第十届美国数学邀请赛试题_AIME_

A 整数 A3 数字问题A3－001 在数3000003中，应把它的百位数字和万位数字0换成什么数字，才能使所得的数能被13整除？【题说】 1950年～1951年波兰数学奥林匹克三试题2．【解】设所求数字为x和y，则有因为106、104、102除以13时，分别得余数1、3、9，所以n≡3＋3x＋9y＋3＝3（2＋x＋3y）（mod 13）当且仅当x＋3y＋2被13整除，即x＋3y＋2＝13m（m为自然数）（1）时，n被13整除．由于x＋3y＋2≤9＋3·9＋2＝38所以m只能取1或2．当m＝1时，由方程（1）及0≤x，y≤9，解得x＝8，y＝1；x＝5，y＝2；x＝2，y＝3当m＝2时，解得x＝9，y＝5；x＝6，y＝6；x＝3，y＝7；x＝0，y＝8．故本题共有7个解：3080103，3050203，3020303，3090503，3060603，3030703，3000803．A3－002 求出所有这样的三位数，使其被11整除后的商数等于该三位数各位数字的平方和．【题说】第二届（1960年）国际数学奥林匹克题1．本题由保加利亚提供．【解】设这个三位数除以11以后的商为10a＋b，其中 a是商的十位数，b是商的个位数．若a＋b≥10，则原数为100（a＋1）＋10（a＋b－10）＋b若a＋b＜10，则原数为100a＋10（a＋b）＋b以下对这两种情形分别讨论．先考虑第一种情形．由题设有（a＋1）2＋（a＋b－10）2＋b2＝10a＋b （1）若a＋b＞10，则有（a＋1）2＋（a＋b－10）2＋b2≥（a＋1）2＋1＋（11－a）2故若（1）式成立，只能有a＋b＝10．将b＝10－a代入（1）解得唯一的一组正整数解a＝7，b＝3再考虑第二种情形．此时由题设有a2＋（a＋b）2＋b2＝10a＋b （2）若a＋b＞5，则有a2＋（a＋b）2＋b2＝2（a＋b）·a＋2b2＞10a＋b故若（2）成立，只能有a＋b≤5．注意在（2）式中左边和10a都是偶数；因此b 也是偶数．若a＋b＜5，则b只能为2，将b＝2代入（2）得不到整数解，因此只能有a＋b＝5．将b＝5－a代入（2）得唯一的一组正整数解a＝5，b＝0综上所述，合乎要求的三位数只有550，803．A3－003 下面是一个八位数除以一个三位数的算式，试求商，并说明理由．【题说】 1958年上海市赛高三题1．【解】原式可写成：其中所有未知数都表示数字，且下标为1的未知数都不等于零．x1x2x3等表示x1·102＋x2·10＋x3等．（1）因为得到商的第一个数字7后，同时移下两个数字a5、a6，所以y2＝0，同理y4＝0．（2）四位数a1a2a3a4与三位数b1b2b3之差为两位数c1c2，所以a1＝1，a2＝0，b1＝9，同理，c1＝1，c2＝0，d1＝9，于是a4＝b3，b2＝9，a3＝0．（3）由7×x1x2x3＝99b3，所以x1＝1，x2＝4．990－7×140＝10，所以x3＝2，b3＝4，从而a4＝b3＝4．（4）由c1＝1，c2＝0可知y3＝7．（5）y5×142是四位数，所以x5≥8．又因y5×142的末位数字是8，所以y5＝9．于是商为70709，除数142，从而被除数为10040678．A3－004 证明：在任意39个连续的自然数中，总能找到一个数，它的数字之和被11整除．【题说】 1961年全俄数学奥林匹克八年级题 3．【证】在任意39个连续自然数中，一定有三个数末位数字为0，而前两个数中一定有一个十位数字不为9，设它为N，N的数字之和为n，则N，N＋1，N＋2，…，N＋9，N＋19这11个数的数字之和依次为n，n＋1，n＋2，…，n＋9，n ＋10，其中必有一个是11的倍数．【注】 39不能改为38．例如999981至1000018这38个连续自然数中，每个数的数字和都不被11整除．本题曾被改编为匈牙利1986年竞赛题、北京市1988年竞赛题．A3－005 求有下列性质的最小自然数n：其十进制表示法以6结尾；当去掉最后一位6并把它写在剩下数字之前，则成为n的四倍数．【题说】第四届（1962年）国际数学奥林匹克题1．本题由波兰提供．【解】设n＝10m＋6，则6×10p＋m＝4（10m＋6），其中p为m的位数．于是m ＝2（10p－4）/13，要使m为整数，p至少为5，此时，n＝153846．A3－006 公共汽车票的号码由六个数字组成．若一张票的号码前三个数字之和等于后三个数字之和，则称它是幸运的．证明：所有幸运车票号码的和能被13整除．【题说】 1965年全俄数学奥林匹克八年级题 4．【证】设幸运车票的号码是A，则A′＝999999－A也是幸运的，且A≠A′．因为A ＋A′＝999999＝999×1001含因数13．而所有幸运号码都能如此两两配对．所以所有幸运号码之和能被13 整除．A3－007 自然数k有如下性质：若n能被k整除，那末把n的数字次序颠倒后得到的数仍能被k整除．证明：k是99的因子．【题说】第一届（1967年）全苏数学奥林匹克十年级题5．【证】 k与10互质．事实上，存在首位为1且能被k整除的数，把它的数字倒过来也能被k整除，而此数的末位数字为1．取以500开头的且被k整除的数：500abc…z，（a，b，c，…，z是这个数的数字），则以下的数均被k整除：（1）z…cba005．（2）和（3）把（2）中的和倒过来z…cba00010abc…z（4）差由此看出，99能被k整除．A3－008 计算由1到109的每一个数的数字之和，得到109个新数，再求每一个新数的数字之和；这样一直进行下去，直到都是一位数为止．那么，最后得到的数中是1多，还是2多？【题说】 1964年全俄数学奥林匹克八年级题3．考虑整数被9除的余数．【解】一个正整数与其数字之和关于9是同余的，故最后所得的一位数为1者，是原数被9除余1的数，即1，10，19，…，999999991及109．同理，最后所得一位数为2者，原数被9除余2，即2，11，20， (999999992)二者相比，余1者多一个数，因此，最后得到的一位数中以1为多．A3－009 求出具有下列性质的所有三位数A：将数A的数字重新排列，得出的所有数的算术平均值等于A．【题说】第八届（1974年）全苏数学奥林匹克九年级题 5．由此可得222（a＋b＋c）＝6（100a＋10b＋c），即7a＝3b＋4c，将这方程改写成7（a－b）＝4（c－b）当0≤b≤2时，a＝b＝c，或a－b＝4且c－b＝7．当7≤b≤9时，b－a＝4，b－c＝7，从而A∈{111，222，…，999，407，518，629，370，481，592}显然这15个三位数都合乎要求．A3－010 当44444444写成十进制数时，它的各位数字之和是A，而B是A的各位数字之和，求B的各位数字之和（所有的数都是十进制数）．【题说】第十七届（1975年）国际数学奥林匹克题4．本题由原苏联提供．【解】因为44444444的位数不超过4×4444＝17776，所以A≤177760B≤1＋5×9＝46，B的数字和C≤4＋9＝13由于一个数与它的数字和mod 9同余，所以C≡B≡A≡44444444≡74444＝（73）1481×7≡11781×7≡7（mod 9）故C＝7，即数B的各位数字之和是7．A3－011 设n是整数，如果n2的十位数字是7，那么n2的个位数字是什么？【题说】第十届（1978年）加拿大数学奥林匹克题1．【解】设n＝10x＋y，x、y为整数，且0≤y≤9，则n2＝100x2＋20xy＋y2＝20A＋y2（A为正整数）因20A的十位数字是偶数，所以要想使n2十位数字是7，必须要y2的十位数字是奇数，这只有y2＝16或36．从而y2的个位数字，即n2的个位数字都是6．A3－013 下列整数的末位数字是否组成周期数列？其中[a]表示数a的整数部分．【题说】第十七届（1983年）全苏数学奥林匹克九年级题 4．由于不循环小数，所以{a2k＋1}从而{a n}不是周期数列．在二进制中的末位数字．显然，b n为偶数时，r n＝0，b n为奇数时，r n＝1．仿（a）可证{r n}不是周期的，从而{b n}也不是周期数列．A3－014 设a n是12＋22＋…＋n2的个位数字，n＝1，2，3，…，试证：0.a1a2…a n…是有理数．【题说】 1984年全国联赛二试题 4．【证】将（n＋1）2，（n＋2）2，…，（n＋100）2这100个数排成下表：（n＋1）2 （n＋2）2 … （n＋10）2（n＋11）2 （n＋12）2 … （n＋20）2… … … …（n＋91）2 （n＋92）2 … （n＋100）2因k2与（k＋10）2的个位数字相同，故表中每一列的10个数的个位数字皆相同．因此，将这100个数相加，和的个位数字是0．所以，a n＋100＝a n对任何n成立．A3－015 是否存在具有如下性质的自然数n：（十进制）数n的数字和等于1000，而数n2的数字和等于10002？【题说】第十九届（1985年）全苏数学奥林匹克八年级题 2．【解】可用归纳法证明更一般的结论：对于任意自然数m，存在由1和0组成的自然数n，它的数字和S（n）＝m，而n2的数字和S（n2）＝m2？当m＝1，n＝1时，显然满足要求．设对自然数m，存在由1和0组成的自然数n，使得S（n）＝m，S（n2）＝m2设n为k位数，取n1＝n×10k＋1＋1，则n1由0，1组成并且S（n1）＝S（n）＋1＝m＋1＝S（n2×102k＋2）＋S（2n×10k＋1）＋S（1）＝S（n2）＋2S（n）＋1＝m2＋2m＋1＝（m＋1）2因此命题对一切自然数m均成立．这说明0.a1a2a3…是循环小数，因而是有理数．A3－017 设自然数n是一个三位数．由它的三个非零数字任意排列成的所有三位数的和减去 n等于1990．求 n．【题说】 1989年芜湖市赛题 3．2090＜222（a＋b＋c）＝1990＋n＜2989而2090＞222×9＝1998，222×10＝2220＝1990＋230222×11＝2442×1990＋452，222×12＝2664＝1990＋674222×13＝2886＝1990＋896，222×14＝3108＞2989经验证：a＋b＋c＝11时，n＝452符合题意．A3－018 定义数列{a n}如下：a1＝19891989，a n等于a n－1的各位数字之和，a5等于什么？【题说】第二十一届（1989年）加拿大数学奥林匹克题 3．【解】由a1＜100001989＝b1，而b1的位数是4×1989＋1＝7957，知a2＜10×8000＝80000，所以a2最多是5位数，从而a3≤5×9＝45，a4≤4＋9＝13，因此a5一定是一位数．另一方面，由9|1989，知9|a1，因而9可整除a1的数字和，即9|a2，又因此有9|a3，9|a4，9|a5．所以a5＝9．A3－019 某州颁发由6个数字组成的车牌证号（由0—9的数字组成），且规定任何两个牌号至少有两个数字不同（因此，证号“027592”与“020592”不能同时使用），试确定车牌证号最多有多少个？【题说】第十九届（1990年）美国数学奥林匹克题1．【解】至多可造出不同的五位证号a1a2a3a4a5105个．令a6是a1＋a1＋a3＋a4＋a5的个位数字，所成的六位数便满足要求．因为如果两个数的前五位中只有一个数字不同，那么第6位数字必然不同．另一方面，任何105＋1个6位数中，总有两个前五位数字完全相同．因此，符合题目要求的车牌证号最多有105个．A3－020 设 A＝99…99（81位全为9），求A2的各位数字之和．【题说】 1991年日本数学奥林匹克预选赛题1．【解】由A＝1081－1知A2＝10162－2·1081＋1＝99...980 (01)↑ ↑162位 82位故A2各位数字之和＝9×（162－82）＋8＋1＝729．4A3－021 如果一个正整数的十进制表示中至少有两个数字，并且每个数字都比它右边的数字小，那么称它为“上升”的．这种“上升”的正整数共有多少个？【题说】第十届（1992年）美国数学邀请赛题2．【解】符合条件的正整数中的数字，都是不同的非零数码，即集合S＝{1，2，3，…，9}的二元或二元以上的子集．反过来，S的每个二元或二元以上的子集，将它的数码从小到大排列，也得到一个符合条件的正整数．S的子集共有29=512个，其中只含一个元素的子集有9个，一个空集．故符合条件的正整数共有512－10=502个．A3－023 求方程的各个正根的乘积的最后三位数字．【题说】第十三届（1995年）美国数学邀请赛题2．【解】令y=1og1995x．由原方程取对数得其最后三位数字为025．A3－024 一个六位数的首位数字是5，是否总能够在它的后面再添加6个数字，使得所得的十二位数恰是一个完全平方数？【题说】1995年城市数学联赛高年级普通水平题3．【解】不．若不然，105个以5为首位数字的六位数可以衍生出105个十二位的完全平方数．即有105个自然数n满足．5×1011≤n2＜6×1011亦即7×105＜n＜8×105由于7×105与8×105之间不存在105个整数，故上式不可能成立．。

A2 整数的求解A2－001 哪些连续正整数之和为1000？试求出所有的解．【题说】 1963年成都市赛高二二试题 3．【解】设这些连续正整数共n个（n＞1），最小的一个数为a，则有a＋（a＋1）＋…＋（a＋n－1）＝1000即n（2a＋n－1）＝2000若n为偶数，则2a＋n－1为奇数；若n为奇数，则2a＋n－1为偶数．因a≥1，故2a＋n－1＞n．同，故只有n＝5，16，25，因此可能的取法只有下列三种：若n＝5，则 a＝198；若n＝16，则 a＝55；若n＝25，则 a＝28．故解有三种：198＋199＋200＋201＋20255＋56＋…＋7028＋29＋…＋52A2－002 N是整数，它的b进制表示是777，求最小的正整数b，使得N是整数的四次方．【题说】第九届（1977年）加拿大数学奥林匹克题3．【解】设b为所求最小正整数，则7b2＋7b＋7＝x4素数7应整除x，故可设x＝7k，k为正整数．于是有b2＋b＋1＝73k4当k＝1时，（b－18）（b＋19）＝0．因此b＝18是满足条件的最小正整数．A2－003 如果比n个连续整数的和大100的数等于其次n个连续数的和，求n．【题说】 1976年美国纽约数学竞赛题 7．s2－s1＝n2＝100从而求得n＝10．A2－004 设a和b为正整数，当a2＋b2被a＋b除时，商是q而余数是r，试求出所有数对（a，b），使得q2＋r＝1977．【题说】第十九届（1977年）国际数学奥林匹克题 5．本题由原联邦德国提供．【解】由题设a2＋b2＝q（a＋b）＋r（0≤r＜a＋b），q2＋r＝1977，所以q2≤1977，从而q≤44．若q≤43，则r＝1977－q2≥1977－432＝128．即（a＋b）≤88，与（a＋b）＞r≥128，矛盾．因此，只能有q＝44，r＝41，从而得a2＋b2＝44（a＋b）＋41（a－22）2＋（b－22）2＝1009不妨设|a－22|≥|b－22|，则1009≥（a－22）2≥504，从而45≤a≤53．经验算得两组解：a＝50，b＝37及a＝50，b＝7．由对称性，还有两组解a＝37，b＝50；a＝7，b＝50．A2－005 数1978n与1978m的最后三位数相等，试求出正整数n和m，使得m＋n 取最小值，这里n＞m≥1．【题说】第二十届（1978年）国际数学奥林匹克题 1．本题由古巴提供．【解】由题设1978n－1978m＝1978m（1978n－m－1）≡0（mod 1000）理注解：设1978n=1000a+c 1978m=1000b+c 1978n－1978m=1000(a-b因而1978m≡2m×989m≡0（mod 8），m≥31978n－m≡1（mod 125）注解：1978m（1978n－m－1）这两式的乘积要为1000整除，显然1978m这式为8的倍数，另一式为125的倍数。

美国数学邀请赛试题解答
美国数学邀请赛试题解答美国数学邀请赛(American Invitational Mathematics Examination，AIME)是一项全国性的高中数学比赛，旨在鼓励高中学生发展其创造性技能，发挥其最大潜能，激发他们发现数学的乐趣。

美国数学邀请赛试题解答首先要掌握一些数学基础知识，比如代数、几何、概率等，只有把握好基础知识，才能在解答试题时发挥出最大的效率。

其次，在解答试题时要细致耐心，一步一步地解决问题，把控住步骤，保证正确性，不能草率回答，要仔细检查自己的答案，确保准确性。

最后，在解答试题时要学会发现问题，仔细阅读题目，理清楚题目的要求，得出正确的答案，不可以死记硬背知识，要能够根据题目灵活运用知识，提出有效答案。

总而言之，解答美国数学邀请赛试题，要掌握基础知识，细致耐心，发现问题，学会运用知识，以及正确检查自己的答案，才能取得优异成绩。

