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Chapter 1 How not to think about numbers（第2页）

&hesinumbers，wehaveourberbase10=2×5，whichisspeethelessbeingtriangularinthat10=1+2+3+4（rememberten-pinbowlihenhaveapairoftwinprimesin11and13，whicharetwosebersthatarebothprime，separatedbythenumber12，whitrasthasmanyfactorsforitssize.Ihefirstso-calledabundahethesumofitsproperfactors，thoselessthantheself，exceedsthenumberiion:1+2+3+4+6=16.Thenumber14=2×7maylookundisti，astheparadoxicalquipgoes，beiundistinguishednumbermakesitdistierall.In15=3×5，wehaveariangularisthefirstoddistheproductoftwoproperfactors.Ofcourse，16=24isnotonlyasquarebutthefirstfourthpower（after1），makingitveryspedeed.Thepair17aherpairoftwinprimes，ahereadertomaketheirowionsaboutthepeatureofthenumbers18，20，andsoon.Foreaakeae.

&otheprimes，thefirsttwentyofthemare:

2，3，5，7，11，13，17，19，23，29，31，37，41，43，47，53，59，61，67，71.

eartheverybeginningofthenumbersequence，primesareohereislittleopportunityforsmallohavefactorizatioheprimesbeerarer.Forexample，thereisoripleofsees:thetrio3，5，7isunique，aseverythirdoddnumberisamultipleof3，andsothiseverhappenagaihinningproeoceis，however，quiteleisurelyandsurprisiiple，thethirtieshaveonlytwoprimes，thosebeing31aimmediatelyafter100therearetwo‘secutive’pairsoftwinprimesin101，103and107，109.

Theprimeshavebeenasourceoffasforthousandsofyearsbecausethey（aclaimthatweshalljustifyier）yettheyariseamouralnumbersihaphazardfashion.ThismysteriousaablefacetoftheirnatureisexploitediographytosafeguardtialunitheI，whichisthesubjectofChapter4.

gforprimality:primedivisibilitytests

&simple-mindedwayoffindingalltheprimesuptoagivennumbersuchas100istowriteallthenumbersdownandcrossoffthebersasyoufiahodbasedonthisideaiscalledtheSieveofEratosthenesandrunsasfollows.Beginbyg2andthencrossoffallthemultiplesof2（theotherevennumbers）inyourlist.Thehebeginninumberyoumeetthathasnotbeencrossedoff（whichwillbe3）andthencrossoffallitsmultiplesintheremaininglist.Byrepeatingthisprocesssuffitlyoften，theprimeswillemergeasthosecrossedout，althoughsomewillbedsomenot.Forexample，Figure1showsthewsofthesieveupto60.

Howdoyouknowwhenyoustopsieviorepeatthisprotilyouumberthatisgreaterthanthesquarerootestnumberinyourlist.Forinstance，ifyoudoyourownsieveforallo120，youwillhthesieveformultiplesof2，3，5，and7，andwhenyoucircle11youstop，as112=121.Atthatpoint，youwillhavecircledasfarasthefirstprimeexgthesquareroestnumber（120inthiscase）withtheremaiiouched.Allberswillnowhavebeeaseachisamultipleofoneormoreof2，3，5，and7.

1.Primesieve:theprimesupto60arethecrossedout

Itisveryeasytotestfordivisibilityby2andby5astheseprimesaretheprimefaberbasetehis，youoochealdigitofthenumberion:nisdivisibleby2exaitsunitsdigitiseven（i.e.0，2，4，6，or8），andnhas5asafadonlyifitendsin0or5.Nomatterhowmanydigitsthenumbernhas，weoocheckthelastdigittodetermiherleof2orof5.Forprimesthatdoo10，weodoabitmoreworkbutherearesimpletestsfordivisibilitythataremuchquirestodoingthefulldivisionsum.

Anumberisdivisibleby3ifandonlyifthesameistrueofthesumofitsdigits.Forexample，thesumofthedigitsofn=145，373，270，099，876，790is87and87=3×29andsonisinthiscasedivisibleby3.Ofcourse，lythetesttotheselfaakingthesumofdigitsoftheouteateachstageuisobvious.Doingthisfivenexampleproducesthefollowingsequence:

145，373，270，099，876，790→87→15→6=2×3.

Youwillseethatallthedivisioedherearesoquickthatyoudlehdozensofdigitswithrelativeeaseeventhoughthesenumbersarebillioerthahwhichyourhandcalculatorcope.

&sgiveheremaio20arebecausetheyareallofthesamegeheseroutinesareallsimpletoapply，althoughitislessobviouswhytheywhthejustifisarenotrecordedhere，theproofsoftheirvalidityarenotespeciallydifficult.

n=27，916，924→2，791，684→279，160→27，916→2，779→259→7

andsonisdivisibleby7.Eachtimewerunthroughtheloopofinstrus，weloseatleasto，sothenumberofpassesthroughtheloopisaboutthesameasthelengthofthehwhichwebegin.

&herornotnhasafactorof11，subtraaldigitfrtrunumbera.Forexample，thenumberisamultipleof11asourmethodreveals:

4，959，746→495，968→49，588→4，950→495→44=4×11.

Tocheckfordivisibilityby13，addfourtimesthefinaldigitttrunumberah7and11.Foriheurnsouttohave13asosprimefactors:

11，264，331→1，126，437→112，671→11，271

→1131→117→39=3×13.For17andfor19，wesubtractfivetimesthefinaldigitinthecaseof17，andaddtwialdigitwheingif19isafaoreapplyingthissteptothetrunumberthatremaiheprocessasoftenasweneed.Forexample，wetest18，905fordivisibilityby17:

18，905→1，865→161→11

soitisnotamultipleof17，butfor19，thetestgivestheopposite:

18，905→1，900=100×19.

&hisbatteryoftests，youreadilychecktheprimalityofallo500（as232=529exceeds500，so19isthelargestpotentialprimefactorthatyouneedyourselfwith）.Forexample，tosettlethematterfor247，wejustocheckfordivisibilityuptotheprime13（asthesquareoftheprime，172=289，exceeds247）.Applyifor13，however，welearnfrom247→（24+28）=52→13，thatleof13:（247=19×13）.

Thedivisibilitytestsforprimesountediofurnishdivisibilitytestsforthosearesquare-freeproductsoftheseprimes（divisiblebythesquareofanyprime）suchas42=2×3×7:anumbernwillbedivisibleby42exarioofdivisibilitytestsfor2，3，asforthosehavesquarefactors，suchas9=32，doically，althoughitisthecasethatnhas9asafadonlyifthatistrueofthesumofthedigitsofn.

Youmightask，afterthousandsofyears，haven’tthoseclevermathematieupwithbetteraicatedmethodsoftestingforprimality?Theanswerisyes.In2002，arelativelyquickwaywasdiscoveredtotestifagivennumberisprime.Theso-called‘AKSprimalitytest’doesnot，however，providethefactorizationofthegivehappee.Theproblemoffindingtheprimefactivehoughinprinciplesolvablebytrial，stillseemspractitraelylargeintegers，andforthatreasonitformsthebasisofmuaryeno，asubjecttoillreturninChapter4.Beforethatweshall，iters，lookalittlemorecloselyatprimesandfactorization.