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project euler problems
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5/2/2015 Archived Problems - Project Euler
https://projecteuler.net/show=all 1/163
Problem1:Multiplesof3and5
Ifwelistallthenaturalnumbersbelow10thataremultiplesof3or5,weget3,5,6and9.Thesumofthesemultiplesis23.
Findthesumofallthemultiplesof3or5below1000.
Problem2:EvenFibonaccinumbers
EachnewtermintheFibonaccisequenceisgeneratedbyaddingtheprevioustwoterms.Bystartingwith1and2,thefirst10termswillbe:
1,2,3,5,8,13,21,34,55,89,...
ByconsideringthetermsintheFibonaccisequencewhosevaluesdonotexceedfourmillion,findthesumoftheevenvaluedterms.
Problem3:Largestprimefactor
Theprimefactorsof13195are5,7,13and29.
Whatisthelargestprimefactorofthenumber600851475143?
Problem4:Largestpalindromeproduct
Apalindromicnumberreadsthesamebothways.Thelargestpalindromemadefromtheproductoftwo2digitnumbersis9009=9199.
Findthelargestpalindromemadefromtheproductoftwo3digitnumbers.
Problem5:Smallestmultiple
2520isthesmallestnumberthatcanbedividedbyeachofthenumbersfrom1to10withoutanyremainder.
Whatisthesmallestpositivenumberthatisevenlydivisiblebyallofthenumbersfrom1to20?
Problem6:Sumsquaredifference
Thesumofthesquaresofthefirsttennaturalnumbersis,
12+22+...+102=385
Thesquareofthesumofthefirsttennaturalnumbersis,
(1+2+...+10)2=552=3025
Hencethedifferencebetweenthesumofthesquaresofthefirsttennaturalnumbersandthesquareofthesumis3025385=2640.
Findthedifferencebetweenthesumofthesquaresofthefirstonehundrednaturalnumbersandthesquareofthesum.
Problem7:10001stprime
Bylistingthefirstsixprimenumbers:2,3,5,7,11,and13,wecanseethatthe6thprimeis13.
Whatisthe10001stprimenumber?
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Problem8:Largestproductinaseries
Thefouradjacentdigitsinthe1000digitnumberthathavethegreatestproductare9989=5832.
7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450
Findthethirteenadjacentdigitsinthe1000digitnumberthathavethegreatestproduct.Whatisthevalueofthisproduct?
Problem9:SpecialPythagoreantriplet
APythagoreantripletisasetofthreenaturalnumbers,a
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Whatisthegreatestproductoffouradjacentnumbersinthesamedirection(up,down,left,right,ordiagonally)inthe2020grid?
Problem12:Highlydivisibletriangularnumber
Thesequenceoftrianglenumbersisgeneratedbyaddingthenaturalnumbers.Sothe7thtrianglenumberwouldbe1+2+3+4+5+6+7=28.Thefirsttentermswouldbe:
1,3,6,10,15,21,28,36,45,55,...
Letuslistthefactorsofthefirstseventrianglenumbers:
1:13:1,36:1,2,3,610:1,2,5,1015:1,3,5,1521:1,3,7,2128:1,2,4,7,14,28
Wecanseethat28isthefirsttrianglenumbertohaveoverfivedivisors.
Whatisthevalueofthefirsttrianglenumbertohaveoverfivehundreddivisors?
Problem13:Largesum
Workoutthefirsttendigitsofthesumofthefollowingonehundred50digitnumbers.