20011The median of the listn,n +3,n +4,n +5,n +6,n +8,n +10,n +12,n +15is 10.What is the mean?(A)4(B)6(C)7(D)10(E)112A number x is 2more than the product of its reciprocal and its additive inverse.In whichinterval does the number lie?(A)−4≤x ≤−2(B)−2<x ≤0(C)0<x ≤2(D)2<x ≤4(E)4<x ≤63The sum of two numbers is S .Suppose 3is added to each number and then each of theresulting numbers is doubled.What is the sum of the final two numbers?(A)2S +3(B)3S +2(C)3S +6(D)2S +6(E)2S +124What is the maximum number of possible points of intersection of a circle and a triangle?(A)2(B)3(C)4(D)5(E)65How many of the twelve pentominoes pictured below have at least one line of symettry?(A)3(B)4(C)5(D)6(E)7[asy]unitsize(5mm);defaultpen(linewidth(1pt));draw(shift(2,0)*unitsquare);draw(shift(2,1)*unitsquare);draw(shift(2,2)*unitsquare);draw(shift(1,2)*unitsq draw(shift(0,2)*unitsquare);draw(shift(2,4)*unitsquare);draw(shift(2,5)*unitsquare);draw(shift(2,6)*unitsquare);draw(shift(1,5)*unitsq draw(shift(0,5)*unitsquare);draw(shift(4,8)*unitsquare);draw(shift(3,8)*unitsquare);draw(shift(2,8)*unitsquare);draw(shift(1,8)*unitsq draw(shift(0,8)*unitsquare);draw(shift(6,8)*unitsquare);draw(shift(7,8)*unitsquare);draw(shift(8,8)*unitsquare);draw(shift(9,8)*unitsq draw(shift(9,9)*unitsquare);draw(shift(6,5)*unitsquare);draw(shift(7,5)*unitsquare);draw(shift(8,5)*unitsquare);draw(shift(7,6)*unitsq draw(shift(7,4)*unitsquare);draw(shift(6,1)*unitsquare);draw(shift(7,1)*unitsquare);draw(shift(8,1)*unitsquare);draw(shift(6,0)*unitsq draw(shift(7,2)*unitsquare);draw(shift(11,8)*unitsquare);draw(shift(12,8)*unitsquare);draw(shift(13,8)*unitsquare);draw(shift(14,8)*u draw(shift(13,9)*unitsquare);draw(shift(11,5)*unitsquare);draw(shift(12,5)*unitsquare);draw(shift(13,5)*unitsquare);draw(shift(11,6)*u draw(shift(13,4)*unitsquare);draw(shift(11,1)*unitsquare);draw(shift(12,1)*unitsquare);draw(shift(13,1)*unitsquare);draw(shift(13,2)*u draw(shift(14,2)*unitsquare);This file was downloaded from the AoPS −MathLinks Math Olympiad Resources PagePage 1http://www.mathlinks.ro/2001draw(shift(16,8)*unitsquare);draw(shift(17,8)*unitsquare);draw(shift(18,8)*unitsquare);draw(shift(17,9)*u draw(shift(18,9)*unitsquare);draw(shift(16,5)*unitsquare);draw(shift(17,6)*unitsquare);draw(shift(18,5)*unitsquare);draw(shift(16,6)*u draw(shift(18,6)*unitsquare);draw(shift(16,0)*unitsquare);draw(shift(17,0)*unitsquare);draw(shift(17,1)*unitsquare);draw(shift(18,1)*u draw(shift(18,2)*unitsquare);[/asy]6Let P (n )and S (n )denote the product and the sum,respectively,of the digits of the integern .For example,P (23)=6and S (23)=5.Suppose N is a two-digit number such thatN =P (N )+S (N ).What is the units digit of N ?(A)2(B)3(C)6(D)8(E)97When the decimal point of a certain positive decimal number is moved four places to theright,the new number is four times the reciprocal of the original number.What is theoriginal number?(A)0.0002(B)0.002(C)0.02(D)0.2(E)28Wanda,Darren,Beatrice,and Chi are tutors in the school math lab.Their schedule is asfollows:Darren works every third school day,Wanda works every fourth school day,Beatriceworks every sixth school day,and Chi works every seventh school day.Today they are allworking in the math lab.In how many school days from today will they next be togethertutoring in the lab?(A)42(B)84(C)126(D)178(E)2529The state income tax where Kristin lives is levied at the rate of p %of the first $28000ofannual income plus (p +2)%of any amount above $28000.Kristin noticed that the stateincome tax she paid amounted to (p +0.25)%of her annual income.What was her annualincome?(A)$28000(B)$32000(C)$35000(D)$42,000(E)$5600010If x ,y ,and z are positive with xy =24,xz =48,and yz =72,then x +y +z is(A)18(B)19(C)20(D)22(E)2411Consider the dark square in an array of unit squares,part of which is shown.The rst ring ofsquares around this center square contains 8unit squares.The second ring contains 16unitsquares.If we continue this process,the number of unit squares in the 100th ring is(A)396(B)404(C)800(D)10,000(E)10,404[asy]unitsize(3mm);defaultpen(linewidth(1pt));2001fill((2,2)–(2,7)–(7,7)–(7,2)–cycle,mediumgray);fill((3,3)–(6,3)–(6,6)–(3,6)–cycle,gray);fill((4,4)–(5,4)–(5,5)–(4,5)–cycle,black);for(real i=0;i¡=9;++i)draw((i,0)–(i,9));draw((0,i)–(9,i));[/asy]12Suppose that n is the product of three consecutive integers and that n is divisible by 7.Whichof the following is not necessarily a divisor of n ?(A)6(B)14(C)21(D)28(E)4213A telephone number has the form ABC −DEF −GHIJ ,where each letter represents adifferent digit.The digits in each part of the numbers are in decreasing order;that is,A >B >C ,D >E >F ,and G >H >I >J .Furthermore,D ,E ,and F are consecutive even digits;G ,H ,I ,and J are consecutive odd digits;and A +B +C =9.Find A .(A)4(B)5(C)6(D)7(E)814A charity sells 140benefit tickets for a total of $2001.Some tickets sell for full price (a wholedollar amount),and the rest sells for half price.How much money is raised by the full-price tickets?(A)$782(B)$986(C)$1158(D)$1219(E)$144915A street has parallel curbs 40feet apart.A crosswalk bounded by two parallel stripes crossesthe street at an angle.The length of the curb between the stripes is 15feet and each stripe is 50feet long.Find the distance,in feet,between the stripes.(A)9(B)10(C)12(D)15(E)2516The mean of three numbers is 10more than the least of the numbers and 15less than thegreatest.The median of the three numbers is 5.What is their sum?(A)5(B)20(C)25(D)30(E)3617Which of the cones listed below can be formed from a 252◦sector of a circle of radius 10byaligning the two straight sides?[asy]import graph;unitsize(1.5cm);defaultpen(fontsize(8pt));draw(Arc((0,0),1,-72,180),linewidth(.8pt));draw(dir(288)–(0,0)–(-1,0),linewidth(.8pt));label(”10”,(-0.5,0),S);draw(Arc((0,0),0.1,-72,180));label(”252◦”,(0.05,0.05),NE);[/asy](A)A cone with slant height of 10and radius 6(B)A cone with height of 10and radius 6(C)A cone with slant height of 10and radius 7(D)A cone with height of 10and radius 7(E)A cone with slant height of 10and radius 818The plane is tiled by congruent squares and congruent pentagons as indicated.The percentof the plane that is enclosed by the pentagons is closest to2001(A)50(B)52(C)54(D)56(E)58[asy]unitsize(3mm);defaultpen(linewidth(0.8pt));path p1=(0,0)–(3,0)–(3,3)–(0,3)–(0,0);path p2=(0,1)–(1,1)–(1,0);path p3=(2,0)–(2,1)–(3,1);path p4=(3,2)–(2,2)–(2,3);path p5=(1,3)–(1,2)–(0,2);path p6=(1,1)–(2,2);path p7=(2,1)–(1,2);path[]p=p1020304050607;for(int i=0;i¡3;++i)for(int j=0;j¡3;++j)draw(shift(3*i,3*j)*p);[/asy]19Pat wants to buy four donuts from an ample supply of three types of donuts:glazed,chocolate,and powdered.How many different selections are possible?(A)6(B)9(C)12(D)15(E)1820A regular octagon is formed by cutting an isosceles right triangle from each of the corners ofa square with sides of length 2000.What is the length of each side of the octagon?(A)13(2000)(B)2000(√2−1)(C)2000(2−√2)(D)1000(E)1000√221A right circular cylinder with its diameter equal to its height is inscribed in a right circularcone.The cone has diameter 10and altitude 12,and the axes of the cylinder and cone coincide.Find the radius of the cylinder.(A)83(B)3011(C)3(D)258(E)7222In the magic square shown,the sums of the numbers in each row,column,and diagonal arethe same.Five of these numbers are represented by v ,w ,x ,y ,and z .Find y +z .(A)43(B)44(C)45(D)46(E)47[asy]unitsize(10mm);defaultpen(linewidth(1pt));for(int i=0;i¡=3;++i)draw((0,i)–(3,i));draw((i,0)–(i,3));label(”25”,(0.5,0.5));label(”z ”,(1.5,0.5));label(”21”,(2.5,0.5));label(”18”,(0.5,1.5));label(”x ”,(1.5,1.5));label(”y ”,(2.5,1.5));label(”v ”,(0.5,2.5));label(”24”,(1.5,2.5));label(”w ”,(2.5,2.5));[/asy]23A box contains exactly five chips,three red and two white.Chips are randomly removed oneat a time without replacement until all the red chips are drawn or all the white chips are drawn.What is the probability that the last chip drawn is white?(A)310(B)25(C)12(D)35(E)71024In trapezoid ABCD ,AB and CD are perpendicular to AD ,with AB +CD =BC ,AB <CD ,and AD =7.What is AB ·CD ?(A)12(B)12.25(C)12.5(D)12.75(E)13200125How many positive integers not exceeding 2001are multiples of 3or 4but not 5?(A)768(B)801(C)934(D)1067(E)1167。

Problem 1The ratio is closest to which of the following numbers?SolutionProblem 2Given that a, b, and c are non-zero real numbers, define. Find.SolutionProblem 3According to the standard convention for exponentiation,.If the order in which the exponentiations are performed is changed, how manyother values are possible?SolutionProblem 4For how many positive integers is there at least 1 positive integer such that ?infinitely manySolutionProblem 5Each of the small circles in the figure has radius one. The innermost circle istangent to the six circles that surround it, and each of those circles is tangent tothe large circle and to its small-circle neighbors. Find the area of the shaded region.SolutionProblem 6From a starting number, Cindy was supposed to subtract 3, and then divide by 9,but instead, Cindy subtracted 9, then divided by 3, getting 43. If the correct instructions were followed, what would the result be?SolutionProblem 7A arc of circle A is equal in length to a arc of circle B. What is the ratio of circle A's area and circle B's area?SolutionProblem 8Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let be the total area of the blue triangles,the total area of the white squares, and the area of the red square. Which of the following is correct?SolutionThere are 