n个正整数的和与积相等条件下的有趣结论

本文讨论了n个正整数的和与积相等的一个必要条件,并证明了两个与素数、合数有关的结论. 结论1:若n(n≥2)个正整数a1,a2,…,an满足条件n∑i=1ai=n∏i=1ai,则ai≤n(i=1,2,…,n). 证明:(1)当n=2时,a1·a2-(a1+a2)=(a1-1)·(a2-1)-1≥0,当且仅当a1=a2=2时等号成立,故a1·a2=(a1+a2)时a1≤2,a2≤2,符合结论1. (2)当n≥3时,设a1≤a2≤…≤an.令a1=a2=…=an-2=1,an-1=2,an=n,则n∑i=1ai=n∏i=1ai=2n.此时ai≤n(i=1,2,…,n). 又设存在n(n≥2)个正整数b1,b2,…,bn满足条件1≤b1≤b2≤…≤bn-1≤bn,bn＞n,且n∑i=1bi=n∏i=1bi.不妨令bi=1+ti(i=1,2,…,n-1,ti∈N),bn=n+tn(n∈N+).     

三角形的个数——特殊与一般的协同作用

认识往往要经历一个从特殊到一般的过程 ,这个过程简述为特殊与一般的协同作用 .下面以探求三角形的个数为例作一说明 .问题　a、b、c表示三角形三边的长 ,它们都是正整数 ,其中a≤b≤c ,若b =n ,这样的三角形有几个 ?分析　三角形的个数与n有关 ,设它的个数为f(n) ,问题变为求函数f(n)的具体表达式 .看来 ,这不是一个垂手可得的问题 ,困难在于n是任意正整数 .如果n是某个特殊正整数 ,问题的难度当然下降 .不妨从n =1,2 ,3试试看 .首先 ,正整数a、b、c受以下两式制约a≤b≤c ,　　①a +b >c .　　②1.当n =1时 ,b=1,由①得a =1;由②得c=1,可…     

一个展开式在求平方根上的应用

设n是某一个正整数,并且n=a^2＋b,其中a为正整数,b为整数且｜b｜〈n则√n=a√n/a^2=a（a^2/n）-^1/2=a（n-b/n）-^1/2,（1）其中t=b/2n.     

关于多重完全数的一个结论

设n为大于 1的正整数 ,ω(n)表示n的不同素因子的个数 ,σ(n)为n的所有正因子之和 .若σ(n) =2n ,则称n为完全数 .若σ(n) =knk≥ 3,则称n为多重完全数 .本文以欧拉定理及费尔马定理为基础讨论了一种特定条件下的多重完全数问题 ,即满足σ(n) =ω(n)·n(ω(n)≥3)的解的情况 ,得到了σ(n) =ω(n)·n(ω(n)≥ 3)的全部解为n =2 3 ·3·5 ,2 5·3·7,2 5·33 ·5·7.     

一类实二次代数整数方幂的整数部分

设a,b是正整数,a b～(1/2)≥2 3～(1/2)且b不是平方数,x=a b～(1/2).给出了所有可使1 [x～n]=x～n x～(-n)对任何正整数n都成立的x,其中[x～n]表示x～n的整数部分.     

第45届IMO预选题(下)

数论部分1.设τ(n)表示正整数n的正因数的个数.证明:存在无穷多个正整数a,使得方程τ(an)=n没有正整数解n.2.已知从正整数集N 到其自身的函数ψ定义为ψ(n)=∑nk=1(k,n),n∈N ,其中(k,n)表示k和n的最大公因数.(1)证明:对于任意两个互质的正整数m、n,有ψ(mn)=ψ(m)ψ(n);(2)证明:对于每一个a∈N ,方程ψ(x)=ax有一个整数解;(3)求所有的a∈N ,使得方程ψ(x)=ax有唯一的整数解.3.一个从正整数集N 到其自身的函数f满足:对于任意的m、n∈N ,(m2 n)2可以被f2(m) f(n)整除.证明:对于每个n∈N ,有f(n)=n.4.设k是一个大于1的固定的整数,m=4k2-5.…     

数学归纳法在证明数列题中的应用

数学归纳法主要用来证明一个与正整数有关的命题,它的步骤如下:1.证明当n取第一个值n0时结论正确;2.假设当n=k(k!N*,且k≥n0)时结论正确,证明当n=k 1时结论也正确.在完成了这两个步骤以后,就可以断定命题对于从n0开始的所有正整数n都正确.例1已知在各项均为正数的数列｛an｝中,它的前n项和Sn满足Sn=12(an a1n).试猜想数列｛an｝的通项公式,并用数学归纳法证明你的猜想.解析∵S1=a1=12(a1 a11),∴a21=1.∵an>0,∴a1=1.∵S2=a1 a2=12(a2 a12),即a22 2a2-1=0,又an>0,∴a2="2-1.∵S3=a1 a2 a3=1 ("2-1) a3=21(a3 a13),即a32 2"2a3-1=0,又an>0…     

灵活应用幂的运算法则

关于幂的运算法则,我们学习了以下四条:(1)am·an=am+n(m、n为正整数);(2)am÷an=am-n(a≠0 m、n为正整数且m>n);(3)(am)n=nmn(m、n为正整数);(4)(ab)n=anbn(n为正整数).并规定了零指数幂和负整数指数幂的意     

