GMAT數(shù)學(xué)考試中余數(shù)精講

GMAT數(shù)學(xué)考試中余數(shù)精講GMAT數(shù)學(xué)考試中余數(shù)精講GMAT數(shù)學(xué)考試中余數(shù)精講Introduction(趕時(shí)間的童鞋能夠掠過。。。不過一些觀點(diǎn)幫助童鞋們回想余數(shù)~~~)DefinitionIfxandyarepositiveintegers,thereexistuniqueintegersqandr,calledthequotientandremainder,respectively,suchthaty=divisor*quotient+remainder=xq+r;and0<=r<x.Forexample,when15isdividedby6,thequotientis2andtheremainderis3since15=6*2+3.Noticethat0<=r<xmeansthatremainderisanon-negativeintegerandalwayslessthandivisor.Thisformulacanalsobewrittenasy/x=q+r/x.PropertiesWhenyisdividedbyxtheremainderis0ifyisamultipleofx.Forexample,12dividedby3yieldstheremainderof0since12isamultipleof3and12=3*4+0.Whenasmallerintegerisdividedbyalargerinteger,thequotientis0andtheremainderisthesmallerinteger.Forexample,7dividedby11hasthequotient0andtheremainder7since7=11*0+7Thepossibleremainderswhenpositiveintegeryisdividedbypositiveintegerxcanrangefrom0tox-1.Forexample,possibleremainderswhenpositiveintegeryisdividedby5canrangefrom0(whenyisamultipleof5)to4(whenyisonelessthanamultipleof5).Ifanumberisdividedby10,itsremainderisthelastdigitofthatnumberisdividedby100thentheremainderisthelasttwodigitsandsoon.Forexample,123dividedby10hastheremainder3and123dividedby100hastheremainderof23.

.Ifit1.CollectionofMethodsMethod1：小數(shù)法（妹紙自己取的名字，包含后邊的方法也都是妹紙取的AwaythattheGMATwilltestremaindersiswhatyouwouldtypicallyjustdividebackintotheproblemtodeterminethedecimals:25/4=6remainder1Dividethat1backby4toget0.25,sotheansweris6.25.Anynumberwitharemaindercouldbeexpressedasadecimal.

~歡迎議論）Theremainderprovidesthedataafterthedecimalpoint,andthequotientgivesyouthenumbertotheleftofthedecimalpoint.Considerthisproblem(whichappearscourtesyofGMAC):Example：Whenpositiveintegerxisdividedbypositiveintegery,theremainderis9.Ifx/y=96.12,whatisthevalueofy?(A)96(B)75(C)48(D)25(E)12Sol：Goingbacktotheconceptoftheremainder,theremainderof9iswhatwillgiveusthat0.12afterthedecimalplace.Theanswertothedivisionproblemx/yiseither:96remainder9Or96.12Therefore,whentheremainderof9isdividedbackovery,weget

0.12.Mathematically,thismeansthat:9/y=0.120.12y=912y=900y=900/12y=300/4y=75ThecorrectanswerisB.方法二：重修法Giventhataninteger"n"whendividedbyaninteger"a"gives"r"asreminderthentheinteger"n"canbewrittenasn=ak+rwherekisaconstantinteger

.Example1：WhatistheremainderwhenBisdividedby6ifBisapositiveinteger?WhenBisdividedby18,theremainderis3WhenBisdividedby12,theremainderis9Sol：STAT1:WhenBisdividedby18,theremainderis3So,wecanwriteBasB=18k+3Now,tocheckthereminderwhenBisdividedby6,weessentiallyneedtocheckthereminderwhen18k+3isdividedby618kgoeswith6sothereminderwill3So,itissufficientSTAT2:WhenBisdividedby12,theremainderis9So,wecanwriteBasB=12k+9Now,tocheckthereminderwhenBisdividedby6,weessentiallyneedtocheckthereminderwhen12k+9isdividedby612kgoeswith6sotheremainderwillbethesameasthereminderfor9dividedby6whichis3So,reminderis3So,itissufficient.AnswerwillbeDPractice:Whatistheremainderwhenpositiveintegertisdividedby5?(1)Whentisdividedby4,theremainderis1(2)Whentisdividedby3,theremainderis1這題請(qǐng)大家自己試一試哦。畢竟嘛。。?？吹某尚允遣蝗缱约鹤鲆活}的成效好哇大片的黃色地區(qū)都是解說哦。。。因此大家不用擔(dān)憂~~~試著做一做吧~~~↗