37107287533902102798797998220837590246510135740250463769376774900097126481248969700780504170182605387432498619952474105947423330951305812372661730962991942213363574161572522430563301811072406154908250230675882075393461711719803104210475137780632466768926167069662363382013637841838368417873436172675728112879812849979408065481931592621691275889832738442742289174325203219235894228767964876702721893184745144573600130643909116721685684458871160315327670386486105843025439939619828917593665686757934951621764571418565606295021572231965867550793241933316490635246274190492910143244581382266334794475817892575867718337217661963751590579239728245598838407582035653253593990084026335689488301894586282278288018119938482628201427819413994056758715117009439035398664372827112653829987240784473053190104293586865155060062958648615320752733719591914205172558297169388870771546649911559348760353292171497005693854370070576826684624621495650076471787294438377604532826541087568284431911906346940378552177792951453612327252500029607107508256381565671088525835072145876576172410976447339110607218265236877223636045174237069058518606604482076212098132878607339694128114266041808683061932846081119106155694051268969251934325451728388641918047049293215058642563049483624672216484350762017279180399446930047329563406911573244438690812579451408905770622942919710792820955037687525678773091862540744969844508330393682126183363848253301546861961243487676812975343759465158038628759287849020152168555482871720121925776695478182833757993103614740356856449095527097864797581167263201004368978425535399209318374414978068609844840309812907779179908821879532736447567559084803087086987551392711854517078544161852424320693150332599594068957565367821070749269665376763262354472106979395067965269474259770973916669376304263398708541052684708299085211399427365734116182760315001271653786073615010808570091499395125570281987460043753582903531743471732693212357815498262974255273730794953759765105305946966067683156574377167401875275889028025717332296191766687138199318110487701902712526768027607800301367868099252546340106163286652636270218540497705585629946580636237993140746255962240744869082311749777923654662572469233228109171419143028819710328859780666976089293863828502533340334413065578016127815921815005561868836468420090470230530811728164304876237919698424872550366387845831148769693215490281042402013833512446218144177347063783299490636259666498587618221225225512486764533677201869716985443124195724099139590089523100588229554825530026352078153229679624948164195386821877476085327132285723110424803456124867697064507995236377742425354112916842768655389262050249103265729672370191327572567528565324825826546309220705859652229798860272258331913126375147341994889534765745501
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184957014548792889848568277260777137214037988797153829820378303147352772158034814451349137322665138134829543829199918180278916522431027392251122869539409579530664052326325380441000596549391598795936352974615218550237130764225512118369380358038858490341698116222072977186158236678424689157993532961922624679571944012690438771072750481023908955235974572318970677254791506150550495392297953090112996751986188088225875314529584099251203829009407770775672113067397083047244838165338735023408456470580773088295917476714036319800818712901187549131054712658197623331044818386269515456334926366572897563400500428462801835170705278318394258821455212272512503275512160354698120058176216521282765275169129689778932238195734329339946437501907836945765883352399886755061649651847751807381688378610915273579297013376217784275219262340194239963916804498399317331273132924185707147349566916674687634660915035914677504995186714302352196288948901024233251169136196266227326746080059154747183079839286853520694694454072476841822524674417161514036427982273348055556214818971426179103425986472045168939894221798260880768528778364618279934631376775430780936333301898264209010848802521674670883215120185883543223812876952786713296124747824645386369930090493103636197638780396218407357239979422340623539380833965132740801111666627891981488087797941876876144230030984490851411606618262936828367647447792391803351109890697907148578694408955299065364044742557608365997664579509666024396409905389607120198219976047599490197230297649139826800329731560371200413779037855660850892521673093931987275027546890690370753941304265231501194809377245048795150954100921645863754710598436791786391670211874924319957006419179697775990283006991536871371193661495281130587638027841075444973307840789923115535562561142322423255033685442488917353448899115014406480203690680639606723221932041495354150312888033953605329934036800697771065056663195481234880673210146739058568557934581403627822703280826165707739483275922328459417065250945123252306082291880205877731971983945018088807242966198081119777158542502016545090413245809786882778948721859617721078384350691861554356628840622574736922845095162084960398013400172393067166682355524525280460972253503534226472524250874054075591789781264330331690
Problem14:LongestCollatzsequence
Thefollowingiterativesequenceisdefinedforthesetofpositiveintegers:
nn/2(niseven)n3n+1(nisodd)
Usingtheruleaboveandstartingwith13,wegeneratethefollowingsequence:
134020105168421
Itcanbeseenthatthissequence(startingat13andfinishingat1)contains10terms.Althoughithasnotbeenprovedyet(CollatzProblem),itisthoughtthatallstartingnumbersfinishat1.
