王進(jìn)明初等數(shù)論習(xí)題及作業(yè)解答兩整數(shù)相除，得商12，余數(shù)26，又知被除數(shù)、除數(shù)、商及余數(shù)之和為454．求被除數(shù).ababbb(121)b被除數(shù)=12+26=360+26=386.ab這題的后面部分是小學(xué)數(shù)學(xué)的典型問(wèn)題之一——“和倍”問(wèn)題。n9qrr當(dāng)Z且r3n27k3n9,；3nk)k)k)19(3k3k1若；3322nk)k)k)1kkk8若33232nnn32的值是整數(shù)。當(dāng)Z326nnn322nnn32326=2nnn是6證3262nnnn(2nn1)(nn(2n1)322(nn(n2nn(nn(nn(n=.|n(nn(nn(n.由!k6|(nn(nn(nn(2n.,若(nn(2nn為1是3為3n為1(n(2n是3.(nn(2n6若n．nnn5(mnl的值總是整數(shù)當(dāng)Nm!n!l!+(mnl(mnl)(mnl(nlnl)(nl(l1)l!=由km(mnl)(mnl(nl,!|(nl)(nll1)ab|(ab)n；當(dāng)Z且annab|(ab)當(dāng)Z且annn．a(chǎn)b|(ab)5a≠nna=，當(dāng)nababbbnnnnabnabb1nabbnn1n1nn1當(dāng)25abaaaaa且212345ii1a2k1,kZ,i1,2,3,4,5iii5555a(2k4k(k52|k(k1)k(k,8|422iiiiiiiii1i1i1i15a8qqZb(2k4k(k1)1bq1k,qZ，，，222**,而2ii1889cax2c0pqpp,p,qa()2bc0qqapbpqcq0即22cq2apbpq若p若q2bpqcq2ap2。這123456789A2或4或8或3或?2和5A2余A5余4和A4余A余8和A8余余3和9A∴A3和9AAy5534．97被5能被3．1至的99?123456789則9.9，．|aaaaaa的特征是：或或整除3210．證明：或或nn1|aaaaaa|nn13210aaaaaaaaaaaaa)．…n3210210n3…21013a，b，c2001200324|62742,求,.．a(chǎn)和ab=?22Z6|．fxbxcfxZ.fffx2c363|62742,8|627423|),且8|0,2,4,6,8,有0,2,4,7,9,50;4;8;2;6.ab+a+b22．fffx,+,b.b.kbf;f=++a+b+,fZn.+當(dāng)nn+n42Z.+32求aa.p和q求.p和859433.qq..n2，n≥.10911,200345.58nn++n+++n+1.n+n=n+n=+±1n+n.424222242++Z,和+11.322+2∵a+是3a+4是3aa為3a,a,1<≤1或。.3得p33|3|33|.33.1第12p12q32p14p24p24p2設(shè)、k,kh,k,得k4qpq2q34q5kp,hq,只4qhq2+11.1p,p,,pN=4pp…pNn12k12k1N1,p為p,p,,pp|12kn,設(shè)222）+2）+…+2+.nstsst-1st-2s+…當(dāng).6.或p,.若則3,,.73nnZ+.=Z.+N+xy是NN.有1號(hào)2233.15542下.設(shè)ba+.A經(jīng)B到ABCB6塊765432倍abanbnnn，1.14a,ba,b,a,b,a,bnnnnna,bna,bna,bna,bnababnn1，。,,a,ba,b,ba,ba==ann=anbn，，n=∴n=b=an，∴||，|||||以。－。12到除9，59元.4500,1237549500,2750,1237524750,24750較小，2475027509.99=米.21621616.5472a+(a,b)d,a,b,t,tx,y1,xtt)(yxt121212122tt,t)tt,t)tt,tt)tt,t)11221211212121(,t)d即，a,b(ab,a,b1212()=(,b)=和米.2個(gè).x歲，,,,,5.將6個(gè).40.若..C.331耀12233nn4，，．4個(gè)2與4個(gè)2個(gè)2與1個(gè)。，．l或2．a(chǎn)bc54或9因得．．．,C,B,B是B是6B|,B|,C231,B23A7,B23,C211,由AC,3,且。121212221,C3,C,B若B2BC與∴237,23,2311,,,1,0,而且與不能同時(shí)為1.ABC12112111和時(shí),各有1,0;0,1種情況,6組2111111和,知形.?!.1373771273,a2,b2,。372222試討論xy與xy的大小關(guān)系。若,,1求證：xyRxy;xyxy1xyx＋xy＋yxyxy＋xy＋顯然xy＋xy＋xy，證得。