~~~下邊加油↖(^ω^)Sol：STAT1:Whentisdividedby4,theremainderis1t=4k+1possiblevaluesoftare1,5,9,13Clearlywecannotfindauniquereminderwhentisdividedby5asinsomecases(t=1)wearegettingthereminderas1andinsome(t=5)wearegettingthereminderas0.So,INSUFFICIENTSTAT2:Whentisdividedby3,theremainderis1t=3s+1possiblevaluesoftare1,4,7,10,13,16,19Clearlywecannotfindauniquereminderwhentisdividedby5asinsomecases(t=1)wearegettingthereminderas1andinsome(t=10)wearegettingthereminderas0.So,INSUFFICIENTSTAT1+STAT21.Writethe

：therearetwoapproachesvaluesoftfromstat1

and

then

from

stat2

andthen

take

thecommonvaluesFromSTAT1t=1,5,9,13,17,21,25,29,33FromSTAT2t=1,4,7,10,13,16,19,22,25,28,31,34Commonvaluesaret=1,13,25,2.Equatet=4k+1tot=3s+1Wehave4k+1=3s+1k=3s/4since,kisanintegersoonlythosevaluesofswhicharemultipleof4willsatisfybothSTAT1andSTAT2so,commonvaluesaregivenbyt=3s+1wheresismultipleof4sot=1,13,25(fors=0,4,8respectively)Clearlywecannotfindauniquereminderwhentisdividedby5asinsomecases(t=1)wearegettingthereminderas1andinsome(t=10)wearegettingthereminderas0.So,INSUFFICIENTSo,answerwillbeEExample2：Ifpandnarepositiveintegersandp>n,whatistheremainderwhenp^2-n^2isdividedby15?Theremainderwhenp+nisdividedby5is1.Theremainderwhenp-nisdividedby3is1Sol:STAT1:Theremainderwhenp+nisdividedby5is1.p+n=5k+1butwecannotsayanythingaboutp^2-n^2justfromthisinformation.So,INSUFFICIENTSTAT2:Theremainderwhenp-nisdividedby3is1p-n=3s+1butwecannotsayanythingaboutp^2-n^2justfromthisinformation.So,INSUFFICIENTSTAT1+STAT2：p^2-n^2=(p+n)*(p-n)=(5k+1)*(3s+1)=15ks+5k+3s+1Thereminderoftheaboveexpressionby15issameasthereminderof5k+3s+1with15as15kswillgowith15.Butwecannotsayanythingaboutthereminderasitsvaluewillchangewiththevaluesofkands.SoINSUFFICIENTHenceanswerwillbeEExample3：Ifnisapositiveintegerandristheremainderwhen4+7nisdividedby3,whatisthevalueofr?n+1isdivisibleby3n>20.Sol:ristheremainderwhen4+7nisdividedby37n+4canwewrittenas6n+n+3+1=3(2n+1)+n+1reminderof7n+4by3willbesameasreminderof3(2n+1)+n+1by33*(2n+1)willgoby3sothereminderwillbethesameasthereminderof(n+1)by3.STAT1:n+1isdivisibleby3n+1=3k(wherekisaninteger)n+1willgive0asthereminderwhendividedby3so,7n+4willalsogive0asthereminderwhenitsdividedby3(asitsreminderissameasthereminderfor(n+1)whendividedby3=>r=0So,SUFFICIENTSTAT2:n>20.wecannotdoanythingbythisinformationastherearemanyvaluesofnso,INSUFFICIENT.Hence,answerwillbeAPractice:Ifxisaninteger,isxbetween27and54?Theremainderwhenxisdividedby7is2.Theremainderwhenxisdividedby3is2.再次~~~請(qǐng)大家來考驗(yàn)下自己吧~~~O(∩_∩)O~~Sol:STAT1:Theremainderwhenxisdividedby7is2.x=7k+2Possiblevaluesofxare2,9,16,...,51,...wecannotsayanythingaboutthevaluesofxso,INSUFFICIENTSTAT2:Theremainderwhenxisdividedby3is2.x=3s+2Possiblevaluesofxare2,5,8,11,...,53,...wecannotsayanythingaboutthevaluesofxso,INSUFFICIENTSTAT1+STAT2：nowtherearetwoapproaches1.writethevaluesoftfromstat1andthenfromstat2andthentakethecommonvaluesfromSTAT1x=2,9,16,23,30,37,44,51,58,...,65,...fromSTAT2x=2,5,8,...,23,...,44,...,59,65,...commonvaluesarex=2,23,44,65,...2.equatex=7k+2tox=3s+2wehave7k+2=3s+2k=3s/7since,kisanintegersoonlythosevaluesofswhicharemultipleof7willsatisfybothSTAT1andSTAT2so,commonvaluesaregivenbyx=3s+2wheresismultipleof7sox=2,23,44,65(fors=0,7,14,21respectively)Clearlytherearevaluesofxwhicharebetween27and54(i.e.44)andthosewhicharenot(2,23,65)So,bothtogetheralsoINSUFFICIENTSo,answerwillbeE方法三：MOD法請(qǐng)大家拜見知之為之之大俠的帖子?。?！