例说用奇偶分析法解数学竞赛题

在各级各类的初中数学竞赛中 ,有许多较困难的问题 ,若能善于运用奇偶分析的方法 ,往往有“四两拨千斤”的效果 .现分类例析当今初中数学竞赛中经常出现的可用奇偶分析法来解决的问题 ,供参考 .1　与正整数有关的问题1 .1　求正整数例 1　一个正整数 ,若分别加上 1 0 0与1 6 8,则可得到两个完全平方数 ,求这个正整数 .( 2 0 0 1年全国初中数学联赛题 )解　设此数为 n,且 n+ 1 6 8=a2 ,n+ 1 0 0= b2 ,则 a2 - b2 =6 8=2 2 × 1 7,即 ( a+ b) ( a-b) =2 2 × 1 7.但 a+ b与 a- b的奇偶性相同 ,故 a+ b=34,a- b=2 .于是 a=1 8,从而 n=1 56 .1 .…     

问题1．9参考答案

问题设n是满足下列条件的最小正整数,它们是75的倍数且恰好有75个正数因数(包括1和本身),求7n5.解由已知条件知n=75k=3×52k,欲使n尽可能小,可设n=2a×3b×5c(c≥2,b≥1),且有(a+1)(b+1)(c+1)=75,所以a+1,b+1,c+1都是奇数,因此a,b,c都是偶数,所以c=2.由(a+1)(b+1)(c+1)=75,得(a+1)(b+1)=25.①a+1=5,b+1=5:a=4,b=4.故n=24×34×52;②a+1=1,b+1=25:a=0,b=24.故n=20×324×52.由①、②知最小的正整数n是24×34×52.故7n5=432.问题1.9参考答案…     

卡思卡特城堡

It is worthy of noting that, whilst Crookston Castle witnessed the earlier and happier portion of Mary's variegated life,     

不同的吃喝玩乐

一、吃和喝吃苹果 eat an apple, 吃药 take medicine,吃糖 have some sweets,吃饭 have one's meals,吃馆子 dine out,吃惊 be surprised/     

漏洞百出的教育

Many college freshmen arrive woefully unprepared to do college work, and as disadvantaged populations continue to grow, the share of the American work force that has made it through college is expected to plummet. Many experts blame that educational failure not just on high schools but also on colleges. School & College, a special report by The Chronicle, looks at efforts to fix the system. What reforms would better prepare students for college? What should schools and colleges be doing differently? How should state and federal officials help?     

只要努力,不要后悔

There are numbers of crossroads on our long and unpredictable life journey where we totally have no idea about which direction to choose. No matter what our decision is, we should not turn back, but face the music and go ahead instead. I am this kind of girl who always does try without regretting, one example is how I dealt with my love.     

浅谈移民问题

Migration occurs behind a variety of reasons and has a great effect on the whole world. People may migrate in order to improve their economic situation, or in order to escape civil strife, persecution, and environmental disasters. The impact of migration is complex, bringing both benefits anddisadvantages. This paper briefly talks about the causes of migration, the allocation of benefits, and the ways in which individual countries and the international community deal with this important subject.     

Sister Carrie, an Adherent of Desires

Sister Carrie is one of the most controversial characters in American literature.Thought as a "fallen woman" firstly,she was defined as a "new woman" by some critics later. However, by digging into the motivaton behind the whole process of Carrie's "success", the relationship between Carrie and her creator (the author), the social conditions of then American, it can be found that Carrie has never been free-standing on her thought and she has never found her real-sdf even after becoming a famous actress. In a society dominated by mass consumerism Carrie is only an adherent of her own desires. She also is a representative of all those country girls flooded into cities, a symbol and a sacrifice of the urbanization of America in a time countryside was overcome by cities.     

Improvement of one-cycle controller by use of proportional integral differential controller

1.IntroductionOne-cyclecontrolmethod,whichwasproposedaboutonedecadeago[1],hasbecomeanattractivemethodinspecialfieldssuchaspowerfactorcorrection[2-6],switchingamplifiers[7,8],etc.Themainideaofthiscontrollerisbasedonintegrationofdiodevoltageinone-cycleandforcesittobeexactlyequaltothereferencevalue.Themainadvantageofthiscontrollerisitsrealtimeabilitytorejectthevariationofinputvoltage[1].Despitethisgreatability,ithasnogoodperformancesinrejectingofloaddisturbanceandfollowingreferencecommands.Espec…     

库克斯顿城堡

城堡, 是欧洲典型的建筑之一, 每个城堡都有一个美丽的传说或是一段历史。从本期开始, 我们将推出“城堡之旅”, 带各位读者去浏览、观赏欧洲各地的城堡, 开始一段文化与历史的旅途!     

The Reader Response Approach to Teaching of English Literature

During interview with English majors, it was found that a majority of students expressed disappointment with their experience of English literature classes. Students' dislike of learning English literature appears to be the major problem of teaching English literature in China. Reader response approach is a good way to solve this problem. In this paper, I explain the rationale behind the reader response approach, such as its origin, definition, features, assumptions, and strategies. Then I illustrate how reader response approach works as a teaching strategy by presenting several teaching models. At last I evaluate its usefuiness for teachers and present the advantages of reader response approach in Chinese context.