初中数学竞赛中的高斯函数问题

第十三讲 高斯函数“续”例1：求证：,2!211--=⇔k n n n 其中k 为某一自然数. （1985年第17届加拿大数学竞赛试题） [证明]2为质数，n!中含2的方次数为∑∞==1].2[)!(2t t n n 若∑∑∞=-=--------=-=++++====1111221111122221]2[]2[)!(2,2t k t k k t k t k k n n n 则故!.|21n n -反之，若n 不等于2的某个非负整数次幕，可设n=2s p ，其中p >1为奇数，这时总可以找出整数t ，使+++=<<--+ ]2[]2[)!(22!,222211p p n n p s s t s t 的方次数为中所含于是 ≤++- 0]2[p t s ].2[]22[])12(2[])222[(21p n p p p p t s t s s t t s t s s s -------+=-=-=+++由于12,2)!(22!,2]2[,221----≤-=-<<n t s ts n n n p 则的方次数中含故则n !.这与已知矛盾，故必要性得证.例2：对任意的∑∞=+*+=∈01].22[,K k kn S N n 计算和 （第10届IMO 试题）【解】因]212[]22[11+=+++k k n n 对一切k =0,1,…成立，因此，].2[]22[]212[111+++-⋅=+k k k nn n 又因为n 为固定数，当k 适当大时,.)]2[]2([,0]2[,1201n nn S n n K k k k k ==-==<∑∞=+ 故从而例3：计算和式.]503305[502的值∑==n nS （1986年东北三省数学竞赛试题）【解】显然有：若.,,1][][][,1}{}{R y x y x y x y x ∈++=+=+则503是一个质数，因此，对n=1,2,…,502, 503305n 都不会是整数，但503305n +,305503)503(305=-n可见此式左端的两数的小数部分之和等于1，于是，[503305n ]+.304]503)503(305[=-n 故 ∑∑===⨯=-+==25115021.76304251304]),503)503(305[]503305([]503305[n n n n n S例4：设M 为一正整数，问方程222}{][x x x =-，在[1，M]中有多少个解？ （1982年瑞典数学竞赛试题）【解】显然x =M 是一个解,下面考察在[1，M]中有少个解.设x 是方程的解.将222}{}{}{2][x x x x x +⋅+=代入原方程，化简得=}]{[2x x,1}{0].}{}]{[2[2<≤+x x x x 由于所以上式成立的充要条件是2[x ]{x }为一个整数. .1)1(],1[,.)1())1(21(2),1[,11.2)1,[),12,,1,0(2}{,][个解中有原方程在因此个解中方程有可知在又由于个解中方程有即在则必有设+--⋅=-+++-≤≤+-==∈=M M M M M M M M m m m m m k mkx N m x例5：求方程.051][4042的实数解=+-x x （第36届美国数学竞赛题） 【解】.0][,1][][不是解又因<+<≤x x x x⎪⎪⎪⎩⎪⎪⎪⎨⎧≤≥>⎪⎪⎪⎩⎪⎪⎪⎨⎧≤≥<⎩⎨⎧≤-->--⎪⎩⎪⎨⎧≤+->+-+∴.217][,23][,211][;217][,23][,25][.07][2)(3][2(.0)11][2)(5][2(.051][4][4,051][40)1]([422x x x x x x x x x x x x x x 或.2269,02694;2229,02294;2189,01894;229,0294:,876][2][2222==-==-==-==-==x x x x x x x x x x 分别代入方程得或或或解得经检验知，这四个值都是原方程的解. 例6：.][3]3[2]2[1][][:,,nnx x x x nx N n R x ++++≥∈+∈*证明 （第10届美国数学竞赛试题）这道题的原解答要极为复杂，现用数学归纳法证明如下. 【证明】.,2,1,][2]2[][ =+++=k kkx x x A k 令 由于.,1],[1命题成立时则==n x A.,,,],[][][][][][][])[])1([(]))2[(]2([])1[(]([][]2[])2[(])1[(][])1[(]2[][][])1[(]2[][][])1[(]2[][)(:].[],2[22,],)1[()1()1(],[,][,][,].)1[(,],2[],[,1122112111221111121证毕均成立故原不等式对一切命题成立时即故相加得所以成立对一切即因为即有时命题成立设*---------∈=≤∴=+++≤++-++-++-+=+++-+-++-+++≤++++++-+++=+-+++=+++-==--=---=-=-=--≤≤≤-≤N n k n kx A kx k kx kx kx kx kx x x k x k x x k x x x x k x k kx x k x x A A A A kx x k x x kA kx x k x x A A A kA x A x A A x k A k A k kx kA kA k kx kA kA kkx A A x k A x A x A k n k k k k k k k k k k k k k k k例7：对自然数n 及一切自然数x ，求证：)].([]1[]2[]1[][苏联数学竞赛题nx nn x n x n x x =-+++++++【证明】则},{][x x x +=]1[]2[]1[][nn x n x n x x -+++++++].[]1[]2[]1[][}].{[]1}[{]2}[{]1}[{}][{.}]{[.1}{,}{11}{1}{.]1}[{]}[{]1}[{]2}[{]1}[{}][{,11}{,1}{,1,.}]{[]1}[{]2}[{]1}[{}][{}],{[][}]{][[][].1}[{]2}[{]1}[{}][{][],1}[{][]2}[{][]1}[{][}][{][]1}{][[]2}{][[]1}{][[}]{][[nx nn x n x n x x x n nn x n x n x x k n x n k n x n k n x n nk x n k x k n n n x n k x n k x n x n x x nk x n k x n k k x n nn x n x n x x x n x n x n x n nx nn x n x n x x x n n n x x n x x n x x x x n n x x n x x n x x x x =-+++++++=-+++++++-=+-<-≥<-+≥+-=-+++++-+++++++<-+≥+≤≤=-++++++++=+=-++++++++=-+++++++++++=-+++++++++++= 从而有知故知且知及由则而使设存在即可故只要证明例8：求出]31010[10020000+的个位数字.（第47届美国普特南数学竞赛试题） 【解】先找出3101010020000+的整数部分与分数部分.3101010020000+=31033103)10(100200100200200100+++-.3108110310910310310]31010[,131093103.310310,3)10(|310310|3)10(,)3(])10[(3)10(1005020000100100200001002002000100200001001001002001002002000022100100200200002210010021002100200200100+-=+-=+-=+<+=++--+---=-知显然是整数知又知其中分母的个位数字为3，分子的个位数字为9，故商的个位数字为3.。