地點(diǎn)：<>RemainderQuestionPatternsBackgroundMostGMATremainderproblemsareencounteredindatasufficiencysection.lAllGMATremainderquestionsarelimitedtopositiveintegersonly.lBothnumberpluggingmethodandalgebraaresuitabletosolveremainderquestions.lSomeremainderquestionscanbedisguisedaswordproblems.Seebelow.lUsuallyyouget1,maximum2questionsonremaindersonthetest(basedonGMATPrepCATs)以下Pattern其實(shí)不是依據(jù)重要性次序來排的喲~~~Pattern6isthemostcommonpattern!妹紙感覺大家都能夠看一看。。。當(dāng)成練習(xí)吧~~~Pattern#1:Theratiooftwointegersisgivenandweareaskedtofindpossiblevalueoftheremainderwhenoneintegerisdividedbyanother

.Q1：OG13diagnostictest,question13Ifsandtarepositiveintegerssuchthats/t=64.12,whichofthefollowingcouldbetheremainderwhensisdividedbyt?2482045Sol：Sdividedbytyieldstheremainderofrcanalwaysbeexpressedas:s/t=q+r/t(whichisthesameass=qt+r),whereqisthequotientandristheremainder.Giventhats/t=64.12=64(12/100)=64(3/25)=64+3/25,soaccordingtotheabover/t=3/25,whichmeansthatrmustbeamultipleof3.OnlyoptionEoffersanswerwhichisamultipleof3Answer.E.Q2：OG13PracticeQuestions,question95Whenpositiveintegerxisdividedbypositiveintegery,theremainderis9.Ifx/y96.12,whatisthevalueofy?(A)96(B)75(C)48(D)25(E)12Sol：Whenpositiveintegerxisdividedbypositiveintegery,theremainderis9-->x=qy+9;x/y=96.12-->x=96y+0.12y(soqaboveequalsto96);0.12y=9-->y=75.Answer:Bfinding

the

remainder

when

an

expression

with

variable

is

divided

bysomeinteger

.OG13PracticeQuestions,question26Ifnisaprimenumbergreater

than

3,

what

is

the

remainder

when

n^2

isdividedby12?(A)0(B)1(C)2(D)3(E)5Sol：Thereareseveralalgebraicwaystosolvethisquestion,buttheeasiestwayisasfollows:sincewecannothavetwocorrectanswersjustpickaprimegreaterthan3,squareitandseewhatwouldbetheremainderupondivisionofitby12.n=5-->n^2=25-->remainderupondivision25by12is1.Answer:B.Pattern#3:min/maxquestioninvolvingremaindersWhenpositiveintegernisdividedby5,theremainderis1.Whennisdividedby7,theremainderis3.Whatisthesmallestpositiveintegerksuchthatk+nisamultipleof35.A.3B.4C.12D.32E.35Sol：Positiveintegernisdividedby5,theremainderis1-->n=5q+1,whereqisthequotient-->1,6,11,16,21,26,31,...Positiveintegernisdividedby7,theremainderis3-->n=7p+3,wherepisthequotient-->3,10,17,24,31,....Youcannotusethesamevariableforquotientsinbothformulas,becausequotientmaynotbethesameupondivisionnbytwodifferentnumbers.Forexample31/5,quotientq=6but31/7,quotientp=4.Thereisawaytoderivegeneralformulaforn(ofatypen=mx+r,wherexisdivisorandrisaremainder)basedonabovetwostatements:Divisorxwouldbetheleastcommonmultipleofabovetwodivisors5and7,hencex=35.Remainderrwouldbethefirstcommonintegerinabovetwopatterns,hencer=31.Thereforegeneralformulabasedonbothstatementsisn=35m+31.Thusthesmallestpositiveintegerksuchthatk+nisamultipleof35is4-->n+4=35k+31+4=35(k+1).Answer:BPattern#4:disguisedPSremainderproblem.Therearebetween100and110cardsinacollectionofcards.Iftheycountedout3atatime,thereare2leftover,butiftheyarecountedout4atatime,thereis1leftover.Howmanycardsareinthecollection?