3 numbers A, B, and C, such that, and. What is the average of A, B, and C?Not uniquely determined SolutionProblem 10What is the sum of all of the roots of?SolutionProblem 11Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB each, 12 of the files take up 0.7 MB each, and the rest take up 0.4 MB each. It is not possible to split a file onto 2 different disks. What is thesmallest number of disks needed to store all 30 files?SolutionProblem 12Mr. Earl E. Bird leaves home every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?SolutionProblem 13Given a triangle with side lengths 15, 20, and 25, find the triangle'ssmallest height.SolutionBoth roots of the quadratic equation are prime numbers. The number of possible values of isSolutionProblem 15Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, usingeach digit only once. What is the sum of the 4 prime numbers?SolutionProblem 16Let. What is?SolutionProblem 17Sarah pours 4 ounces of coffee into a cup that can hold 8 ounces. Then she pours4 ounces of cream into a second cup that can also hold 8 ounces. She then pours half of the contents of the first cup into the second cup, completely mixes the contents of the second cup, then pours half of the contents of the second cup back into the first cup. What fraction of the contents in the first cup is cream?SolutionProblem 18A 3x3x3 cube is made of 27 normal dice. Each die's opposite sides sum to 7.What is the smallest possible sum of all of the values visible on the 6 faces of the large cube?SolutionProblem 19Spot's doghouse has a regular hexagonal base that measures one yard oneach side. He is tethered to a vertex with a two-yard rope. What is the area, insquare yards, of the region outside of the doghouse that Spot can reach?SolutionProblem 20Points and lie, in that order, on , dividing it into fivesegments, each of length 1. Point is not on line . Point lies on ,and point lies on . The line segments and are parallel.Find .SolutionProblem 21The mean, median, unique mode, and range of a collection of eight integersare all equal to 8. The largest integer that can be an element of this collection isSolutionProblem 22A set of tiles numbered 1 through 100 is modified repeatedly by the followingoperation: remove all tiles numbered with a perfect square , and renumber theremaining tiles consecutively starting with 1. How many times must the operationbe performed to reduce the number of tiles in the set to one?SolutionProblem 23Points and lie on a line, in that order, with and. Point is not on the line, and. The perimeter ofis twice the perimeter of. Find.SolutionProblem 24Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina?SolutionProblem 25, we have , In trapezoid, , andwith bases and(diagram notto scale). The area of is Solution。

目 录2018年亚太地区数学奥林匹克 (1)2018年波罗的海地区数学奥林匹克 (2)2018年第10届Benelux数学奥林匹克 (5)2018年巴尔干地区数学奥林匹克 (6)2018年巴尔干地区初中数学奥林匹克 (7)2018年高加索地区数学奥林匹克 (8)2018年中美洲及加勒比地区数学奥林匹克 (10)2018年Cono Sur数学奥林匹克 (11)2018年捷克-波兰-斯洛伐克联合数学竞赛 (12)2018年捷克和斯洛伐克数学奥林匹克 (13)2018年多瑙河地区数学奥林匹克 (14)2018年欧洲女子数学奥林匹克 (16)2018年欧洲数学杯奥林匹克 (18)2018年拉丁美洲数学奥林匹克 (20)2018年国际大都市数学竞赛(IOM) (21)2018年第2届IMO复仇赛 (22)2018年第5届伊朗几何奥林匹克 (23)2018年第17届基辅数学节竞赛 (27)2018年地中海地区数学竞赛 (29)2018年中欧数学奥林匹克 (30)2018年北欧数学奥林匹克 (32)2018年泛非数学奥林匹克 (33)2018年罗马尼亚大师杯数学奥林匹克 (35)2018年第14届Sharygin几何奥林匹克 (36)2018年丝绸之路数学奥林匹克 (42)2018年Tuymaada国际数学奥林匹克 (43)2018年乌克兰几何奥林匹克 (45)2018年第14届Zhautykov国际数学奥林匹克 (47)2018年ARML数学竞赛 (48)2018年美国数学邀请赛(AIME) I (57)2018年美国数学邀请赛(AIME) II (60)2018年美国数学奥林匹克 (63)2018年美国初中数学奥林匹克 (64)2018年美国IMO代表队选拔考试 (65)2018年美国TSTST (67)2018年美国第20届ELMO (69)2018年第20届美国旧金山湾区数学奥林匹克 (71)2017-2018年度USAMTS (74)2018年美国女子数学奖学金竞赛(决赛) (79)2017-2018年度威斯康星数学、科学与工程学人才选拔 (80)2018年奥地利数学奥林匹克 (84)2018年澳大利亚、英国IMO国家队联合训练考试 (87)2018年波黑数学奥林匹克(地区级) (88)2018年波黑EGMO代表队选拔考试 (90)2018年波黑JBMO代表队选拔考试 (91)2018年巴西数学奥林匹克 (92)2018年巴西数学奥林匹克复仇赛 (94)2017/2018英国数学竞赛 (95)2018年保加利亚数学奥林匹克 (97)2018年保加利亚JBMO代表队选拔考试 (98)2018年加拿大数学奥林匹克 (99)2018年塞浦路斯IMO代表队选拔考试 (100)2018年塞浦路斯JBMO代表队选拔考试 (102)2018年丹麦数学奥林匹克(第二轮) (105)2018年德国数学奥林匹克(12年级决赛) (106)2018年希腊数学奥林匹克 (107)2018年香港数学奥林匹克 (109)2018年香港IMO代表队选拔考试 (110)2018年匈牙利库尔沙克数学竞赛 (112)2018年印度全国数学奥林匹克 (113)2018年印度IMO代表队选拔考试 (114)2018年伊朗数学奥林匹克 (117)2018年伊朗IMO代表队选拔考试 (120)2018年爱尔兰数学奥林匹克 (123)2018年意大利数学奥林匹克 (125)2018年哈萨克斯坦数学奥林匹克(11年级决赛) (126)2018韩国数学奥林匹克 (127)2018年韩国数学冬令营训练题 (130)2018年科索沃IMO培训考试 (131)2018年马其顿数学奥林匹克 (132)2018年墨西哥数学奥林匹克 (133)2018年摩尔多瓦EGMO代表队选拔考试 (135)2018年摩尔多瓦IMO代表队选拔赛 (136)2018年摩尔多瓦JBMO代表队选拔考试 (138)2018年摩洛哥IMO代表队选拔考试 (139)2017-2018年度挪威数学奥林匹克(决赛) (140)2017-2018年度波兰数学奥林匹克 (141)2017-2018年度波兰初中数学奥林匹克 (145)2018年罗马尼亚数学奥林匹克 (147)2018年罗马尼亚IMO代表队选拔考试 (149)2018年罗马尼亚JBMO代表队选拔考试 (150)2018年全俄数学奥林匹克 (154)2018年圣彼得堡数学奥林匹克 (158)2018年塞尔维亚数学奥林匹克 (161)2018年塞尔维亚JBMO代表队选拔考试 (162)2018年斯洛文尼亚IMO代表队选拔考试 (163)2018年南非数学奥林匹克 (164)2018年西班牙数学奥林匹克 (165)2018年塔吉克斯坦IMO代表队选拔考试 (166)2018年土耳其数学奥林匹克 (168)2018年乌克兰数学奥林匹克 (169)2018年越南数学奥林匹克 (171)2018年越南IMO代表队选拔考试 (173)2018年国际大学生数学竞赛(IMC) (175)2018年V ojtěch Jarník国际大学生数学竞赛 (177)2018年Putnam数学竞赛 (179)2018年哈佛大学-麻省理工学院数学竞赛春季赛 (181)2018年哈佛大学-麻省理工学院数学邀请赛 (189)2018年哈佛大学-麻省理工学院数学竞赛冬季赛 (190)2018年Berkeley数学竞赛 (197)2018年卡内基梅隆大学数学竞赛 (213)2018年普林斯顿大学数学竞赛 (226)2018年斯坦福大学数学竞赛 (237)2018年哈维穆德学院数学竞赛 (254)2018年MMATHS数学竞赛 (259)2018年Duke大学数学竞赛 (264)2018年亚太地区数学奥林匹克试题比赛时间: 2018年3月13日1. 设H 为△ABC 的垂心. 点M , N 分别为边AB , AC 的中点, 点H 位于四边形BMNC 的内部. △BMH 与△CNH 的外接圆相外切. 过H 作BC 的平行线, 与△BMH 与△CNH 的外接圆分别相交于点K , L (均不同于点H ). 直线MK 与NL 相交于点F . 设△MNH 的内心为J . 证明: FJ = F A .2. 对实数x , 定义函数f (x ), g (x )如下:20181...41211)(-++-+-+=x x x x x f , 20171...513111)(-++-+-+-=x x x x x g . 证明: 对任意满足0 < x < 2018的非整数的实数x , 有|f (x ) – g (x )| > 2成立.3. 我们称平面上n 个正方形的摆放方式为"三足鼎立"的, 如果它们同时满足以下三个条件:i) 所有正方形均全等.ii) 如果两个正方形有公共点P , 则P 同时为这两个正方形的顶点.iii) 每一个正方形都恰好与其他三个正方形有公共点.求在2018 ≤ n ≤ 3018范围内, 有多少个整数n , 使得存在n 个正方形为"三足鼎立"的.4. 一束光线从正△ABC 的顶点A 出发, 在三角形内部遵循光反射定律(即入射角等于出射角)不断反射, 但当光线到达△ABC 的任一顶点处时, 反射停止. 求所有可能的正整数n , 使得光线在△ABC 内经过n 次反射后, 恰在顶点A 处停止.5. 求所有的整系数多项式P (x ), 使得对任意的实数s , t , 如果P (s ), P (t )均为整数, 则P (st )也是整数.2018年波罗的海地区数学奥林匹克试题1. 称一个由有限个正实数(不必互异)构成的集合为"平衡"的, 如果其中每一个数都小于其余各数之和. 求所有的整数m ≥ 3, 使得任何由m 个正实数构成的平衡集均可被划分为3个无公共元素的子集, 满足每个子集的各元素之和均小于另两个子集的各元素的总和.2. 考虑一个100 ⨯ 100的表格. 对每一个整数1 ≤ k ≤ 100, 该表格的第k 行填有按自左向右递增顺序排列的数1, 2, …, k (但不一定位于连续的格子内); 而该行其余的100 – k 个格子均填0. 证明: 该表格中存在两列, 使得其中一列的各数之和至少是另一列各数之和的19倍.3. 设正实数a , b , c , d 满足abcd = 1. 证明:110321≤+++∑cyc c b a . 4. 求所有具有下述特点的函数f : [0, ∞) → [0, ∞): 对所有的正整数n 及非负实数x 1, x 2, …, x n , 有2222122221)(...)()()...(n n x f x f x f x x x f +++=+++成立. 5. 称一个实系数多项式f (x )为"生成"的, 如果对每一个实系数多项式ϕ(x ), 均存在正整数k 及实系数多项式g 1(x ), g 2(x ), …, g k (x ), 使得ϕ(x ) = f (g 1(x )) + f (g 2(x )) + … + f (g k (x ))成立. 求所有的生成多项式.6. 设n 为正整数. 精灵Elfie 从原点(0, 0, 0)开始, 在三维空间里旅行. 每一步, 她可以瞬移至距她当前所在点距离恰为n 的任意整点. 但是, 瞬移是一件复杂的事情: Elfie 最初处于正常状态, 但是第一次瞬移后变为怪异状态, 第二次瞬移后恢复为正常状态, 以后则如此交替变化. 求所有的n , 使得对所有整点, Elfie 都能够以正常状态访问过该点.7. 一个16 ⨯ 16圆环体有512条边(如图), 将每条边染为红色或蓝色之一. 称一种染色方式为"好"的, 如果每一个顶点都是偶数条红色边的顶点. 定义一步"转换"为将任一格的四条边均改变颜色(红变蓝, 蓝变红). 问最多有多少个"好"的染色方式, 使得其中任意一个染色方式都不能够通过一系列的"转换"而变为另一个.8. 一个图具有N个顶点. 在某一顶点处有一只不可见的兔子. 一群猎人计划猎杀这只兔子. 在每一步, 每个猎人都瞄准某一个顶点同时开枪射击, 他们可以事先商量好每人瞄准哪一个顶点. 如果兔子恰在被瞄准射击的顶点之一, 则打猎活动结束. 否则, 兔子在接下来的一步中可以选择继续停留在原顶点处或跳至某个相邻顶点处. 假设已知有一种方案可以使猎人至多经N!步就可以猎杀兔子. 证明: 存在一种方案, 可以使得猎人至多经2N步就可以猎杀兔子.9. Olga和Sasha在一个无限六边形网格上玩游戏. 他们轮流选择一个空的六边形,并在其上放置一张骨牌, 由Olga先行. 恰在第2018张骨牌放置之前, 一条新规则开始起效: 从此时起, 只能在和至少两个已被放置骨牌的六边形相邻的空六边形上放置骨牌. 如果一个玩家无法继续放置骨牌, 或者放置骨牌后会出现呈菱形分布的四个相邻六边形均被放置骨牌的情况(如图所示, 但方向可以不同), 则判该玩家输. 确定是否某个玩家有获胜策略; 如果有, 赢家是谁?10. 将整数1, 2, …, n写在n张卡片上, 每张上写一个不同的数. 首先, 由玩家1取走一张卡片. 接下来, 玩家2取走写有连续正整数的两张卡片. 然后, 再由玩家1取走写有连续正整数的三张卡片. 最后, 由玩家2取走写有连续正整数的四张卡片. 求最小的n, 使得玩家2能确保完成他的两次取卡片的操作. 11. 给定一圆w及圆上依A, B, C, D顺序排列的四点, 且AD为圆w的直径. 假设AB = BC = a , CD = c , 其中a 和c 为互质正整数. 证明: 如果圆w 的直径长d 也是正整数, 则d 及2d 中必有一个完全平方数.12. 锐角△ABC 的高BB 1, CC 1相交于点H . 点B 2, C 2分别位于线段BH , CH 上, 且BB 2 = B 1H , CC 2 = C 1H . △B 2HC 2的外接圆与△ABC 的外接圆相交于点D 和E . 证明: △DEH 为直角三角形.13. 在△ABC 中, ∠A 的内角平分线与直线BC 交于点D , 与△ABC 的外接圆交于点E . 设K , L , M , N 分别为线段AB , BD , CD , AC 的中点. 点P , Q 分别为△EKL , △EMN 的外心. 证明: ∠PEQ = ∠BAC .14. 设四边形ABCD 有内切圆w . 令圆w 与AC 的交点中较靠近点A 的那个为E . 设F 为E 关于圆w 的对径点. 经点F 作圆w 的切线, 分别交直线AB , BC 于A 1, C 1, 并与直线AD , CD 分别交于A 2, C 2. 证明: A 1C 1 = A 2C 2.15. 考虑平面内相离的两个圆. 分别选取两个圆的直径A 1B 1和A 2B 2, 使得线段A 1A 2与B 1B 2相交于点C . 设A 1A 2, B 1B 2的中点分别为A , B . 证明: 不管如何选取直径A 1B 1和A 2B 2, △ABC 的垂心总位于一条直线上.16. 设p 为奇质数. 求所有的正整数n , 使得np n -2为正整数.17. 证明: 对所有满足q p >11的正整数p , q , 不等式pqq p 2111>-成立. 18. 设整数n ≥ 3满足4n + 1为质数. 证明: 4n + 1整除12-n n .19. 