高斯函数一、基本知识定义：设R x ∈,用[]x 表示不超过x 的最大整数,则[]x y =称为高斯函数.函数的定义域为R ,值域为Z .任一实数都能写成整数部分与非负纯小数之和,即[]α+=x x ()10<≤α,因此. [][]1+<≤x x x 。

我们称[]x 为x 的整数部分，称{}[]x x x -=为x 的小数部分。

函数{}x y =的定义域为R ，值域为[)1,0。

二、性质1. 函数[]x y =是不减函数,即当21x x ≤时,有[][]21x x ≤；2. [][]11+<≤<-x x x x ；3. [][]n x n x Z n +=+⇔∈；4. [][][]y x y x +≤+，{}{}{}y x y x +≥+； 推广：（1）[][][][]n n x x x x x x +++≤+++ 2121 （2）[][]nx x n ≤ ()N n ∈5. 若0,0≥≥y x ，则[][][]y x xy ≥；6. 若0,1>≥y x ，则[][]x y x y ≤⎥⎦⎤⎢⎣⎡；7. []⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡n x n x （其中*N n ∈） 8. （1）若121≥-x x ，则存在整数k ，使[][]12121x k x x k x ≤≤+⇒≤<；（2）[][][][]110212121+==⇒<-≤x x x x x x 或； （3）[][]12121<-⇒=x x x x9. [][]()[]()⎩⎨⎧∉--∈-=-Z x x Z x x x 110. [][]1,,-+=+⇒∈+∉y x y x Z y x Z y x ；11. 若整数b a ,满足r bq a += ()b r r q b <≤>0,,,0是整数，则q b a =⎥⎦⎤⎢⎣⎡；12. [][]x x x 221=⎥⎦⎤⎢⎣⎡++；13. 设1>x ，m 为正整数，则从1到x 的整数中，m 的倍数有⎥⎦⎤⎢⎣⎡m x 个；14. 设为p 任一质数，在!n 中含p 的最高乘方次数记为()!n p ，则()⎥⎦⎤⎢⎣⎡++⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡=m p n p n p n n p 2! ()1+<≤m m p n p例1．求!30的标准分解式。