Whichstartingnumber,underonemillion,producesthelongestchain?
NOTE:Oncethechainstartsthetermsareallowedtogoaboveonemillion.
Problem15:Latticepaths
Startinginthetopleftcornerofa22grid,andonlybeingabletomovetotherightanddown,thereareexactly6routestothebottomrightcorner.
Howmanysuchroutesaretherethrougha2020grid?
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Problem16:Powerdigitsum
215=32768andthesumofitsdigitsis3+2+7+6+8=26.
Whatisthesumofthedigitsofthenumber21000?
Problem17:Numberlettercounts
Ifthenumbers1to5arewrittenoutinwords:one,two,three,four,five,thenthereare3+3+5+4+4=19lettersusedintotal.
Ifallthenumbersfrom1to1000(onethousand)inclusivewerewrittenoutinwords,howmanyletterswouldbeused?
NOTE:Donotcountspacesorhyphens.Forexample,342(threehundredandfortytwo)contains23lettersand115(onehundredandfifteen)contains20letters.Theuseof"and"whenwritingoutnumbersisincompliancewithBritishusage.
Problem18:MaximumpathsumI
Bystartingatthetopofthetrianglebelowandmovingtoadjacentnumbersontherowbelow,themaximumtotalfromtoptobottomis23.
374
2468593
Thatis,3+7+4+9=23.
Findthemaximumtotalfromtoptobottomofthetrianglebelow:
759564
17478218358710
2004824765190123750334
880277730763679965042806167092
41412656834080703341487233473237169429
5371446525439152975114701133287773177839681757
917152381714914358502729486366046889536730731669874031
046298272309709873933853600423
NOTE:Asthereareonly16384routes,itispossibletosolvethisproblembytryingeveryroute.However,Problem67,isthesamechallengewithatrianglecontainingonehundredrowsitcannotbesolvedbybruteforce,andrequiresaclevermethod!o)
Problem19:CountingSundays
Youaregiventhefollowinginformation,butyoumayprefertodosomeresearchforyourself.
1Jan1900wasaMonday.ThirtydayshasSeptember,April,JuneandNovember.Alltheresthavethirtyone,SavingFebruaryalone,Whichhastwentyeight,rainorshine.Andonleapyears,twentynine.Aleapyearoccursonanyyearevenlydivisibleby4,butnotonacenturyunlessitisdivisibleby400.
HowmanySundaysfellonthefirstofthemonthduringthetwentiethcentury(1Jan1901to31Dec2000)?
Problem20:Factorialdigitsum
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n!meansn(n1)...321
Forexample,10!=109...321=3628800,andthesumofthedigitsinthenumber10!is3+6+2+8+8+0+0=27.
Findthesumofthedigitsinthenumber100!
Problem21:Amicablenumbers
Letd(n)bedefinedasthesumofproperdivisorsofn(numberslessthannwhichdivideevenlyinton).Ifd(a)=bandd(b)=a,whereab,thenaandbareanamicablepairandeachofaandbarecalledamicablenumbers.
Forexample,theproperdivisorsof220are1,2,4,5,10,11,20,22,44,55and110therefored(220)=284.Theproperdivisorsof284are1,2,4,71and142sod(284)=220.
Evaluatethesumofalltheamicablenumbersunder10000.
Problem22:Namesscores
Usingnames.txt(rightclickand'SaveLink/TargetAs...'),a46Ktextfilecontainingoverfivethousandfirstnames,beginbysortingitintoalphabeticalorder.Thenworkingoutthealphabeticalvalueforeachname,multiplythisvaluebyitsalphabeticalpositioninthelisttoobtainanamescore.
Forexample,whenthelistissortedintoalphabeticalorder,COLIN,whichisworth3+15+12+9+14=53,isthe938thnameinthelist.So,COLINwouldobtainascoreof93853=49714.
Whatisthetotalofallthenamescoresinthefile?
Problem23:Nonabundantsums
Aperfectnumberisanumberforwhichthesumofitsproperdivisorsisexactlyequaltothenumber.Forexample,thesumoftheproperdivisorsof28wouldbe1+2+4+7+14=28,whichmeansthat28isaperfectnumber.