设无限正整数集合B 满足以下条件: 对任意的a , b ∈ B 且a > b , 有),gcd(b a b a - ∈ B . 证明: B 是由所有正整数构成的集合.20. 求所有的正整数(a , b , c ), 使得ba c a cbc b a 444)()()(+++++为整数, 且a + b + c 为质数.2018年第10届Benelux 数学奥林匹克试题比赛时间: 2018年4月28日1. a) 设x , y 为正实数. 求⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛++⎪⎪⎭⎫ ⎝⎛-+⎪⎪⎭⎫ ⎝⎛+201811201811x y x y y x y x 的最小值. b) 设x , y 为正实数. 求⎪⎭⎫ ⎝⎛++⎪⎭⎫ ⎝⎛++⎪⎪⎭⎫ ⎝⎛++⎪⎪⎭⎫ ⎝⎛+201811201811x y x y y x y x 的最小值. 2. 在七星岛上, 共使用4种不同的硬币和3种不同的纸币, 它们的面额分别为7个不同的正整数, 且最小额纸币的面额大于4种不同硬币的面额之和. 一位游客恰好有不同面额的硬币各1枚及不同面额的纸币各1张, 但是这些钱的总额不够支付他想购买的一本关于钱币学的书. 幸运的是, 爱好数学的书店老板同意将此书按这位游客所提出的价格卖给他, 但前提是该游客可以用超过一种方式支付此价格.(游客可以用超过一种方式支付某价格, 指的是在由他的硬币与纸币构成的集合中, 存在两个不同的子集, 每个子集里钱的面额之和均等于该价格.) a) 证明: 如果每张纸币的面额均小于49, 则该游客能够购买到这本书. b) 证明: 如果最大额纸币的面额等于49, 则该游客有可能空手而归.3. 设H 为三角形ABC 的垂心, D , E , F 分别为AB , AC , AH 的中点. 点B , C 关于点F 的对称点分别为P , Q .a) 证明: 直线PE 和QD 的交点位于三角形ABC 的外接圆上.b) 证明: 直线PD 和QE 的交点位于AH 上.4. 我们称一个恰有s 个正因数1 = d 1 < d 2 < … < d s = n 的正整数n > 2为"好"的, 如果存在整数2 ≤ k ≤ s , 满足d k > 1 + d 1 + … + d k –1.如果一个整数n > 2不是"好"的, 则称之为"坏"的.a) 证明: 存在无穷多个"坏"的整数.b) 证明: 在均大于2的任意7个连续整数中, 至少有4个整数为"好"的. c) 证明: 存在无穷多个由连续7个整数构成的序列, 其中每个序列里的数都是"好"的.2018年巴尔干地区数学奥林匹克试题比赛时间: 2018年5月9日1. 凸四边形ABCD内接于圆k, 其中AB > CD, 且AB不平行于CD. 点M为对角线AC与BD的交点, 自M作AB的垂线, 与AB相交于点E.如果EM平分 CED, 证明: AB为圆k的一条直径.2. 设q为正有理数. 两只蚂蚁最初均位于平面上的同一点X处. 在第n分钟(n = 1, 2, …), 每只蚂蚁各自在北, 东, 南, 西四个方向中选择一个方向, 并沿此方向移动q n米. 经过整数个分钟后, 它们再次位于平面上的同一点处(不一定是点X), 但是在此时间段内它们的移动路径并不完全相同.求q所有可能的取值.3. Alice和Bob一起玩如下的移硬币游戏. 他们从两堆均非空的硬币开始, 首先由Alice开始, 轮流进行以下操作: 该轮玩家选择数目为偶数的一堆硬币, 将该堆的一半硬币移到另一堆里. 如果某位玩家无法进行上述操作, 则游戏结束, 并判对方获胜.求所有的正整数对(a, b), 使得如果最初两堆分别有a和b枚硬币, 则Bob有获胜策略.4. 求所有的素数对(p, q), 使得3p q–1 + 1整除11p + 17p.2018年巴尔干地区初中数学奥林匹克(JBMO)试题比赛时间: 2018年6月21日1. 求满足方程m 5 – n 5 = 16mn 的所有整数(m , n ).2. 设有n 个三位正整数同时满足以下条件:i) 每个数均不含数字0;ii) 每个数的数字和为9;iii) 任意两个数的个位数均不同;iv) 任意两个数的十位数均不同;v) 任意两个数的百位数均不同.求n 的最大可能值.3. 设k > 1为正整数, n > 2018为正奇数. 不全相等的非零有理数x 1, x 2, …, x n 满足如下关系式:11433221...x k x x k x x k x x k x x k x n n n +=+==+=+=+-. i) 求乘积n x x x ⋅⋅⋅...21(用关于k 和n 的函数表示).ii) 求最小的k , 使得存在满足所给条件的n , x 1, x 2, …, x n .4. 设A', B', C'分别为△ABC 的顶点A , B , C 关于对边的对称点. △ABB'的外接圆交△ACC'的外接圆于A 1 (A 1 ≠ A ). 类似定义点B 1和C 1. 证明: AA 1, BB 1, CC 1三线共点.2018年高加索地区数学奥林匹克试题初级组第1天(2018年3月16日)1. 设a, b, c为不全为0的实数. 证明: a + b + c = 0的充分必要条件为a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.2. 在8 ⨯ 8国际象棋棋盘上, 放置了n > 6只马, 使得对其中任意6只马, 均存在2只马可以互相攻击. 求n的最大可能值.3. 正整数a, b, c满足条件: a b整除b c, a c整除c b. 证明: a2整除bc.4. 我们定义四边形的重心为连接对边中点的两条直线的交点. 设六边形ABCDEF内接于以O为圆心的圆Ω, 且AB = DE, BC = EF. 令X, Y, Z分别为四边形ABDE, BCEF, CDFA的重心. 证明: O为△XYZ的垂心.第2天(2017年3月17日)5. Munсhausen男爵发现了如下"定理": 对任意正整数a和b, 总存在正整数n, 使得an为完全平方数, 而bn为完全立方数. 请确定该男爵的"定理"是否正确.6. 凸四边形ABCD中, ∠BCD = 90o, E为AB的中点. 证明: 2EC≤AD + BD.7. 给定正整数n > 1. 考虑一个n⨯n棋盘. 最初棋盘上没有玻璃球, 按照以下规则逐个地往棋格里放入玻璃球: 如果一个空棋格与至少2个空棋格相邻(指有一公共边), 则该棋格内可以放入一个玻璃球. 问在此规则下, 棋盘内最多可以放入多少个玻璃球?8. 设a, b, c为一个三角形的三边长. 证明:2)()()()(2cbacaacbccbabba++≥+++++.高级组第1天(2018年3月16日)1. 给定一个四面体. 是否能够将10个连续正整数分别放置在该四面体的四个顶点及六条棱的中点上, 使得每条棱中点上的数等于该棱两端点上的数的算术平均值?2. 设I为锐角△ABC的内心. 点P, Q, R分别在边AB, BC, CA上, 满足AP = AR, BP = BQ, ∠PIQ = ∠BAC. 证明: QR⊥AC.3. 我们称2n个正整数的一个匹配(即分成n对)为"非平方"的, 如果每一对中的2个数之积均不是完全平方数. 证明: 如果存在一个"非平方"匹配, 则至少存在n!个"非平方"匹配.4. Morteza在n⨯n棋盘的每一个棋格内放置一个[0, 1] → [0, 1]的函数(即定义域为[0, 1], 值域为[0, 1]的函数). Pavel计划在棋盘每一行的左边及每一列的下边分别放置一个[0, 1] → [0, 1]的函数(共放置2n个函数), 使得棋盘的每一格均满足以下条件:如果h为该棋格内的函数, f为该棋格所在列下边的函数, g为该棋格所在行左边的函数, 则h(x) = f(g(x))对所有的x∈ [0, 1]成立.证明: Pavel总是可以实现他的计划.第2天(2018年3月17日)5. Munсhausen男爵发现了如下"定理": 对任意正整数a和b, 总存在正整数n, 使得an为完全立方数, 而bn为完全五次方数. 请确定该男爵的"定理"是否正确.6. 在坐标平面内, 两个二次多项式的图像G1, G2的交点为A, B. 设O为G1的顶点. 直线OA, OB分别与G2再次相交于点C, D. 证明: CD平行于x轴.7. 锐角△ABC中, 经过顶点A, B, C的高分别交对边于A1, B1, C1, 并分别交△ABC的外接圆于A2, B2, C2. 直线A1C1分别交△AC1C2, △CA1A2的外接圆于点P, Q (P≠C1, Q≠A1). 证明: △PQB1的外接圆与AC相切.8. 考虑一个8 ⨯8棋盘. 最初棋盘上没有玻璃球, 按照以下规则逐个地往棋格里放入玻璃球: 如果一个空棋格与至少3个空棋格相邻(指有一公共边), 则该棋格内可以放入一个玻璃球. 问在此规则下, 棋盘内最多可以放入多少个玻璃球?2018年中美洲及加勒比地区数学奥林匹克试题第1天1. 在2018张卡片上分别标记数1, 2, …, 2018, 每张卡片上标记一个数. 卡片上的数始终可见. Tito 和Pepe 一起玩游戏. 由Tito 首先开始, 他们轮流选取一张卡片, 已选过的卡片不能再选, 直到所有卡片均被选取. 然后, 每个人计算自己选取卡片上所标记数的和, 判和为偶数者获胜. 确定谁有获胜策略, 并描述该策略.2. △ABC 的外接圆为w , 外心为O . 设T 为C 关于点O 的对称点, T'为T 关于直线AB 的对称点. 直线BT'与圆w 再次相交于点R . 过O 作CT 的垂线, 交直线AC 于点L . 直线TR 与AC 相交于点N . 证明: CN = 2AL .3. 设x , y 为实数, 使得x – y , x 2 – y 2, x 3 – y 3均为素数. 证明: x – y = 3.第2天4. 求所有的3元正整数组(p , q , r ), 其中p , q 为素数, 满足215222=--p q r . 5. 设1 < n < 2018为正整数. 对i = 1, 2, …, n , 定义多项式S i (x ) = x 2 – 2018x + l i , 其中l 1, l 2, …, l n 为互不相同的正整数. 证明: 如果多项式S 1(x ) + S 2(x ) + … + S n (x )至少有一个整数根, 则l 1, l 2, …, l n 中至少有一个数不小于2018.6. 2018对夫妻参加在哈瓦那举行的一场舞会. 舞会中, 将一个圆周上2018个互异的点分别标记为0, 1, …, 2017, 每一对夫妻位于一个点上(不同夫妻位于不同的点). 对整数i ≥ 1, 令s i ≡ i (mod 2018), r i ≡ 2i (mod 2018). 舞会从第0分钟开始, 在第i 分钟, 位于点s i 的夫妻(如果存在的话)移至点r i , 而位于点r i 的夫妻(如果存在的话)则退场, 舞会由剩下的夫妻继续进行. 在20182分钟后, 舞会结束. 请确定舞会结束时还剩下多少对夫妻在场上注: 如果r i = s i , 则位于点s i 的夫妻留在原位, 不退场.2018年Cono Sur 数学奥林匹克试题第1天1. 设ABCD 为凸四边形, 点R , S 分别位于边DC , AB 上, 且满足AD = RC , BC = SA . 点P , Q , M 分别是RD , BS , CA 的中点. 设∠MPC + ∠MQA = 90o . 证明: ABCD 为圆内接四边形.2. 证明: 每一个正整数都可以表示成3, 4和7的若干幂的和, 其中同一个数不允许重复出现相同的幂次.例如: 2 = 70 + 70和22 = 32 + 32 + 41就是不允许出现的表示方式; 但是, 2= 30 + 70和22 = 32 + 30 + 41 + 40 + 71则是允许出现的表示方式.3. 考虑乘积P n = 1!⋅2!⋅3!⋅…⋅n !.i) 求所有的正整数m , 使得!2020m P 为完全平方数. ii) 证明: 存在无穷多个正整数n , 使得至少对2个正整数m ,!m P n 为完全平方数.第2天4. 对每一个正整整n ≥ 4, 考虑{1, 2, …, n }的m 个子集A 1, A 2, …, A m , 使得A 1恰含1个元素, A 2恰含2个元素, …, A m 恰含m 个元素; 且这些子集中没有一个子集是另一个子集的子集. 求m 可能取的最大值.5. 锐角△ABC 中, ∠BAC = 60o , I 为内心, O 为外心. 设H 为O 在△BOC 外接圆上的对径点. 证明: IH = BI + IC .6. 称正整数序列a 1, a 2, …, a n 为"好"的, 如果对所有的正整数n , 以下两个条件同时成立:i) n n a a a a a ...321!=.ii) a n 为某个正整数的n 次幂.求所有"好"的序列.2018年捷克-波兰-斯洛伐克联合数学竞赛试题第1天 (2018年6月25日)1. 求所有的函数f : R → R , 使得对所有的实数x , y , 成立等式:)()()()()(2y x xf x yf y f x f xy x f +++=+.2. 设△ABC 为锐角非等边三角形. 点D , E 分别在边AB , AC 上, 满足BD = CE . 设O 1, O 2分别为△ABE , △ACD 的外心. 证明: △ABC , △ADE 及△AO 1O 2的外接圆有一个异于点A 的公共点.3. 2018个玩家围桌而坐. 在游戏开始时, 我们将一摞共K 张牌任意地分发给玩家(有些玩家可能没有得到牌). 定义一轮操作如下: 如果一名玩家的左右邻居的牌数均非零, 则选他为这一轮的幸运玩家(如果有多名玩家符合条件, 则由我们任意选取一个), 让他从左右邻居那儿各拿一张牌给自己. 如果找不出这样的玩家, 则游戏结束. 求K 的最大可能值, 使得无论我们如何发牌及如何挑选幸运玩家, 该游戏总能在有限轮次后结束.第2天 (2018年6月26日)4. 设锐角△ABC 的周长为2s . 分别以A , B , C 为圆心, 作3个两两之间无公共内点的圆(不包括边界). 证明: 存在一个半径为s 的圆, 将上述三个圆同时覆盖.5. 在一个2 ⨯ 3矩形的内部, 有一个长度为36的折线(允许折线自交). 证明: 存在一条平行于矩形两边的直线, 与矩形的另两条边的内部相交, 且与折线的交点数少于10个.6. 我们称正整数n 为"奇妙"的, 如果存在正有理数a 和b , 使得bb a a n 11+++=. a) 证明: 存在无穷多个质数p , 使得p 的倍数均不是"奇妙"的.b) 证明: 存在无穷多个质数p , 使得p 的某个倍数是"奇妙"的.2018年捷克和斯洛伐克数学奥林匹克试题第1天1. 在一群人中, 存在一些两人对, 这两人相互为朋友. 对正整数k ≥ 3, 我们称该群人为"k -佳"的, 如果该群人中每k 个人(不计顺序)组成的一组人都可以围桌而坐, 使得每个人的邻座均为其朋友. 证明: 如果一群人是"6-佳"的, 则该群人必是"7-佳"的.2. 设x , y , z 为实数, 且数|2|12yz x +, |2|12zx y +, |2|12xy z +构成一非退化三角形的三边长. 求xy + yz + zx 的所有可能值.3. 三角形ABC 中, 点D 为∠A 内角平分线与边BC 的交点. 点E , F 分别是三角形ABD , ACD 的外心. 设三角形AEF 的外心位于直线BC 上. 求∠BAC 的所有可能值.第2天4. 设整数a , b , c 为某一三角形的三边长, 满足gcd(a , b , c ) = 1, 且c b a c b a -+-+222, a c b a c b -+-+222, ba cb ac -+-+222的值也均为整数. 证明: (a + b – c )(b + c – a )(c + a – b )和2(a + b – c )(b + c – a )(c + a – b )中, 至少有一个为完全平方数.5. 设ABCD 为等腰梯形, AB 为较长的底边. 令I △ABC 的内心, J 为△ACD 对应于顶点C 的旁心. 证明: IJ // AB .6. 求具有下述性质的最小正整数n : 无论用三种颜色对整数1, 2, …, n 如何染色(每个数染三种颜色之一), 从中总能够找到互异的两个数a , b , 它们染有相同的颜色, 并且|a – b |为完全平方数.2018年多瑙河地区数学奥林匹克试题比赛时间: 2018年10月27日初级组1. 求所有同时满足以下条件的正整数对(n , m ):i) n 是合数;ii) 如果d 1, d 2, …, d k (k ∈ Z +)为n 的所有真因数, 则d 1 + 1, d 2 + 1, …, d k + 1为m 的所有真因数.2. 设在△ABC 内部存在一点D , 使得∠DAC = ∠DCA = 30o , ∠DBA = 60o . E 为BC 的中点. 点F 位于线段AC 上, 且AF = 2FC . 证明: DE ⊥EF .3. 求所有具有下述性质的正整数n : 存在正整数k ≥ 2及正有理数a 1, a 2, …, a k , 使得a 1 + a 2 + … + a k = a 1a 2…a k = n 成立.4. 设M 为由全体正奇数构成的集合. 对每一个正整数n , 定义A (n )为满足元素和为n 的M 的子集的个数. 例如, A (9) = 2, 因为恰有M 的两个子集满足其元素和为9, 分别是{9}, {1, 3, 5}.a) 证明: 对每一个正整数n ≥ 2, A (n ) ≤ A (n + 1).b) 求满足A (n ) = A (n + 1)的所有正整数n ≥ 2.高级组1. 假设我们有一个由n 颗珍珠构成的项链. 在每一颗珍珠上标记一个整数, 使得所有珍珠上的数之和为n – 1. 证明: 我们可以将此项链从某处切断, 形成一根所标记整数依次为x 1, x 2, …, x n 的珍珠链, 满足11-≤∑=k x ki i 对所有k = 1, 2, …,n 成立.2. 证明: 存在无穷多组正整数(m , n )同时满足以下条件: m 整除n 2 + 1, n 整除m 2 + 1.3. 设△ABC 为非等腰锐角三角形. ∠A 的内角平分线与△ABC 的外接圆再次相交于点D . 设O 为△ABC 的外心. ∠AOB , ∠AOC 的角平分线分别与以AD 为直径的圆γ相交于点P , Q . 直线PQ 与AD 的垂直平分线相交于点R . 证明: AR // BC .4. 设n≥ 3为奇数. 将n⨯n方格纸的每一单元格都染为红色或蓝色之一. 称两个单元格为"相邻"的, 如果它们同色且至少有一个公共顶点. 称两个单元格a, b 为"连通"的, 如果存在若干个单元格c1, c2, …, c k, 满足c1 = a, c k = b, 且对每一个i = 1, 2, …, k – 1, c i与c i+1均相邻; 否则, 就称a, b为"不连通"的. (例如, 两个染色不同的单元格就是不连通的). 求最大的正整数M, 使得存在一种染色方案, 其中有M个两两不连通的单元格.2018年欧洲女子数学奥林匹克试题第1天 (2018年4月11日)1. 三角形ABC 中, CA = CB , ∠ACB = 120o , M 为AB 的中点. 设P 为三角形ABC 外接圆上一动点, Q 为线段CP 上一点, 且满足QP = 2QC . 已知经过点P 且垂直于AB 的直线与直线MQ 相交于唯一的一点N . 证明: 对点P 的所有可能位置, 点N 均位于一个固定圆上.2. 考虑集合A = ⎭⎬⎫⎩⎨⎧=+,...3,2,1:11k k . a) 证明: 每一个整数x ≥ 2均可以表示成A 中至少1个元素之积(各元素不必互异).b) 对每一个整数x ≥ 2, 设f (x )为最小的整数, 使得x 可以表示成A 中f (x )个元素之积(各元素不必互异). 证明: 存在无穷多组整数对(x , y ), 满足x ≥ 2, y ≥ 2, 且f (xy ) < f (x ) + f (y ).(如果x 1 ≠ x 2或y 1 ≠ y 2, 则认为整数对(x 1, y 1)与(x 2, y 2)是不同的.)3. 设某一届EGMO 的n 个参赛者为C 1, C 2, …, C n . 在比赛结束后, 所有参赛者在餐厅门口按照以下规则排成一个队列候餐:i) 由组委会确定各位参赛者在队列中的最初位置.ii) 每一分钟, 组委会选择一个整数i , 其中1 ≤ i ≤ n .-- 如果在参赛者C i 前面至少有i 名其他参赛者, 她将付给组委会1欧元, 并在队列中向前移动i 个位置.-- 如果在参赛者C i 前面的其他参赛者少于i 名, 则餐厅门打开, 候餐结束. a) 证明: 不管组委会如何选择, 上述候餐过程总会结束.b) 对每一个n , 求在经过精巧地选择最初位置及移动顺序下, 组委会能够得到的欧元数的最大值.第2天 (2018年4月12日)4. 定义多米诺骨牌指的是1 ⨯ 2或2 ⨯ 1的骨牌. 设n ≥ 3为整数. 在n ⨯ n 棋盘内放置若干多米诺骨牌, 使得每一个多米诺骨牌恰好覆盖两个棋格, 且多米诺骨。