2yZ時(shí)，xyxy0,有xyxy;＋而x1.2,y2.1,xy2.52,xy0.52,xy0.02,xyxy;又x1.9,y1.8,xy3.42,xy0.42,xy0.72,xyxy.3x5x0）3x5yxyx,.533x8x8x8x.8又xx,0x08x555120yx.3505y5055亦即6,13y14,838882x(xx)0x515151xx。x2223×5×763212）5152522212515259697212229697975151952053853955893129797979797975785969797111111考慮11）23200821223200720082211111111223111123111123122120082200822222個(gè).人.盞.2x11121xxxx2x1x1,x2,2x1；1x1,x2,即x01x0,即22x0,2xZ,1x,x0;x1,x2x2∴≤1或≤或或,77xyx8且可知等式左邊從右向左10057x8x7.43.由等式左邊從右向左第項(xiàng)100100x若和d,與4.和d,d|－d|和d,d|－d|∴d|,或22由.設(shè)n|4的加15.…,n:2,4,8,16,32,64,128,256,512,1024,2048,k64,又624124k215.4+即Z8.222+則+)≡y，y.2aa332421122.33331?50nmmnmm11.r,.22220129i>j222,，ijjj|j|jφ=102222為0129m為φm最小簡(jiǎn)化剩余系φ11133777944245568911111122233796977778335559896xxxm12k∴i≠jixxi=ji∴xxx.12k又ixi≠j≤xji∴xxxφm12k∴xxxm.12k7.∴799同.當(dāng)m.===，====若m.)m=1或m當(dāng)m.m=pp…mα>1pααα12ij∴p-p1-1p-pk-1αααα11kk1ppppip1.1iiiiiiiip.mii(m)mm為p,ik這時(shí)(m)p1,顯然是偶數(shù).ii1m.=14==p-1pp11kk11=m.m=pp…-p-1-kαk=1121km=m=2或m=m32n=.若m.nn解=φ=φ.∵m=1.∵=φ..++…φ..；；；；當(dāng)m=1.∴94×8）×8≡×8≡設(shè)q3p≡q.22設(shè)p5p.4=(3)=(3)=pq.∴p≡q.2222又∵q第92.∴2≡.∴2≡.=(3)=(5)=ppp）≡.4422又p22+8∴.==∴.paa≡p-1aaa≡.q是兩相異的質(zhì)a≡a≡p-1q-1≡.p是質(zhì)=1a為奇≡1.∴a+≡.ap-1a,()≡p.）n∵a≡a≡)≡a≡a≡)≡a≡ppqqqpp=pq≡.≡()1≡.m設(shè)ABAB.當(dāng)Z|x-|.2m∵()1()1≡a()b∵=a()=1mmm.設(shè)B=<A≤=B<1+5=C≤4+6=又=1，=≡7≡4≡4≡∴B×××x-x-2|x或2|x12≡-≡.又-+∴當(dāng)7|x或7|x-6+≡.而--+x+∴當(dāng)5|x或5|x+x+1≡.又-2++x+1當(dāng)3|x或3|x++x+≡.∴|x-4=+,x=2|x-p|9999p2和5kp.p1個(gè)9-p-1-≡.又1≡-.2|x或2|x-≡當(dāng)3|x或3|x≡故-≡6|-.10p1.=≡.kp-1p1k個(gè)9|9999∴≡.∴p.p1個(gè)9m()+n().nm=1n()≡m而m()≡0mm()()≡1mnnm()≡.∴m()≡..mm()n()nmam.11171x；；；；；∴1=，∴x，y00x62tt.31ty∴1=+，∴x，y00x28ty2111t.28040y86yx162y.1717yx=+x40ty10t∴.；；；；），；），；），；），；；；.買(mǎi)2元61521把.5..+=z2中.，z.．459y48z5設(shè)a，bca+b=cb222222.x=z.442b,b,2222ab或a則2222222或則。z=)=xy+=+)+z)44428448444444422=z2=2.nn=x+y-zz.222x=1d=d=d<1.22+y.2222n-x=my=m+y-m，22得=y－.2當(dāng)0y和-1y+y與z若y與z+y被822222被8y與z令x+2211+=x和tx和t2211112+2+2=2t<.11111.22x+y.22x.2求.2求x+.2求.設(shè)nn3.由=21=得22，，即，，，，由x+yy.22yx+y無(wú)422yxy=k=k=h=6或k=h=61,x13,x7,x11,x13;yy1y7.yx35xx是2x6或5是6±2x5或x+y+z=則z由x2222y+z=6即z.22x+y=p22.得-入x+y+z=333z6x+y+z被8-1.222x+y+z被94.333....第2題ax5(mod18),x5(mod18),x8(mod21).xa(mod35).xa(mod35),x8(mod21).即x=(tx3(mod7),x6(mod13),,x5(mod11).x7(mod24).5x7(mod11),2x4(mod8