Anumberniscalleddeficientifthesumofitsproperdivisorsislessthannanditiscalledabundantifthissumexceedsn.
As12isthesmallestabundantnumber,1+2+3+4+6=16,thesmallestnumberthatcanbewrittenasthesumoftwoabundantnumbersis24.Bymathematicalanalysis,itcanbeshownthatallintegersgreaterthan28123canbewrittenasthesumoftwoabundantnumbers.However,thisupperlimitcannotbereducedanyfurtherbyanalysiseventhoughitisknownthatthegreatestnumberthatcannotbeexpressedasthesumoftwoabundantnumbersislessthanthislimit.
Findthesumofallthepositiveintegerswhichcannotbewrittenasthesumoftwoabundantnumbers.
Problem24:Lexicographicpermutations
Apermutationisanorderedarrangementofobjects.Forexample,3124isonepossiblepermutationofthedigits1,2,3and4.Ifallofthepermutationsarelistednumericallyoralphabetically,wecallitlexicographicorder.Thelexicographicpermutationsof0,1and2are:
012021102120201210
Whatisthemillionthlexicographicpermutationofthedigits0,1,2,3,4,5,6,7,8and9?
Problem25:1000digitFibonaccinumber
TheFibonaccisequenceisdefinedbytherecurrencerelation:
Fn=Fn1+Fn2,whereF1=1andF2=1.
Hencethefirst12termswillbe:
F1=1
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F2=1F3=2F4=3F5=5F6=8F7=13F8=21F9=34F10=55F11=89F12=144
The12thterm,F12,isthefirsttermtocontainthreedigits.
WhatistheindexofthefirsttermintheFibonaccisequencetocontain1000digits?
Problem26:Reciprocalcycles
Aunitfractioncontains1inthenumerator.Thedecimalrepresentationoftheunitfractionswithdenominators2to10aregiven:
1/2 = 0.51/3 = 0.(3)1/4 = 0.251/5 = 0.21/6 = 0.1(6)1/7 = 0.(142857)1/8 = 0.1251/9 = 0.(1)1/10 = 0.1
Where0.1(6)means0.166666...,andhasa1digitrecurringcycle.Itcanbeseenthat1/7hasa6digitrecurringcycle.
Findthevalueofd
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21222324252078910196121118543121716151413
Itcanbeverifiedthatthesumofthenumbersonthediagonalsis101.
Whatisthesumofthenumbersonthediagonalsina1001by1001spiralformedinthesameway?
Problem29:Distinctpowers
Considerallintegercombinationsofabfor2a5and2b5:
22=4,23=8,24=16,25=3232=9,33=27,34=81,35=24342=16,43=64,44=256,45=102452=25,53=125,54=625,55=3125
Iftheyarethenplacedinnumericalorder,withanyrepeatsremoved,wegetthefollowingsequenceof15distinctterms:
4,8,9,16,25,27,32,64,81,125,243,256,625,1024,3125
Howmanydistincttermsareinthesequencegeneratedbyabfor2a100and2b100?
Problem30:Digitfifthpowers
Surprisinglythereareonlythreenumbersthatcanbewrittenasthesumoffourthpowersoftheirdigits:
1634=14+64+34+44
8208=84+24+04+84
9474=94+44+74+44
Thesumofthesenumbersis1634+8208+9474=19316.
Findthesumofallthenumbersthatcanbewrittenasthesumoffifthpowersoftheirdigits.
Problem31:Coinsums
InEnglandthecurrencyismadeupofpound,,andpence,p,andthereareeightcoinsingeneralcirculation:
1p,2p,5p,10p,20p,50p,1(100p)and2(200p).
Itispossibletomake2inthefollowingway:
11+150p+220p+15p+12p+31p
Howmanydifferentwayscan2bemadeusinganynumberofcoins?
Problem32:Pandigitalproducts
Weshallsaythatanndigitnumberispandigitalifitmakesuseofallthedigits1tonexactlyonceforexample,the5digitnumber,15234,is1through5pandigital.