4. 试题的编排顺序应作一些调整. 建议第 7 题放在第 9 题之后 ,第 17 题放在第 18 题之后. 如 果仅就试卷的知识特点 ,逻辑结构来看按试卷本 身的设置是有道理的. 但高考的主体是考生 , 如 果一切从实际出发 , 以考生为本 , 考虑到考生的 心理承受能力和实际水平作适当调整是应当的.
17 2
和
x
=
37 是原方程的 2
根.
例 4 用[ x ] 表示不大于实数 x 的最大整
数 ,方程 lg2 x - [lg x ] - 2 = 0 的实根个数是
. (1995 年全国高中数学联赛)
分析 : 这是一个含高斯函数 [lg x ] 的方程 ,
利用高斯函数性质 [lg x ] ≤lg x , 将解方程问题
2. 考察逻辑思维 , 演绎推理论证的能力应 把握一个度. 不可否认 , 数学是逻辑化最突出的 学科 ,推理能力的水平是学生数学水平的标志 之一. 但试题连续在第 18 题 (1) ,20 (2) ,21 (3) 三个大题都出现证明题 , 实在是有点太过头了. 更何况考证明题的弱点是给有的考生以投机取
巧的机会 , 省略中间过程 , 东拉西扯 , 最后还似 乎得到结果 ,给教师评卷也带来一些困难. 笔者 评阅第 20 题就发现考生的证明五花八门 , 让评
阅老师如同雾里看花 , 看得头晕眼花 , 苦不堪 言. 时间紧 ,任务重 ,评判失误就在所难免.
3. 以平面几何模型为背景的应用题第 17 题这本来是一个很好的问题 , 因为几何学的发 展就是以测量土地等为基础的 , 但是命题者对 学生的基本情况缺乏深刻的把握. 问题出在 :其 一 ,很多学生由于初中平面几何知识不过关 , 作 不出辅助线 , 导致无从下手 ; 其二 , 此题只需初 中水平的学生就可以解决. 高考后我找了居住 在自己小区的三个初中毕业生作此题 , 有两个 学生就用巧作一根垂直于 CD 的直线构成直角 三角形 ,用勾股定理建立方程就解决了.