USA AMC 10 20001In the year , the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product . What's the largest possible value of the sum ?SolutionThe sum is the highest if two factors are the lowest.So, and .2Solution.3Each day, Jenny ate of the jellybeans that were in her jar at the beginning of the day. At the end of the second day, remained. How many jellybeans were in the jar originally?Solution4Chandra pays an online service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixxed monthly fee?SolutionLet be the fixed fee, and be the amount she pays for the minutes she used in the first month.We want the fixed fee, which is5Points and are the midpoints of sides and of . As moves along a line that is parallel to side , how many of the four quantities listed below change?(a) the length of the segment(b) the perimeter of(c) the area of(d) the area of trapezoidSolution(a) Clearly does not change, and , so doesn't change either.(b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base and its corresponding height remain the same.(d) The bases and do not change, and neither does the height, so the trapezoid remains the same.Only quantity changes, so the correct answer is .6The Fibonacci Sequence starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence?SolutionThe pattern of the units digits areIn order of appearance:.is the last.7In rectangle , , is on , and and trisect . What is the perimeter of ?Solution.Since is trisected, .Thus,.Adding, .8At Olympic High School, of the freshmen and of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?There are five times as many sophomores as freshmen.There are twice as many sophomores as freshmen.There are as many freshmen as sophomores.There are twice as many freshmen as sophomores.There are five times as many freshmen as sophomores.SolutionLet be the number of freshman and be the number of sophomores.There are twice as many freshmen as sophomores.9If , where , thenSolution, so ...10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ?SolutionFrom the triangle inequality, and . The smallest positive number not possible is , which is .11Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?SolutionTwo prime numbers between and are both odd.Thus, we can discard the even choices.Both and are even, so one more than is a multiple of four.is the only possible choice.satisfy this, .12Figures , , , and consist of , , , and nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?SolutionSolution 1We have a recursion:.I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we add.Solution 2We can divide up figure to get the sum of the sum of the firstodd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is , so for figure , there are unit squares. We plug in to get , which is choice13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?SolutionIn each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were , , , , and . What was the last score Mrs. Walter entered? SolutionThe sum of the first scores must be even, so we must choose evens or the odds to be the first two scores.Let us look at the numbers in mod .If we choose the two odds, the next number must be a multiple of , of which there is none.Similarly, if we choose or , the next number must be a multiple of , of which there is none.So we choose first.The next number must be 1 in mod 3, of which only remains.The sum of the first three scores is . This is equivalent to in mod .Thus, we need to choose one number that is in mod . is the only one that works.Thus, is the last score entered.15Two non-zero real numbers, and , satisfy . Which of the following is a possible value of ?SolutionSubstituting , we get16The diagram shows lattice points, each one unit from its nearest neighbors. Segment meets segment at . Find the length of segment .SolutionSolution 1Let be the line containing and and let be the line containing and . If we set the bottom left point at , then , , , and .The line is given by the equation . The -intercept is , so . We are given two points on , hence we cancompute the slope, to be , so is the lineSimilarly, is given by . The slope in this case is , so . Plugging in the point gives us , so is the line .At , the intersection point, both of the equations must be true, soWe have the coordinates of and , so we can use the distance formula here:which is answer choiceSolution 2Draw the perpendiculars from and to , respectively. As it turns out, . Let be the point on for which ., and , so by AA similarity,By the Pythagorean Theorem, we have ,, and . Let , so , thenThis is answer choiceAlso, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.17Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?SolutionConsider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by cents.This implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is18Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?SolutionThe area she sees looks at follows:The part inside the walk has area . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area . The four arcs together form a circle with radius . Therefore the total area she can see is, which rounded to the nearest integer is .19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is SolutionLet the square have area , then it follows that the altitude of one of the triangles is . The area of the other triangle is .By similar triangles, we haveThis is choice(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle times changes each of the areas times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)20Let , , and be nonnegative integers such that . What is the maximum value of ? SolutionThe trick is to realize that the sum is similar to the product .If we multiply , we get.We know that , therefore.Therefore the maximum value of is equal to the maximum value of . Now we will find this maximum.Suppose that some two of , , and differ by at least . Then this triple is surely not optimal.Proof: WLOG let . We can then increase the value ofby changing and .Therefore the maximum is achieved in the cases where is a rotation of . The value of in this case is . And thus the maximum of is.21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.SolutionWe interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as , , and .We got the following information:▪If is an , then is an .▪There is some that is a and at the same time an .We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are s, but only Johnny is a "meets both conditions, but the first statement is false.We CAN conclude that the second statement is true. We know that there is some that is a and at the same time an . Pick one such and call it Bobby. Additionally, we know that if is an , then is an. Bobby is an , therefore Bobby is an . And this is enough to prove the second statement -- Bobby is an that is also a .We CAN NOT conclude that the third statement is true. For example, consider the situation when , and are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.Therefore the answer is .22One morning each member of Angela's family drank an -ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?SolutionThe exact value "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.Let be the total number of ounces of milk drank by the family and the total number of ounces of coffee. Thus the whole family drank a total of ounces of fluids.Let be the number of family members. Then each family member drank ounces of fluids.We know that Angela drank ounces of fluids.As Angela is a family member, we have .Multiply both sides by to get .If , we have .If , we have .Therefore the only remaining option is .23When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ? SolutionAs occurs three times and each of the three other values just once, regardless of what we choose the mode will always be .The sum of all numbers is , therefore the mean is .The six known values, in sorted order, are . From this sequence we conclude: If , the median will be . If , the median will be . Finally, if , the median will be .We will now examine each of these three cases separately.In the case , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.In the case we have , because. Therefore our three values inorder are . We want this to be an arithmetic progression. From the first two terms the difference must be . Therefore thethird term must be .Solving we get the only solution for this case: . The case remains. Once again, we have ,therefore the order is . The only solution is when , i. e., .The sum of all solutions is therefore .24Let be a function for which . Find the sum of all values of for which .SolutionIn the definition of , let . We get: . As we have , we must have , in other words .One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots ofis . In our case this is .