are(A)101(B)103(C)106(D)107(E)109Sol：Ifthecardsarecountedout3atatime,thereare2leftover:x=3q+2.Fromthenumbersfrom100to110followingthreegivetheremainderof2upondivisionby3:101,104and107;Ifthecardsarecountedout4atatime,thereare1leftover:x=4p+1.Fromthenumbersfrom100to110followingthreegivetheremainderof1upondivisionby4:101,105and109.Sincex,thenumberofcards,shouldsatisfybothconditionsthenitequalsto101.Answer:APattern#5:weneedtoanswersomequestionaboutaninteger,whenthestatementsgiveinfoinvolvingremainders.Q1：OG13PracticeQuestions,question58Whatisthetensdigitofpositiveintegerx?xdividedby100hasaremainderof30.xdividedby110hasaremainderof30.Sol：(1)xdividedby100hasaremainderof30-->x=100q+30,soxcanbe:30,130,230,...Eachhasthetensdigitof3.Sufficient.xdividedby110hasaremainderof30-->x=110p+30,soxcanbe:30,250,...Wealreadyhavetwovaluesforthetensdigit.Notsufficient.Answer:A.Q2：OG13PracticeQuestions,question83Ifkisanintegersuchthat56<k<66,whatisthevalueofk?Ifkweredividedby2,theremainderwouldbe1.Ifk+1weredividedby3,theremainderwouldbe0.Sol：(1)Ifkweredividedby2,theremainderwouldbe1-->kisanoddnumber,thusitcouldbe57,59,61,63,or65.Notsufficient.Ifk+1weredividedby3,theremainderwouldbe0-->kis1lessthanamultipleof3,thusitcouldbe59,62,or65.Notsufficient.(1)+(2)kcouldstilltakemorethanonevalue:59or65.Notsufficient.Answer:E.Pattern#6:weneedtofindtheremainderwhensomevariableoranexpressionwithvariable(s)isdividedbysomeinteger.Usuallythestatementsgivedivisibility/remainderinfo.Mostcommonpatter.Q1：Whatistheremainderwhenthepositiveintegernisdividedby6?nismultipleof5nisamultipleof12Sol：nismultipleof5.Ifn=5,thennyieldstheremainderof5whendividedby6butifn=10,thennyieldstheremainderof4whendividedby6.Wealreadyhavetwodifferentanswers,whichmeansthatthisstatementisnotsufficient.nisamultipleof12.Everymultipleof12isalsoamultipleof6,thusndividedby6yieldstheremainderof0.Sufficient.Answer:B.Q2：Ifxandyareinteger,whatistheremainderwhenx^2+y^2isdividedby5?Whenx-yisdividedby5,theremainderis1Whenx+yisdividedby5,theremainderis2Sol：(1)Whenx-yisdividedby5,theremainderis1-->x-y=5q+1,sox-ycanbe1,6,11,...Now,x=2andy=1(x-y=1)thenx^2+y^2=5andthustheremainderis0,butifx=3andy=2(x-y=1)thenx^2+y^2=13andthustheremainderis3.Notsufficient.(2)Whenx+yisdividedby5,theremainderis2-->x+y=5p+2,sox+ycanbe2,7,12,...Now,x=1andy=1(x+y=2)thenx^2+y^2=2andthustheremainderis2,butifx=5andy=2(x+y=7)thenx^2+y^2=29andthustheremainderis4.Notsufficient.(1)+(2)Squarebothexpressions:x^2-2xy+y^2=25q^2+10q+1andx^2+2xy+y^2=25p^2+20p+4-->addthemup:2(x^2+y^2)=5(5q^2+2q+5p^2+4p+1)-->so2(x^2+y^2)isdivisibleby5(remainder0),whichmeansthatsoisx^2+y^2.Sufficient.Answer:C.Q3：Iftisapositiveintegerandristheremainderwhent^2+5t+6isdividedby7,whatisthevalueofr?Whentisdividedby7,theremainderis6.Whent^2isdividedby7,theremainderis1.Sol：Firstofallfactort^2+5t+6-->t^2+5t+6=(t+2)(t+3)(1)Whentisdividedby7,theremainderis6

-->t=7q+6