Chartered Accountants Canadian Instituteof ActuariesGreat West Lifeand London Life Sybasei Anywhere Solutionsc 2004permitted.the contest bookleton your5.Scoring:6.Diagrams7.When yourPlease seeon publicationsScoring: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.The value of 3×46is(A)1(B)2(C)3(D)4(E)62.0.8−0.07equals(A)0.1(B)0.71(C)0.793(D)0.01(E)0.733.Contestants on “Gauss Reality TV”are rated by an applause metre.In the diagram,the arrow for one of the con-testants is pointing to a rating that is closest to (A)9.4(B)9.3(C)9.7(D)9.9(E)9.54.Twelve million added to twelve thousand equals(A)12012000(B)12120000(C)120120000(D)12000012000(E)120120000005.The largest number in the set {0.109,0.2,0.111,0.114,0.19}is(A)0.109(B)0.2(C)0.11(D)0.114(E)0.196.At a class party,each student randomly selects a wrapped prize from a bag.The prizesinclude books and calculators.There are 27prizes in the bag.Meghan is the first to choose a prize.If the probability of Meghan choosing a book for her prize is 23,how many books are in the bag?(A)15(B)9(C)21(D)7(E)187.Karen has just been chosen the new “Math Idol”.A total of 1480000votes were cast andKaren received 83%of them.How many people voted for her?(A)830000(B)1228400(C)1100000(D)251600(E)17831328.In the diagram,the size of ∠ACB is (A)57◦(B)37◦(C)47◦(D)60◦(E)17◦D CB A93o130o9.A movie theatre has eleven rows of seats.The rows are numbered from 1to 11.Odd-numbered rows have 15seatsand even-numbered rows have 16seats.How many seats are there in the theatre?(A)176(B)186(C)165(D)170(E)17110.In relation to Smiths Falls,Ontario,the local time in St.John’s,Newfoundland,is 90minutesahead,and the local time in Whitehorse,Yukon,is 3hours behind.When the local time in St.John’s is 5:36p.m.,the local time in Whitehorse is(A)1:06p.m.(B)2:36p.m.(C)4:06p.m.(D)12:06p.m.(E)10:06p.m.Part B:Each correct answer is worth 6.11.The temperature range on a given day is the difference between the daily high and the daily low temperatures.On the graph shown,which day has the greatest temperature range?(A)Monday (B)Tuesday (C)Wednesday (D)Thursday (E)FridayT em p e r a t u r e(o C)12.A bamboo plant grows at a rate of 105cm per day.On May 1st at noon it was 2m tall.Approximately how tall,in metres,was it on May 8th at noon?(A)10.40(B)8.30(C)3.05(D)7.35(E)9.3513.In the diagram,thelength of DC is twice the length of BD .The area of the triangle ABC is(A)24(B)72(C)12(D)18(E)3614.The numbers on opposite sides of a die total 7.What isthe sum of the numbers on the unseen faces of the two diceshown?(A)14(B)20(C)21(D)24(E)3015.In the diagram,the area of rectangle P QRS is 24.If T Q =T R ,the area of quadrilateral P T RS is(A)18(B)20(C)16(D)6(E)15P S RTQ16.Nicholas is counting the sheep in a flock as they cross a road.The sheep begin to cross theroad at 2:00p.m.and cross at a constant rate of three sheep per minute.After counting 42sheep,Nicholas falls asleep.He wakes up an hour and a half later,at which point exactly half of the total flock has crossed the road since 2:00p.m.How many sheep are there in the entire flock?(A)630(B)621(C)582(D)624(E)61817.The symbol 3456is evaluated as 3×6+4×5=38.If 261is evaluated as 16,then the number that should be placed in the empty space is(A)1(B)2(C)3(D)4(E)518.A game is said to be fair if your chance of winning is equal to your chance of losing.How many of the following games,involving tossing a regular six-sided die,are fair?•You win if you roll a 2•You win if you roll an even number•You win if you roll a number less than 4•You win if you roll a number divisible by 3(A)0(B)1(C)2(D)3(E)419.Chris and Pat are playing catch.Standing 1m apart,Pat first throws the ball to Chris andthen Chris throws the ball back to Pat.Next,standing 2m apart,Pat throws to Chris and Chris throws back to Pat.After each pair of throws,Chris moves 1m farther away from Pat.They stop playing when one of them misses the ball.If the game ends when the 29th throw is missed,how far apart are they standing and who misses catching the ball?(A)15m,Chris (B)15m,Pat (C)14m,Chris (D)14m,Pat (E)16m,Pat20.While driving at 80km/h,Sally’s car passes a hydro pole every four seconds.Which of thefollowing is closest to the distance between two neighbouring hydro poles?(A)50m (B)60m (C)70m (D)80m (E)90mPart C:Each correct answer is worth 8.21.Emily was at a garage sale where the price of every item was reduced by 10%of its currentprice every 15minutes.At 9:00a.m.,the price of a carpet was $10.00.At 9:15a.m.,the price was reduced to $9.00.As soon as the price of the carpet fell below $8.00,Emily bought it.At what time did Emily buy the carpet?(A)9:45a.m.(B)9:15a.m.(C)9:30a.m.(D)10:15a.m.(E)10:00a.m.22.In a bin at the Gauss Grocery,the ratio of the number of apples to the number of orangesis 1:4,and the ratio of the number of oranges to the number of lemons is 5:2.What is the ratio of the number of apples to the number of lemons?(A)1:2(B)4:5(C)5:8(D)20:8(E)2:1ing an equal-armed balance,ifbalancesof thefollowing would not(B)(C)(E)24.On a circular track,Alphonse is at point A and Beryl is diametrically opposite at point B .Alphonse runscounterclockwise and Beryl runs clockwise.They run atconstant,but different,speeds.After running for a while theynotice that when they pass each other it is always at the samethree places on the track.What is the ratio of their speeds?(A)3:2(B)3:1(C)4:1(D)2:1(E)5:225.How many different combinations of pennies,nickels,dimes and quarters use 48coins tototal $1.00?(A)3(B)4(C)5(D)6(E)8。