Theproduct7254isunusual,astheidentity,39186=7254,containingmultiplicand,multiplier,andproductis1through9pandigital.
Findthesumofallproductswhosemultiplicand/multiplier/productidentitycanbewrittenasa1through9pandigital.
HINT:Someproductscanbeobtainedinmorethanonewaysobesuretoonlyincludeitonceinyoursum.
As1=14isnotasumitisnotincluded.
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Problem33:Digitcancellingfractions
Thefraction49/98isacuriousfraction,asaninexperiencedmathematicianinattemptingtosimplifyitmayincorrectlybelievethat49/98=4/8,whichiscorrect,isobtainedbycancellingthe9s.
Weshallconsiderfractionslike,30/50=3/5,tobetrivialexamples.
Thereareexactlyfournontrivialexamplesofthistypeoffraction,lessthanoneinvalue,andcontainingtwodigitsinthenumeratoranddenominator.
Iftheproductofthesefourfractionsisgiveninitslowestcommonterms,findthevalueofthedenominator.
Problem34:Digitfactorials
145isacuriousnumber,as1!+4!+5!=1+24+120=145.
Findthesumofallnumberswhichareequaltothesumofthefactorialoftheirdigits.
Problem35:Circularprimes
Thenumber,197,iscalledacircularprimebecauseallrotationsofthedigits:197,971,and719,arethemselvesprime.
Therearethirteensuchprimesbelow100:2,3,5,7,11,13,17,31,37,71,73,79,and97.
Howmanycircularprimesaretherebelowonemillion?
Problem36:Doublebasepalindromes
Thedecimalnumber,585=10010010012(binary),ispalindromicinbothbases.
Findthesumofallnumbers,lessthanonemillion,whicharepalindromicinbase10andbase2.
Problem37:Truncatableprimes
Thenumber3797hasaninterestingproperty.Beingprimeitself,itispossibletocontinuouslyremovedigitsfromlefttoright,andremainprimeateachstage:3797,797,97,and7.Similarlywecanworkfromrighttoleft:3797,379,37,and3.
Findthesumoftheonlyelevenprimesthatarebothtruncatablefromlefttorightandrighttoleft.
Problem38:Pandigitalmultiples
Takethenumber192andmultiplyitbyeachof1,2,and3:
1921=1921922=3841923=576
Byconcatenatingeachproductwegetthe1to9pandigital,192384576.Wewillcall192384576theconcatenatedproductof192and(1,2,3)
Thesamecanbeachievedbystartingwith9andmultiplyingby1,2,3,4,and5,givingthepandigital,918273645,whichistheconcatenatedproductof9and(1,2,3,4,5).
Whatisthelargest1to9pandigital9digitnumberthatcanbeformedastheconcatenatedproductofanintegerwith(1,2,...,n)wheren>1?
Note:as1!=1and2!=2arenotsumstheyarenotincluded.
(Pleasenotethatthepalindromicnumber,ineitherbase,maynotincludeleadingzeros.)
NOTE:2,3,5,and7arenotconsideredtobetruncatableprimes.
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Problem39:Integerrighttriangles
Ifpistheperimeterofarightangletrianglewithintegrallengthsides,{a,b,c},thereareexactlythreesolutionsforp=120.
{20,48,52},{24,45,51},{30,40,50}
Forwhichvalueofp1000,isthenumberofsolutionsmaximised?
Problem40:Champernowne'sconstant
Anirrationaldecimalfractioniscreatedbyconcatenatingthepositiveintegers:
0.123456789101112131415161718192021...
Itcanbeseenthatthe12thdigitofthefractionalpartis1.
Ifdnrepresentsthenthdigitofthefractionalpart,findthevalueofthefollowingexpression.
d1d10d100d1000d10000d100000d1000000
Problem41:Pandigitalprime
Weshallsaythatanndigitnumberispandigitalifitmakesuseofallthedigits1tonexactlyonce.Forexample,2143isa4digitpandigitalandisalsoprime.
Whatisthelargestndigitpandigitalprimethatexists?