(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that and .)25In year , the day of the year is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the of year occur?SolutionClearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.Let be the day of year , the day of year and the day of year .If year is not a leap year, the day will bedays after . As , that would be a Monday.Therefore year must be a leap year. (Then is days after .) As there can not be two leap years after each other, is not a leap year. Therefore day is days after . We have . Therefore is weekdays before , i.e., is a.(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for we have=Tuesday, October 26th 2004, =Tuesday, July 19th, 2005 and =Thursday, April 10th 2003.)。

2020AMC10B（美国数学竞赛）真题加详解2020 AMC 10B Solution Problem1What is the value ofSolutionWe know that when we subtract negative numbers, .The equation becomesProblem2Carl has cubes each having side length , and Kate has cubes each having side length . What is the total volume of these cubes?SolutionA cube with side length has volume , so of these will have a total volume of .A cube with side length has volume , so of these will have a total volume of .~quacker88Problem 3The ratio of to is , the ratio of to is , and the ratioof to is . What is the ratio of toSolution 1WLOG, let and .Since the ratio of to is , we can substitute in the value of toget .The ratio of to is , so .The ratio of to is then so our answeris ~quacker88Solution 2We need to somehow link all three of the ratios together. We can start by connecting the last two ratios together by multiplying the last ratio by two., and since , we can link themtogether to get .Finally, since , we can link this again to get: ,so ~quacker88Problem4The acute angles of a right triangle are and , where andboth and are prime numbers. What is the least possible value of ?SolutionSince the three angles of a triangle add up to and one of the anglesis because it's a right triangle, .The greatest prime number less than is . If ,then , which is not prime.The next greatest prime number less than is . If ,then , which IS prime, so we have our answer ~quacker88 Solution 2Looking at the answer choices, only and are coprime to . Testing , the smaller angle, makes the other angle which is prime, therefore our answerisProblem5How many distinguishable arrangements are there of 1 brown tile, 1 purple tile, 2 green tiles, and 3 yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)SolutionLet's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.There are ways to order objects. However, since there's ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and ways to order the green tiles, we have to divide out these possibilities.~quacker88SolutionWe can repeat chooses extensively to find the answer. Thereare choose ways to arrange the brown tiles which is . Then from the remaining tiles there are choose ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answerofProblem6Driving along a highway, Megan noticed that her odometershowed (miles). This number is a palindrome-it reads the same forward and backward. Then hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this -hour period?SolutionIn order to get the smallest palindrome greater than , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.So we raise to the next largest value, , but obviously, that's not how place value works, so we're in the s now. To keep this a palindrome, our number is now .So Megan drove miles. Since this happened over hours, she drove at mph. ~quacker88 Problem7How many positive even multiples of less than are perfect squares?SolutionAny even multiple of is a multiple of , so we need to find multiples of that are perfect squares and less than . Any solution that we want will be in theform , where is a positive integer. The smallest possible value isat , and the largest is at (where the expression equals ). Therefore, there are a total of possible numbers.-PCChess Problem8 Points and lie in a plane with . How many locations forpoint in this plane are there such that the triangle with vertices , ,and is a right triangle with area square units?Solution 1There are options here:1. is the right angle.It's clear that there are points that fit this, one that's directly to the rightof and one that's directly to the left. We don't need to find the length, we just need to know that it is possible, which it is.2. is the right angle.Using the exact same reasoning, there are also solutions for this one.3. The new point is the right angle.(Diagram temporarily removed due to asymptote error)The diagram looks something like this. We know that the altitude tobase must be since the area is . From here, we must see if there are valid triangles that satisfy the necessary requirements. First of all, because of the area.Next, from the Pythagorean Theorem.From here, we must look to see if there are valid solutions. There are multiple ways to do this:We know that the minimum value of iswhen . In this case, the equationbecomes , which is LESSthan . . The equationbecomes , which is obviously greater than . We canconclude that there are values for and in between that satisfy the Pythagorean Theorem.And since , the triangle is not isoceles, meaning we could reflectit over and/or the line perpendicular to for a total of triangles this case.Solution 2Note that line segment can either be the shorter leg, longer leg or thehypotenuse. If it is the shorter leg, there are two possible points for that cansatisfy the requirements - that being above or below . As such, thereare ways for this case. Similarly, one can find that there are also ways for point to lie if is the longer leg. If it is a hypotenuse, then thereare possible points because the arrangement of the shorter and longer legs can be switched, and can be either above or below the line segment. Therefore, the answer is .Problem9How many ordered pairs of integers satisfy theequationSolutionRearranging the terms and and completing the square for yields theresult . Then, notice that can onlybe , and because any value of that is greater than 1 will causethe term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , , , and gives a totalof ordered pairs.Solution 2Bringing all of the terms to the LHS, we see a quadraticequation in terms of . Applying the quadratic formula, weget In order for to be real, which it must be given the stipulation that we are seekingintegral answers, we know that the discriminant, must benonnegative. Therefore, Here, we see that we must split the inequality into a compound, resultingin .The only integers that satisfy this are . Plugging thesevalues back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only 1 solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .Thus, the answer is .