奇完全数至少有6个素因数≡1(3)的条件

本文证明了如下结果:设N=π~1m~2是一个奇完全数,这里π是奇素数且π≡l≡1(4)。如果3~(11)|σ(m~2),则N至少有6个素因数≡1(3),由此结果证明了若n是一个恰有8个不同素因数的奇完全数,且3·5·11|n,则3~4||n或3~6||n。     

奇完全数问题探讨

本文目的是探讨一个自然数如果是奇完全数 ,则其应具有的一些性质 .主要引理在本文中ｎ =ｐ1α1ｐ2 α2 … ｐｋαｋ,其中 ｐ1,ｐ2 ,… ,ｐｋ 为ｎ的不同素因子 .引理 1　若ａ >ｂ >0 ,ｍ >0 ,则 ａｂ>ａ ｍｂ ｍ .引理 2　 ｐ≠ 2 (1 -1ｐ2 ) -1=π28(此处 ｐ经过一切奇素数 ) .证明 :由文 [1 ]知 ｐ(1 -1ｐ2 ) -1=π26 (ｐ经过一切素数 ) ,∴　 ｐ≠ 2 (1 -1ｐ2 ) -1=π26 × (1 -12 2 ) =π28.引理 3　 (1 )若 (1 1ｐｉ-1 )≤ 2 (ｐｉ 为ｎ的不同素因子 ) ,则ｎ不是奇完全数 .(2 )若 (1 1ｐｉ)≥ 2 (ｐｉ 为ｎ的不同素因子 ) …     

奇完全数的Euler因子

设n是奇完全数,p是r的Euler因子.此时n=P4r+1m2,其中m,r是适合m≠0 (mod p)的正整数.本文证明了:τ(m2)≥15p4r+1,其中σ(m2)是m2的不同约数之和.     

关于一个整除性问题

先观察一例:若n为非负整数,则3~(4??+2)+5~(2n+1)能被14整除. 证明:由二项式定理(a+b)~n=am+b~n,(m∈N)则3~(4n+2)+5~(2n+1)=9·81~n+5·25~n =9·(56+25)~n+5·25~n =56m_1+9·25~n+5·25~n(m_1∈N) =14m_2+14·25~n(m_2∈N) =14(m_2+25~n)=14m_3.(m_3∈N) 故3~(4n+2)+5~(2n+1)能被14整除. 考察3~(4n+2)+5~(2n+1)=9·81~n+5.25~n有     

对一个猜测的探讨

在众多的组合式中 ,有一个相当漂亮的结构 :∑rn=1·Crn- 1 + 2·Crn- 2 + 3· Crn- 3+…+ ( n- r) Crr.(规定 Cnm中 ,若 m相似文献   

一类三角函数的最小正周期

我们熟悉了g(x) =Asin(ωx φ) B的最小正周期T =2π｜ω｜,那么｜g(x) ｜的最小正周期呢 ?定理 1　已知f(x) =｜Asin(ωx φ) B｜ ,A、B、ω、φ为常数且A、ω≠ 0 .1.1　若B =0 ,则f(x)最小正周期为T =π｜ω｜;1.2　若B≠ 0 ,则f(x)最小正周期为T =2π｜ω｜.定理 2　已知f(x) =｜Acos(ωx φ) B｜ ,A、B、ω、φ为常数且A、ω≠ 0 .2 .1　若B =0 ,则f(x)最小正周期为T =π｜ω｜;2 .2　若B≠ 0 ,则f(x)最小正周期为T =2π｜ω｜.定理 3　已知f(x) =｜Atan(ωx φ) B｜ ,A、B、ω、φ为常数且A、ω≠ 0 ,则f(x)最…     

例析二项式定理问题类型及解法

二项式定理的问题相对独立 ,题型繁多 ,解法灵活且较难掌握 .本文结合近年来的高考试题 ,根据二项式定理的不同问题 ,进行分类 ,并作出解法探讨 .一、确定二项式中的有关元素此类问题一般是根据已知条件 ,列出等式 ,从而可解得所要求的二项式中的有关元素 .【例 1】　已知 ( ax -x2 ) 9的展开式中x3的系数为 94,常数a的值为　　　　 .解 :Tr+1 =Cr9( ax) 9-r( -x2 ) r=Cr9( -1 ) r· 2 - r2 ·a9-r·x32 r- 9令32 r-9=3 ,即r=8.依题意 ,得C89( -1 ) 8· 2 - 4·a9- 8=94.解得a=1【例 2】　若在 ( 5x-1x) n 的展开式中 ,第 4项是常数项 ,则…     

一个不等式的证明

定理nn-1[(m+1)n-1n-1]<∑mi=11niαn-αn-1(α>1,n∈N,n≥2).证明由二项式定理得(α-1n)n=∑nr=0(-1)rCrn1nrαn-r,∵Crn(1n)r-Cr+1n(1n)r+1=Cr+1n(1n)r+1·nr+rn-r≥0,∴Crn(1n)r≥Cr+1n(1n)r+1(当且仅当r=0时等号成立).若n为偶数时,(α-1n)n=αn-αn-1+(C2n1n2αn-2-C3n1n3·αn-3)+…+(Cn-2n1nn-2α2-Cn-1n1nn-1α)+Cnn1nn>αn-αn-1;若n为奇数时,(α-1n)n=αn-αn-1+(C2n1n2αn-2-C3n1n3·αn-3)+…+(Cn-1n1nn-1α-Cnn1nn)>αn-αn-1.2定理的证明(1)∑m…     

3、4、5、6重完全数

在本文里,我们发现了一批多重完全数;2个3重完全数;16个4重完全数;11个5重完全数;1个6重完全数.     

数学单元测试题(四) (整式乘法)

一、填空题1.若(9“)。一3。,则m=== .2.化简(3口。)。一7口。瞄。一日(4口。+1)]得 .3.已知n为正整数,,州·z一。·(一∥)。的值是 .4.4¨×5弘的计算结果是 位数.5.梯形的上底长为(3口+26),下底长为(2口+6),高为(2n+26),则梯形的面积为6.已知口一55船,6:66匏,则n与6的大小关系是 .7.长为3n+26,宽为5口一6的长方形的面积为 .8.圆环的外圆半径为7a+26,内圆半径为6n一6,则它的面积是 .9.已知(3z+2)。=dz。+Pz+,,则d—e一 .10.已知z满足(z一5)×(6一z)一4,则(z一5)(z一6)一 . 。二、选择题1.若m≠n,则下式中不能成立的是 ( )A.(m—n)。===(”…     

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.