Problem42:Codedtrianglenumbers
Thenthtermofthesequenceoftrianglenumbersisgivenby,tn=n(n+1)sothefirsttentrianglenumbersare:
1,3,6,10,15,21,28,36,45,55,...
Byconvertingeachletterinawordtoanumbercorrespondingtoitsalphabeticalpositionandaddingthesevaluesweformawordvalue.Forexample,thewordvalueforSKYis19+11+25=55=t10.Ifthewordvalueisatrianglenumberthenweshallcallthewordatriangleword.
Usingwords.txt(rightclickand'SaveLink/TargetAs...'),a16KtextfilecontainingnearlytwothousandcommonEnglishwords,howmanyaretrianglewords?
Problem43:Substringdivisibility
Thenumber,1406357289,isa0to9pandigitalnumberbecauseitismadeupofeachofthedigits0to9insomeorder,butitalsohasaratherinterestingsubstringdivisibilityproperty.
Letd1bethe1stdigit,d2bethe2nddigit,andsoon.Inthisway,wenotethefollowing:
d2d3d4=406isdivisibleby2d3d4d5=063isdivisibleby3d4d5d6=635isdivisibleby5d5d6d7=357isdivisibleby7d6d7d8=572isdivisibleby11d7d8d9=728isdivisibleby13d8d9d10=289isdivisibleby17
Findthesumofall0to9pandigitalnumberswiththisproperty.
Problem44:Pentagonnumbers
Pentagonalnumbersaregeneratedbytheformula,Pn=n(3n1)/2.Thefirsttenpentagonalnumbersare:
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1,5,12,22,35,51,70,92,117,145,...
ItcanbeseenthatP4+P7=22+70=92=P8.However,theirdifference,7022=48,isnotpentagonal.
Findthepairofpentagonalnumbers,PjandPk,forwhichtheirsumanddifferencearepentagonalandD=|PkPj|isminimisedwhatisthevalueofD?
Problem45:Triangular,pentagonal,andhexagonal
Triangle,pentagonal,andhexagonalnumbersaregeneratedbythefollowingformulae:
Triangle Tn=n(n+1)/2 1,3,6,10,15,...Pentagonal Pn=n(3n1)/2 1,5,12,22,35,...Hexagonal Hn=n(2n1) 1,6,15,28,45,...
ItcanbeverifiedthatT285=P165=H143=40755.
Findthenexttrianglenumberthatisalsopentagonalandhexagonal.
Problem46:Goldbach'sotherconjecture
ItwasproposedbyChristianGoldbachthateveryoddcompositenumbercanbewrittenasthesumofaprimeandtwiceasquare.
9=7+212
15=7+222
21=3+232
25=7+232
27=19+222
33=31+212
Itturnsoutthattheconjecturewasfalse.
Whatisthesmallestoddcompositethatcannotbewrittenasthesumofaprimeandtwiceasquare?
Problem47:Distinctprimesfactors
Thefirsttwoconsecutivenumberstohavetwodistinctprimefactorsare:
14=2715=35
Thefirstthreeconsecutivenumberstohavethreedistinctprimefactorsare:
644=2723645=3543646=21719.
Findthefirstfourconsecutiveintegerstohavefourdistinctprimefactors.Whatisthefirstofthesenumbers?
Problem48:Selfpowers
Theseries,11+22+33+...+1010=10405071317.
Findthelasttendigitsoftheseries,11+22+33+...+10001000.
Problem49:Primepermutations
Thearithmeticsequence,1487,4817,8147,inwhicheachofthetermsincreasesby3330,isunusualintwoways:(i)eachofthethreetermsareprime,and,(ii)eachofthe4digitnumbersarepermutationsofoneanother.
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Therearenoarithmeticsequencesmadeupofthree1,2,or3digitprimes,exhibitingthisproperty,butthereisoneother4digitincreasingsequence.
What12digitnumberdoyouformbyconcatenatingthethreetermsinthissequence?
Problem50:Consecutiveprimesum
Theprime41,canbewrittenasthesumofsixconsecutiveprimes:
41=2+3+5+7+11+13
Thisisthelongestsumofconsecutiveprimesthataddstoaprimebelowonehundred.