~TiblisSolution 3, x firstSet it up as a quadratic in terms of y:Then the discriminant is This will clearly only yield real solutionswhen , because it is always positive. Then . Checking each one: and are the same when raised to the 2020th power:This has only has solutions , so are solutions. Next, if :Which has 2 solutions, so andThese are the only 4 solutions, soSolution 4, y firstMove the term to the other side toget . Because for all , then . If or , the right side is and therefore . When , the right side become , therefore . Our solutions are , , , . There are solutions, so the answer is - wwt7535Problem 10A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubicinches?SolutionNotice that when the cone is created, the radius of the circle will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.We can calculate that the intact circumference of the circle is . Since that is also equal to the circumference of the cone, the radius of the cone is . We also have that the slant height of the cone is . Therefore, we use the Pythagorean Theorem to calculate that the height of the coneis . The volume of the coneis -PCChessSolution 2 (Last Resort/Cheap)Using a ruler, measure a circle of radius 4 and cut out the circle and then the quarter missing. Then, fold it into a cone and measure the diameter to be 6cm . You can form a right triangle with sides 3, 4, and then through the Pythagorean theorem the height is found tobe . The volume of a cone is . Plugging in we findProblem11Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?SolutionWe don't care about which books Harold selects. We just care that Bettypicks books from Harold's list and that aren't on Harold's list.The total amount of combinations of books that Betty can selectis .There are ways for Betty to choose of the books that are on Harold's list.From the remaining books that aren't on Harold's list, thereare ways to choose of them.~quacker88Problem12The decimal representation of consists of a string of zeros after the decimal point, followed by a and then several more digits. How many zeros are in that initial string of zeros after the decimal point?Solution 1Now we do some estimation. Notice that , which meansthat is a little more than . Multiplying itwith , we get that the denominator is about . Notice that whenwe divide by an digit number, there are zeros before the first nonzero digit. This means that when we divide by the digit integer , there are zeros in the initial string after the decimal point. -PCChessSolution 2First rewrite as . Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to findthe number of digits in .and memming (alternatively use the factthat ),digits.Our answer is .Solution 3 (Brute Force)Just as in Solution we rewrite as We thencalculate entirely by hand, first doing then multiplying that product by itself, resulting in Because this is digits,after dividing this number by fourteen times, the decimal point is beforethe Dividing the number again by twenty-six more times allows a stringof zeroes to be formed. -OreoChocolateSolution 4 (Smarter Brute Force)Just as in Solutions and we rewrite as We can then look at the number of digits in powersof . , , , , ,, and so on. We notice after a few iterations that every power of five with an exponent of , the number of digits doesn't increase. This means should have digits since thereare numbers which are from to , or digits total. This means our expression can be written as , where is in therange . Canceling gives , or zeroes before the since the number should start on where the one would be in . ~aop2014 Solution 5 (Logarithms)Problem13Andy the Ant lives on a coordinate plane and is currently at facingeast (that is, in the positive -direction). Andy moves unit and thenturns degrees left. From there, Andy moves units (north) and thenturns degrees left. He then moves units (west) and againturns degrees left. Andy continues his progress, increasing his distance each time by unit and always turning left. What is the location of the point at which Andy makes the th leftturn?Solution 1You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you theanswer of ~happykeeperProblem14As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?Solution 1Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on,and , as a regular hexagon has angles of 120,and is half of any angle in this hexagon. Now, using the sinelaw, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagonis . Since the area of an equilateral triangle can be foundwith the formula , where is the side length of the equilateral triangle,the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixthof a circle with radius 1 is . In each sixth of the hexagon, thereare two equilateral triangles colored white, each with an area of , and onesixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixthof the hexagon is , which equals , and the total areacolored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white,the area colored gray is , whichequals .Solution 2First, subdivide the hexagon into 24 equilateral triangles with side length1:Now note that the entire shadedregion is just 6 times this part:The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: The arc that is not included has an area of:Hence, the area ofthe shaded region in that section is For a final areaof:Problem15Steve wrote the digits , , , , and in order repeatedly from left to right, forming a list of digits, beginning He thenerased every third digit from his list (that is, the rd, th, th, digits from the left), then erased every fourth digit from the resulting list (that is, the th, th, th, digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions ?Solution 1After erasing every third digit, the list becomes repeated. After erasing every fourth digit from this list, the listbecomes repeated. Finally, after erasing every fifth digit from this list, the list becomes repeated. Since this list repeats every digits andsince are respectively in we have that the th, th, and st digits are the rd, th, and thdigits respectively. It follows that the answer is~dolphin7Problem16Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in theinterval . Thereafter, the player whose turn it is chooses a real numberthat is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?SolutionNotice that to use the optimal strategy to win the game, Bela must select themiddle number in the range and then mirror whatever number Jennselects. Therefore, if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so theanswer is .Solution 2 (Guessing)First of all, realize that the value of should have no effect on the strategy at all. This is because they can choose real numbers, not integers, so even if is odd, for example, they can still go halfway. Similarly, there is no reason the strategy would change when .So we are left with (A) and (B). From here it is best to try out random numbers and try to find the strategy that will let Bela win, but if you can't find it, realize thatit is more likely the answer is since Bela has the first move and thus has more control.Problem17There are people standing equally spaced around a circle. Each person knows exactly of the other people: the people standing next to her or him, as well as the person directly across the circle. How many ways are there forthe people to split up into pairs so that the members of each pair know each other?SolutionLet us use casework on the number of diagonals.Case 1: diagonals There are ways: either pairs with , pairs with , and so on or pairs with , pairs with , etc.Case 2: diagonal There are possible diagonals to draw (everyone else pairs with the person next to them.Note that there cannot be 2 diagonals.Case 3: diagonalsNote that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.Case 4: diagonals There is way to do this.Thus, in total there are possible ways. Problem18An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?SolutionLet denote that George selects a red ball and that he selects a blue one. Now, in order to get balls of each color, he needs more of both and .There are 6cases:(wecan confirm that there are only since ). However we canclump , ,and together since they are equivalent by symmetry.andLet's find the probability that he picks the balls in the order of .。

