和思维能力的机会。

每年，成千上万的学生参与其中，争夺数学竞赛的荣誉和奖项。

以下是一些澳大利亚数学竞赛的数学试题，以展示这项活动的难度和挑战。

1. 解方程：解方程：$3x + 4 = 2x + 9$。

2. 几何题：在一个矩形的底部，有一段长为10cm的线段。

在矩形的顶部，有一段长为6cm的线段。

连接这两段线段的是一条直线，它与矩形左边的边框相交于一个点。

求这个交点与矩形顶部线段的距离。

3. 概率题：一枚硬币抛掷三次，每次都有两个可能的结果：正面或反面。

求至少出现两次正面的概率。

4. 数列题：求$1^2 + 2^2 + 3^2 + 4^2 + ... + 100^2$的和。

5. 几何题：在一个等边三角形ABC中，点D是BC边的中点。

连接AD并延长至E，使得AE与BC平行相交于E。

如果三角形ABC的边长为10cm，求$\angle BAE$的度数。

以上是一些典型的澳大利亚数学竞赛的数学试题。

这些题目涉及到各种不同的数学概念和解题技巧，考察了学生的数学思维能力和逻辑推理能力。

通过参与这样的竞赛，学生们能够提高他们的解题能力和思考问题的能力，并且在竞争中不断进步。

数学竞赛是培养学生数学兴趣和才能的一个重要渠道。

它不仅仅是为了选拔出数学方面的天才，更是为了培养每个学生对于数学的兴趣和热爱，同时也培养他们解决问题的能力。

参加数学竞赛的学生需要掌握一定的数学知识，但更重要的是他们需要学会运用这些知识来解决各种类型的问题。

这就要求学生们在学习过程中注重理论知识的学习，同时也要进行大量的实践和练习。

数学竞赛的数学试题往往涉及到一些常见且基础的数学概念，但同时也包含了一定的难度和挑战。

这些题目不仅仅要求学生掌握基础知识，还需要他们具备灵活运用知识的能力，能够从不同的角度思考问题并得出合理的解决方案。

澳大利亚数学竞赛的数学试题可以对学生进行全面而有针对性的考察。