Thelongestsumofconsecutiveprimesbelowonethousandthataddstoaprime,contains21terms,andisequalto953.
Whichprime,belowonemillion,canbewrittenasthesumofthemostconsecutiveprimes?
Problem51:Primedigitreplacements
Byreplacingthe1stdigitofthe2digitnumber*3,itturnsoutthatsixoftheninepossiblevalues:13,23,43,53,73,and83,areallprime.
Byreplacingthe3rdand4thdigitsof56**3withthesamedigit,this5digitnumberisthefirstexamplehavingsevenprimesamongthetengeneratednumbers,yieldingthefamily:56003,56113,56333,56443,56663,56773,and56993.Consequently56003,beingthefirstmemberofthisfamily,isthesmallestprimewiththisproperty.
Findthesmallestprimewhich,byreplacingpartofthenumber(notnecessarilyadjacentdigits)withthesamedigit,ispartofaneightprimevaluefamily.
Problem52:Permutedmultiples
Itcanbeseenthatthenumber,125874,anditsdouble,251748,containexactlythesamedigits,butinadifferentorder.
Findthesmallestpositiveinteger,x,suchthat2x,3x,4x,5x,and6x,containthesamedigits.
Problem53:Combinatoricselections
Thereareexactlytenwaysofselectingthreefromfive,12345:
123,124,125,134,135,145,234,235,245,and345
Incombinatorics,weusethenotation,5C3=10.
Ingeneral,
nCr=n!
r!(nr)! ,wherern,n!=n(n1)...321,and0!=1.
Itisnotuntiln=23,thatavalueexceedsonemillion:23C10=1144066.
Howmany,notnecessarilydistinct,valuesofnCr,for1n100,aregreaterthanonemillion?
Problem54:Pokerhands
Inthecardgamepoker,ahandconsistsoffivecardsandareranked,fromlowesttohighest,inthefollowingway:
HighCard:Highestvaluecard.OnePair:Twocardsofthesamevalue.TwoPairs:Twodifferentpairs.ThreeofaKind:Threecardsofthesamevalue.
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Straight:Allcardsareconsecutivevalues.Flush:Allcardsofthesamesuit.FullHouse:Threeofakindandapair.FourofaKind:Fourcardsofthesamevalue.StraightFlush:Allcardsareconsecutivevaluesofsamesuit.RoyalFlush:Ten,Jack,Queen,King,Ace,insamesuit.
Thecardsarevaluedintheorder:2,3,4,5,6,7,8,9,10,Jack,Queen,King,Ace.
Iftwoplayershavethesamerankedhandsthentherankmadeupofthehighestvaluewinsforexample,apairofeightsbeatsapairoffives(seeexample1below).Butiftworankstie,forexample,bothplayershaveapairofqueens,thenhighestcardsineachhandarecompared(seeexample4below)ifthehighestcardstiethenthenexthighestcardsarecompared,andsoon.
Considerthefollowingfivehandsdealttotwoplayers:
Hand Player1 Player2 Winner1 5H5C6S7SKDPairofFives
2C3S8S8DTDPairofEights
Player2
2 5D8C9SJSACHighestcardAce 2C5C7D8SQHHighestcardQueen
Player1
3 2D9CASAHACThreeAces 3D6D7DTDQDFlushwithDiamonds
Player2
44D6S9HQHQC
PairofQueensHighestcardNine
3D6D7HQDQS
PairofQueensHighestcardSeven
Player1
52H2D4C4D4S
FullHouseWithThreeFours
3C3D3S9S9D
FullHousewithThreeThrees
Player1
Thefile,poker.txt,containsonethousandrandomhandsdealttotwoplayers.Eachlineofthefilecontainstencards(separatedbyasinglespace):thefirstfivearePlayer1'scardsandthelastfivearePlayer2'scards.Youcanassumethatallhandsarevalid(noinvalidcharactersorrepeatedcards),eachplayer'shandisinnospecificorder,andineachhandthereisaclearwinner.