A2 整数的求解A2－001 哪些连续正整数之和为1000？试求出所有的解．【题说】 1963年成都市赛高二二试题 3．【解】设这些连续正整数共n个（n＞1），最小的一个数为a，则有a＋（a＋1）＋…＋（a＋n－1）＝1000即n（2a＋n－1）＝2000若n为偶数，则2a＋n－1为奇数；若n为奇数，则2a＋n－1为偶数．因a≥1，故2a＋n－1＞n．同，故只有n＝5，16，25，因此可能的取法只有下列三种：若n＝5，则 a＝198；若n＝16，则 a＝55；若n＝25，则 a＝28．故解有三种：198＋199＋200＋201＋20255＋56＋…＋7028＋29＋…＋52A2－002 N是整数，它的b进制表示是777，求最小的正整数b，使得N是整数的四次方．【题说】第九届（1977年）加拿大数学奥林匹克题3．【解】设b为所求最小正整数，则7b2＋7b＋7＝x4素数7应整除x，故可设x＝7k，k为正整数．于是有b2＋b＋1＝73k4当k＝1时，（b－18）（b＋19）＝0．因此b＝18是满足条件的最小正整数．A2－003 如果比n个连续整数的和大100的数等于其次n个连续数的和，求n．【题说】 1976年美国纽约数学竞赛题 7．s2－s1＝n2＝100从而求得n＝10．A2－004 设a和b为正整数，当a2＋b2被a＋b除时，商是q而余数是r，试求出所有数对（a，b），使得q2＋r＝1977．【题说】第十九届（1977年）国际数学奥林匹克题 5．本题由原联邦德国提供．【解】由题设a2＋b2＝q（a＋b）＋r（0≤r＜a＋b），q2＋r＝1977，所以q2≤1977，从而q≤44．若q≤43，则r＝1977－q2≥1977－432＝128．即（a＋b）≤88，与（a＋b）＞r≥128，矛盾．因此，只能有q＝44，r＝41，从而得a2＋b2＝44（a＋b）＋41（a－22）2＋（b－22）2＝1009不妨设|a－22|≥|b－22|，则1009≥（a－22）2≥504，从而45≤a≤53．经验算得两组解：a＝50，b＝37及a＝50，b＝7．由对称性，还有两组解a＝37，b＝50；a＝7，b＝50．A2－005 数1978n与1978m的最后三位数相等，试求出正整数n和m，使得m＋n 取最小值，这里n＞m≥1．【题说】第二十届（1978年）国际数学奥林匹克题 1．本题由古巴提供．【解】由题设1978n－1978m＝1978m（1978n－m－1）≡0（mod 1000）理注解：设1978n=1000a+c 1978m=1000b+c 1978n－1978m=1000(a-b因而1978m≡2m×989m≡0（mod 8），m≥31978n－m≡1（mod 125）注解：1978m（1978n－m－1）这两式的乘积要为1000整除，显然1978m这式为8的倍数，另一式为125的倍数。

B4二项式，数学归纳法，概率B4－001求(1＋x)3＋(1＋x)4＋(1＋x)5＋…＋(1＋x)n＋2展开式里的x2的系数．【题说】1963年北京市赛高三一试题3．【解】因为(1＋x)3＋(1＋x)4＋(1＋x)5＋…＋(1＋x)n＋2所以展开式中x2的系数为【别解】x2的系数为B4－002设f是具有下列性质的函数：(1)f(n)对每个正整数n有定义；(2)f(n)是正整数；(3)f(2)＝2；(4)f(mn)＝f(m)f(n)，对一切m，n成立；(5)f(m)＞f(n)，当m＞n时．试证：f(n)＝n．【题说】第一届(1969年)加拿大数学奥林匹克题8．【证】先用数学归纳法证明f(2k)＝2k(k＝1，2，…)．事实上，由(3)，k＝1时，f(2)＝2成立．假设k＝j成立，则由(4)f(2j＋1)＝f(2·2j)＝f(2)f(2j)＝2·2j＝2j＋1．故对所有自然数k，f(2k)＝2k．现考虑自然数n＝1．由(5)函数f的严格递增性知：f(2)＝2＞f(1)．由(2)，f(1)＝1．再考虑自然数n：2k＜n＜2k＋1．由(5)有2k＝f(2k)＜f(2k＋1)＜f(2k＋2)＜…＜f(2k＋1－1)＜f(2k ＋1)＝2k＋1，故必有f(2k＋1)＝2k＋1，f(2k＋2)＝2k＋2，…，f(2k＋1－1)＝2k＋1－1综上所述，对任何正整数n，都有f(n)＝nB4－003证明：对任何自然数n，一定存在一个由1和2组成的n位数，能被2n整除．【题说】第五届(1971年)全苏数学奥林匹克八年级题1．【证】用归纳法．(1)当n＝1时，取该数为2即可；(2)设A＝2n B是一个能被2n整除的n位数，则2·10n＋A和1·10n＋A中必有一个能被2n＋1整除．从而，命题得证．B4－004假设一个随机数选择器只能从1，2，…，9这九个数字中选一个，并且以等概率作这些选择，试确定在n次选择(n＞1)后，选出的n个数的乘积能被10整除的概率．【题说】第一届(1972年)美国数学奥林匹克题3．【解】要使n个数之积被10整除，必须有一个数是5，有一个数是偶数．n次选择的方法总共有9n种，其中A．每一次均不取5的取法，有8n种；B．每一次均不取偶数的取法，有5n种；C．每一次均在{1，3，7，9}中取数的方法有4n种，显然C中的取法既包含于A，也包含于B，所以，取n个数之积能被10整除的概率是B4－005一副纸牌共有N张，其中有三张A，现随机地洗牌(假定纸牌一切可能的分布都有相等机会)．然后从顶上开始一张接一张地翻牌，直至翻到第二张A出现为止．求证：翻过的纸牌数的期望(平均)值是(N＋1)/2．【题说】第四届(1975年)美国数学奥林匹克题5．【证】设三张A的序号分别是x1、x2、x3．若将牌序颠倒过来，则第二张A的序号为N＋1－x2．在这两副纸牌中，第二张A的平均位置(即翻过的纸牌数的期望值)为[x2＋(N＋1)－x2]/2＝(N＋1)/2【别证】由题设，除了第1张和最后一张外，其余各张皆可能是第2张A，且是等可能的．因此第2张A所在序号的平均期望值是[2＋3＋…＋(N—1)]/(N－2)＝(N＋1)/2．B4－006某艘渔船未经允许在A国领海上捕鱼．每撒一次网将使A国的捕鱼量蒙受一个价值固定并且相同的损失．在每次撒网期间渔船被A国海岸巡逻队拘留的概率等于1/k，这里k 是某个固定的正整数．假定在每次撒网期间由渔船被拘留或不被拘留所组成的事件是与其前的捕鱼过程无关的．若渔船被巡逻队拘留，则原先捕获的鱼全被没收，并且今后不能再来捕鱼．船长打算捕完第n网后离开A国领海．因为不能排除渔船被巡逻队拘留的可能性，所以捕鱼所得的收益是一个随机变量．求n，使捕鱼收益的期望值达到最大．【题说】1975年～1976年波兰数学奥林匹克三试题5．这里ω是撒一次网的收益．由(1)可知f(n)达到最大值．B4－007大于7公斤的任何一种整公斤数的重量都可以用3公斤和5公斤的两种砝码来称，而用不着增添其他不同重量的砝码．试用数学归纳法加以证明．【题说】1978年重庆市赛二试选作题1(3)．数a，b，使得n＝3a＋5b．事实上(1)当n＝8，9，10，11时，不难验证命题成立．(2)设k＞11并且当8≤n＜k时，命题成立，则当n＝k时，由归纳假设k－3＝3l＋5m，m，n为非负整数所以 k＝(k－3)＋3＝3l＋5m＋3＝3(l＋1)＋5m故命题对k成立．B4－008给定三只相同的n面骰子，它们的对应面标上同样的任意整数．证明：如果随机投掷它们，那么向上的三个面上的数的和被3整除的概率大于或等于1/4．【题说】第八届(1979年)美国数学奥林匹克题3．【证】因为问题只涉及和是否被3整除，所以不妨假定，每个面上的数是被3除后的余数；0、1、2．设每个骰子上标“0”的有a个，标“1”的有b个，标“2”的有c个．这里a，b，c 是适合下列条件的整数：0≤a，b，c≤n， a＋b＋c＝n (1)随机地投掷三只骰子，总共有n3种等可能情形．其中朝上三个数的和被3整除的情形有以下四种类型：0，0，0；1，1，1； 2，2，2；0，1，2第一类共有a3种，第二类共有b3种，第三类有c3种，第四类有3！abc＝6abc种．因此，原问题转化为在条件(1)下，证明不等式即 4(a3＋b3＋c3＋6abc)≥(a＋b＋c)3上式可化简为等价的不等式a3＋b3＋c3＋6abc≥a2b＋a2c＋b2a＋b2c＋c2a＋c2b (2)不妨设a≥b≥c，则a3＋b3＋2abc－a2b－ab2－a2c－b2c＝a2(a－b)＋b2(b－a)＋ac(b－a)＋bc(a－b)＝(a－b)(a2－b2－ac＋bc)＝(a－b)2(a＋b－c)≥0， (3)c3＋abc－c2a－c2b＝bc(a－c)＋c2(c－a)＝c(a－c)(b－c)≥0 (4)(3)、(4)相加得a3＋b3＋c3＋3abc≥a2b＋a2c＋b2a＋b2c＋c2a＋c2b从而(2)成立．B4－009抛掷一枚硬币，每次正面出现得1分，反面出现得2分．试【题说】第十二届(1980年)加拿大数学奥林匹克题4．【证】令得到n分的概率为P n．因为得不到n分的情况只可能是：先得n－1分，再掷出一次反面．所以有由于 P1＝1/2B4－010某个国王的25位骑士围坐在一张圆桌旁．他们中的三位被选派去杀一条恶龙(设三次挑选都是等可能的)，令P是被挑到的三人中至少有两人是邻座的概率．若P写成一个既约分数，其分子与分母之和是多少？【题说】第一届(1983年)美国数学邀请赛题7．【解】选二相邻的骑士有25种方法．再随着选第三位，有23种，故共有25×23种方法．但其中三者相邻的25种情况重复，应减去．故因此，所求之分子、分母之和为57．【别解】所选3人分两种情况：3人皆相邻，或2人相邻、1人不邻，故有25＋25×(25－4)种．B4－011在给定的圆周上随机地选择A、B、C、D、E、F六点，这些点的选择是独立的，对于弧长而言是等可能的．求ABC、DEF这两个三角形不相交(即没有公共点)的概率．【题说】第十二届(1983年)美国数学奥林匹克题1．【解】设圆周上给定6个点，从这6点中取3个点作为△ABC的顶B4－012一个园丁把三棵枫树、四棵橡树和五棵白桦树种成一行．十二棵树的排列次序是随机的，每一种排列都是等可能的．把没有两棵白桦树相邻的概率写成既约分数m/n．试求m ＋n．【题说】第二届(1984年)美国数学邀请赛题11．【解】先把三棵枫树和四棵橡树排好，有7！种排法，中间6个空所以，m＋n＝106为所求．B4－013设A、B、C、D是一个正四面体的顶点，每条棱长1米．一只小虫从顶点A出发，遵照下列规则爬行：在每一个顶点相交的三条棱中选一条(三条棱选到的可能性相等)，然后从这条棱爬到另一个点．设小虫爬了7米路之后，又回到顶点A的概率为P＝m/729，求m的值．【题说】第三届(1985年)美国数学邀请赛题12．【解】设从A出发走过n米回到A点的走法为a n种．由于从A出发走n－1米的走法共3n－1种，其中a n－1种走到A的，下一步一定离开A．除去这an－1种，其余的每一种都可以再走1米到达A点．因此有a n＝3n－1－a n－1B4－014某商店有10台电视机，排成一排．已知其中有三台是次品，如果我们对这批电视机作一次随机抽查，那么在前5台电视机中出现所有次品的概率是多少？【题说】1988年新加坡数学奥林匹克(A组)题9．原题为选择题．品的概率是B4－015把一个质地不均匀的硬币抛掷5次，正面朝上恰为一次的可能性不为0，而且与正面朝上恰为二次的概率相同．令既约分数i/j为硬币在5次抛掷中有3次正面朝上的概率．求i＋j．【题说】第七届(1989年)美国数学邀请赛题5．【解】令r是掷一次硬币正面朝上的概率，则在n次投掷中k次正面朝上的概率为由已知，有由此得r＝0，1或1/3．但r＝0，1都不可能，故r＝1/3．于是5次投掷3次正面朝上的概率为因此 i＋j＝283B4－016 n(n＋1)/2个不同的数随机排成一个三角阵：设M k是从上往下数第k行中的最大数，求M1＜M2＜…＜M n的概率．【题说】第二十二届(1990年)加拿大数学奥林匹克题2．【解】设所求概率为p n，显然p1＝1，p2＝2/3假设 p k＝2k/(k＋1)！对于n＝k＋1，最大数在最下一行的概率为因此，对所有自然数n，都有p n＝2n/(n＋1)！B4－017在吐姆巴利亚仅有总统与发言人两名诚实的人．其它人均以概率p(0＜P＜1)说谎．总统决定再次竞选，并告诉他身边的第一个人，这个人再告诉他身边的人，如此继续下去，直到这链上第n个人将总统的决定告诉发言人．发言人在这以前未听到有关总统的决定的信息，在n＝19与n＝20中，哪一种情况，发言人宣布的结果与总统决定相符的可能性较大？【题说】1990年匈牙利数学奥林匹克第二轮较高水平题1．【解】设发言人宣布结果与总统决定相符的概率为Q n，则有递推公式Q n＋1＝P(1－Q n)＋(1－P)Q n＝P＋(1－2P)Q n将n＋1换为n得Q n＝P＋(1－2P)Q n－1所以Q n＋1－Q n＝(1－2P)(Q n－Q n－1)由于Q0＝1，Q1＝1－P，所以Q n＋1－Q n＝(1－2P)n·(－P)时，Q20＜Q19．B4－018某生物学家想要计算湖中鱼的数目，在5月1日他随机地捞出60条鱼并给它们做了记号，然后放回湖中．在9月1日他又随机捞出70条鱼，发现其中有3条有标记．他假定5月1日时湖中的鱼有25％在9月1日时已不在湖中了(由于死亡或移居)，9月1日湖中40％的鱼在5月1日时不在湖里(由于新出生或刚刚迁入湖中)，并且在9月1日捞的鱼能代表整个湖中鱼的情况．问5月1日湖中有多少条鱼？【题说】第八届(1990年)美国数学邀请赛题6．【解】设5月1日湖中有x条鱼因此x＝840．【注】题中条件25％可改为任一百分数，不影响结果．B4－019用二项式定理展开(1＋0.2)1000，有(1＋0.2)1000＝A0＋A1＋…＋A1000【题说】第九届(1991年)美国数学邀请赛题3．比较A k－1与A k．B4－020有两串字母aaa与bbb要在电讯线上传送．每一串都是一个一个字母地传送．由于设备的毛病，这些字母的每一个都以1/3的概率被错误地接收到，即该收到a的都收到b，该收到b的都收到a．但每一个字母是否被正确收到与接收其他字母的状况互相独立．以S a记传送aaa时收到的一串3个字母，以S b记传送bbb时收到的一串3个字母，按词典顺序，S a在S b之前的概率记为P，将P写成既约分数，它的分子是多少？【题说】第九届(1991年)美国数学邀请赛题10．【解】设S a＝x1x2x3，S b＝y1y2y3．因此所求的数是532．B4－021一只抽屉内装有红袜子和蓝袜子，袜子至多有1991只．现在的情况是：不放回地随机取两只袜子，它们都是红色或都是蓝色的概率恰为1/2，按此情况，抽屉中红袜子的数目最多可能是几只？【题说】第九届(1991年)美国数学邀请赛题13．【解】设红、蓝袜子数分别为x和y．由已知，任取两只袜子其颜色不同的概率是1/2．故有即 (x－y)2＝x＋y令n＝x－y，则 n2＝x＋y≤1991B4－022一位网球选手的“赢率”是她赢的场数比参赛的场数．在一个周末开始时，她的赢率恰好是0.500．在这个周末期间她比赛了四场，赢了三场，输了一场，到这个周末结束时，她的赢率大于0.503．在这个周末开始之前，她最多可能赢几场？【题说】第十届(1992年)美国数学邀请赛题3．【解】设W是这网球运动员在周末开始时已赢的局数，M是她已若W＝164，M＝328，则W/M＝0.500．而(W＋3)/(M＋4)＞0.503．因此，在周末开始前，这运动员最多可赢164场．B4－023在贾宪－杨辉三角形中，每一个数值是它上面的二个数值之和，这三角形开头几行如下：在贾宪－杨辉三角形中的哪一行中会出现三个相邻的数，它们的比是3∶4∶5？【题说】第十届(1992年)美国数学邀请赛题4．n组成．如果第n行中有那么 3n－7k＝－3，4n－9k＝5解这个联立方程组，得k＝27，n＝62．即第62行有三个相邻的数B4－024从集合{1，2，3，…，1000}中随机地、不放回地取出3个数a1、a2、a3，然后再从剩下的997个数中同样随机地、不放回地取出3个数b1、b2、b3．令p为a1×a2×a3的砖能放在b1×b2×b3的盒子中的概率．若将p写成既约分数，那么分子和分母的和是多少？【题说】第十一届(1993年)美国数学邀请赛题7．【解】不妨设a1＜a2＜a3，b1＜b2＜b3，当且仅当a1＜b1，a2＜b2，a3＜b3时砖可放入盒中．设c1＜c2＜c3＜c4＜c5＜c6是从｛1，2，…，1000｝中选出的6个数，再从中选出3个有种方法．这3个作为a1、a2、a3，剩下3个作为b1、b2、b3．符合要求的a1只能是c1．a2若为c2，则a3可为c3或c4或c5；a2若为c3，则求分子、分母的和为1＋4＝5．B4－024从集合{1，2，3，…，1000}中随机地、不放回地取出3个数a1、a2、a3，然后再从剩下的997个数中同样随机地、不放回地取出3个数b1、b2、b3．令p为a1×a2×a3的砖能放在b1×b2×b3的盒子中的概率．若将p写成既约分数，那么分子和分母的和是多少？【题说】第十一届(1993年)美国数学邀请赛题7．【解】不妨设a1＜a2＜a3，b1＜b2＜b3，当且仅当a1＜b1，a2＜b2，a3＜b3时砖可放入盒中．设c1＜c2＜c3＜c4＜c5＜c6是从｛1，2，…，1000｝中选出的6个数，再从中选出3个有种方法．这3个作为a1、a2、a3，剩下3个作为b1、b2、b3．符合要求的a1只能是c1．a2若为c2，则a3可为c3或c4或c5；a2若为c3，则求分子、分母的和为1＋4＝5．B4－025 A和B轮流掷一个均匀的硬币，谁先掷出人头的一面谁获胜，他们玩了n次，而且前一场的输家下一场先掷．若A第一场先掷，数码是什么？【题说】第十一届(1993年)美国数学邀请赛题11．【解】任一场比赛，先掷的人赢的概率为令P k为A赢第k场比赛的概率，则P1＝．对k≥2，有所以，m＋n＝1093，其最后三个数码为093．B4－026一种单人纸牌游戏，其规则如下：将6对不相同的纸牌放入一个书包中，游戏者每次随机地从书包中抽牌并放回，不过当抽到成对的牌时，就将其放到一边，如果游戏者每次总取三张牌，若抽到的三张牌中两两互不成对，游戏就结束，否则抽牌继续进行直到书包中没【题说】第十二届(1994年)美国数学邀请赛题9．【解】设书包中有n(≥2)对互不相同的牌，p(n)为按所说规则抽牌使书包空的概率．则P(2)＝1．由于前三张牌中有两张成对的概率为所以，对n≥3，有反复利用这个递推公式，得当n＝6时，有所以，p＋q＝9＋385＝394．B4－027质点x按下列规则(1)，(2)在p、q两点之间移动：(1)x在q处时，1秒后必移到p处；(2)x在p处时，1秒p处的概率．【题说】1995年日本数学奥林匹克预选赛题5．【解】设n秒后x在p处的概率为p n，x在q处的概率为q n．则B4－028在重复掷一枚均匀硬币的过程中，在连得2个反面之前的正整数，求m＋n．【题说】第十三届(1995年)美国数学邀请赛题15．【解】设掷k次，不出现连续2个反面的情况有b k种，易知b1＝2，b2＝3，约定b0＝1．由于第一次为正面，再掷k－1次不出现连续2个反面的情况有b k－1种．第一次为反面，第2次必须为正面，再掷k－2次不出现连续2个反面的情况有b k－2种，所以b k＝b k－1＋b k－2(1)又设掷k次，无连续2个反面，而有5个连续正面，并且最后一次为正面的情况有a k种．这a k种，倒数1～5次均为正面的情况有b k－5种，倒数1～4次均正、第5次为反面的情况有a k－5种，倒数1～3次均正、第4次为反面的情况有a k－4种，依此类推，从而有递推关系a k＝b k－5＋a k－5＋a k－4＋a k－3＋a k－2(2)又显然a1＝a2＝a3＝a4＝0，a5＝1，a6＝2．掷k＋2次，最后2次为反面，而且在这前面已有5个连续正面，没利用递推关系(2)有再利用(1)所以m＋n＝3＋34＝37B4－029一目标在坐标平面上一步步移动．它从(0，0)出发，每一步移动一个单位长度，可以向左、向右、向上、向下，四个方向是等可能的．设p为该目标移动6步或更少的步数到达(2，2)的概率．p【题说】第十三届(1995年)美国数学邀请赛题3．【解】到达(2，2)需4步或6步．6步到达有两类情况，一类一下三上两右，另一类一左三右两上．概率为4步到达后再走两步仍回到(2，2)的概率为所以B4－030在五个队参加的比赛中，每个队与别的队都比赛一场．一场比赛中每个参加的队有50％赢的机会(没有平局)．整个比赛既没有m＋n．【题说】第十四届(1996年)美国数学邀请赛题6．所以m＋n＝17＋32＝49。