-->(t+2)(t+3)=(7q+8)(7q+9).Now,noneedtoexpandandmultiplyalltheterms,justnoticethatwhenweexpandalltermsbutthelastone,whichwillbe8*9=72,willhave7asafactorand72yieldstheremainderof2upondivisionby7.Sufficient.(2)Whent^2isdividedby7,theremainderis1-->differentvaluesoftpossible:forexamplet=1ort=6,whichwhen

substituted

in

(t+2)(t+3)

will

yielddifferentremainderupondivisionby7.Notsufficient.Answer:A.Q4：Ifpisapositiveoddinteger,whatistheremainderwhenpisdividedby4?:Whenpisdividedby8,theremainderis5.pisthesumofthesquaresoftwopositiveintegers.Sol：(1)Whenpisdividedby8,theremainderis5-->p=8q+5=(8q+4)+1=4(2q+1)+1-->sotheremainderupondivisionofpby4is1(sincefirsttermisdivisibleby4andsecondtermyieldsremainderof1upon

division

by4).Sufficient.pisthesumofthesquaresoftwopositiveintegers-->sincepisanoddintegerthenoneoftheintegersmustbeevenandanotherodd:p=(2n)^2+(2m+1)^2=4n^2+4m^2+4m+1=4(n^2+m^2+m)+1-->thesamewayasabove:theremainderupondivisionofpby4is1(sincefirsttermisdivisibleby4andsecondtermyieldsremainderof1upondivisionby4).Sufficient.Answer:D.Q5：Ifnisapositiveintegerandristheremainderwhen(n-1)(n+1)isdividedby24,whatisthevalueofr?nisnotdivisibleby2nisnotdivisibleby3Sol：Plug-inmethod:(n-1)(n+1)=n^2-1nisnotdivisibleby2-->picktwooddnumbers:let'ssay1and3-->ifn=1,thenn^2-1=0andaszeroisdivisibleby24(zeroisdivisiblebyanyintegerexceptzeroitself)soremainderis0butifn=3,thenn^2-1=8and8dividedby24yieldsremainderof8.Twodifferentanswers,hencenotsufficient.(2)nisnotdivisibleby3-->picktwonumberswhicharenotdivisibleby3:let'ssay1and2-->ifn=1,thenn^2-1=0,soremainderis0butifn=2,thenn^2-1=3and3dividedby24yieldsremainderof3.Twodifferentanswers,hencenotsufficient.(1)+(2)Let'scheckforseveralnumberswhicharenotdivisibleby2or3:n=1-->n^2-1=0-->remainder0;n=5-->n^2-1=24-->remainder0;n=7-->n^2-1=48-->remainder0;n=11-->n^2-1=120-->remainder0.Wellitseemsthatallappropriatenumberswillgiveremainderof0.Sufficient.Algebraicapproach:nisnotdivisibleby2.Insufficientonitsown,butthisstatementsaysthatn=odd-->n-1andn+1areconsecutiveevenintegers-->(n-1)(n+1)mustbedivisibleby8(asbothmultiplesareevenandoneofthemwillbedivisibleby4.Fromconsecutiveevenintegersoneisdivisibleby4:(2,4);(4,6);(6,8);(8,10);(10,12),...).nisnotdivisibleby3.Insufficientonitsown,butformthisstatementeithern-1orn+1mustbedivisibleby3(asn-1,n,andn+1areconsecutiveintegers,sooneofthemmustbedivisibleby3,wearetoldthatit'snotn,henceeithern-1orn+1).(1)+(2)From(1)(n-1)(n+1)isdivisibleby8,from(2)it'salsodivisibleby3,thereforeitmustbedivisibleby8*3=24,whichmeansthatremainderupondivision(n-1)(n+1)by24willbe0.Sufficient.Answer:C.Pattern#7:disguisedDSremainderproblem.Apersoninheritedfewgoldcoinsfromhisfather.Ifheput9coinsineachbagthen7coinsareleftover.Howeverifheputs7coinsineachbagthen3coinsareleftover.Whatisthenumberofcoinsheinheritedfromhisfather?Thenumberofcoinsliesbetween50to120.Ifheput13coinsinonebagthennocoinisleftoverandnumberofcoinsbeinglesserthan200.Sol：Ifheputs9coinsineachbagthen7coinsareleftover-->c=9q+7,so#ofcoinscanbe:7,16,25,34,43,52,61,...Ifheputs7coinsineachbagthen3coinsareleftover-->c=7p+3,so#ofcoinscanbe:3,10,17,24,31,38,45,52,59,...Generalformulaforcbasedonabovetwostatementswillbe:c=63k+52(thedivisorshouldbetheleastcommonmultipleofabovetwodivisors9and7,so63andtheremaindershouldbethefirstcommonintegerinabovetwopatterns,hence52).C=63k+52meansthat#ofcoinscanbe:52,115,178,241,...(1)Thenumberofcoinsliesbetween50to120-->#ofcoinscanbe52or115.Notsufficient.Ifheput13coinsinonebagthennocoinisleftoverandnumberofcoinsbeinglesserthan200-->#ofcoinsisamultipleof13andlessthan200:only52satisfiesthiscondition.Sufficient.Answer:B.Pattern#8:C-Trapremainderproblem."Ctrap"isaproblemwhichisVERYOBVIOUSLYsufficientifbothstatementsaretakentogether.WhenyouseesuchquestionyoushouldbeextremelycautiouswhenchoosingCforananswer.Ifaandbarepositiveintegers,whatistheremainderwhen4^(2a+1+b)isdividedby10?(1)a=1(2)b=2Sol：4inpositiveintegerpowercanhaveonly2lastdigits:4,whenthepowerisoddor6whenthepoweriseven.Hence,togettheremainderof4^x/10weshouldknowwhetherthepowerisoddoreven:ifit'soddtheremainderwillbe4andifit'seventheremainderwillbe6.a=1-->4^(2a+1+b)=4^(3+b)dependingonbthepowercanbeevenorodd.Notsufficient.(2)b=2-->4^(2a+1+b)=4^(2a+3)=4^(even+odd)=4^odd-->theremainderupondivisionof4^oddby10is4.Sufficient.Answer:B.CollectionofGMATRemainderProblems這些題目妹紙感覺仍是自己做一遍會(huì)很實(shí)用哦?。?！因此妹紙把剖析所有隱蔽在黃色highlight里面啦。。。大家加油~~~~1.Ifristheremainderwhenthepositiveintegernisdividedby7,whatisthevalueofrWhennisdividedby21,theremainderisanoddnumberWhennisdividedby28,theremainderis3Sol：Thepossiblereminderscanbe1,2,3,4,5and6.Wehavethepinpointtheexactremainderfromthis6numbers.St1:whennisdividedby21(7and3)theremainderisanoddnumber.Butitcannotbe7,3or9.Hencethepossibilitiesare:1and5.Hencetherecanbetworemainders,1and5,whendividedby7.NOTSUFFICIENTSt2:whennisdividedby28theremainderis3.As7isafactorof28,theremainderwhendividedby7willbe3SUFFICIENTAnswer:B2.Ifnandmarepositiveintegers,whatistheremainderwhen3^(4n+2+m)isdividedby10?n=2m=1Sol：TheConcepttestedhereiscyclesofpowersof3.Thecyclesofpowersof3are:3,9,7,1StI)n=2.Thismakes3^(4*2+2+m)=3^(10+m).wedonotknowmandhencecannotfigureouttheunitdigit.StII)m=1.Thismakes3^(4*n+2+1).4ncanbe4,8,12,16...3^(4*n+2+1)willbe3^7,3^11,3^15,3^19.....ineachcasetheunitdigitwillbe7.SUFFHenceB3.Ifpisapositiveoddinteger,whatistheremainderwhenpisdividedby4?Whenpisdividedby8,theremainderis5.pisthesumofthesquaresoftwopositiveintegers.Sol：st1.takemultiplesof8....dividethemby4...remainder=1ineachcase...st2.pisodd,sincepissquareof2integers...onewillbeodd....nowwhenwedivideanyevensquareby4vllgt0remainderdivideoddsquarevllget1asremainder......sointotalremainder=1Ans:D

even

andother..andwhen4.Ifpandnarepositiveintegersandp>n,whatistheremainderwhenp^2-n^2isdividedby15?Theremainderwhenp+nisdividedby5is1.Theremainderwhenp-nisdivide