HowmanyhandsdoesPlayer1win?
Problem55:Lychrelnumbers
Ifwetake47,reverseandadd,47+74=121,whichispalindromic.
Notallnumbersproducepalindromessoquickly.Forexample,
349+943=1292,1292+2921=42134213+3124=7337
Thatis,349tookthreeiterationstoarriveatapalindrome.
Althoughnoonehasprovedityet,itisthoughtthatsomenumbers,like196,neverproduceapalindrome.AnumberthatneverformsapalindromethroughthereverseandaddprocessiscalledaLychrelnumber.Duetothetheoreticalnatureofthesenumbers,andforthepurposeofthisproblem,weshallassumethatanumberisLychreluntilprovenotherwise.Inadditionyouaregiventhatforeverynumberbelowtenthousand,itwilleither(i)becomeapalindromeinlessthanfiftyiterations,or,(ii)noone,withallthecomputingpowerthatexists,hasmanagedsofartomapittoapalindrome.Infact,10677isthefirstnumbertobeshowntorequireoverfiftyiterationsbeforeproducingapalindrome:4668731596684224866951378664(53iterations,28digits).
Surprisingly,therearepalindromicnumbersthatarethemselvesLychrelnumbersthefirstexampleis4994.
HowmanyLychrelnumbersaretherebelowtenthousand?
NOTE:Wordingwasmodifiedslightlyon24April2007toemphasisethetheoreticalnatureofLychrelnumbers.
Problem56:Powerfuldigitsum
Agoogol(10100)isamassivenumber:onefollowedbyonehundredzeros100100isalmostunimaginablylarge:onefollowedbytwohundredzeros.Despitetheirsize,thesumofthedigitsineachnumberisonly1.
Consideringnaturalnumbersoftheform,ab,wherea,b
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Problem57:Squarerootconvergents
Itispossibletoshowthatthesquarerootoftwocanbeexpressedasaninfinitecontinuedfraction.
2=1+1/(2+1/(2+1/(2+...)))=1.414213...
Byexpandingthisforthefirstfouriterations,weget:
1+1/2=3/2=1.51+1/(2+1/2)=7/5=1.41+1/(2+1/(2+1/2))=17/12=1.41666...1+1/(2+1/(2+1/(2+1/2)))=41/29=1.41379...
Thenextthreeexpansionsare99/70,239/169,and577/408,buttheeighthexpansion,1393/985,isthefirstexamplewherethenumberofdigitsinthenumeratorexceedsthenumberofdigitsinthedenominator.
Inthefirstonethousandexpansions,howmanyfractionscontainanumeratorwithmoredigitsthandenominator?
Problem58:Spiralprimes
Startingwith1andspirallinganticlockwiseinthefollowingway,asquarespiralwithsidelength7isformed.
37363534333231381716151413303918543122940196121128412078910274221222324252643444546474849
Itisinterestingtonotethattheoddsquaresliealongthebottomrightdiagonal,butwhatismoreinterestingisthat8outofthe13numberslyingalongbothdiagonalsareprimethatis,aratioof8/1362%.
Ifonecompletenewlayeriswrappedaroundthespiralabove,asquarespiralwithsidelength9willbeformed.Ifthisprocessiscontinued,whatisthesidelengthofthesquarespiralforwhichtheratioofprimesalongbothdiagonalsfirstfallsbelow10%?
Problem59:XORdecryption
EachcharacteronacomputerisassignedauniquecodeandthepreferredstandardisASCII(AmericanStandardCodeforInformationInterchange).Forexample,uppercaseA=65,asterisk(*)=42,andlowercasek=107.
Amodernencryptionmethodistotakeatextfile,convertthebytestoASCII,thenXOReachbytewithagivenvalue,takenfromasecretkey.TheadvantagewiththeXORfunctionisthatusingthesameencryptionkeyontheciphertext,restorestheplaintextforexample,65XOR42=107,then107XOR42=65.
Forunbreakableencryption,thekeyisthesamelengthastheplaintextmessage,andthekeyismadeupofrandombytes.Theuserwouldkeeptheencryptedmessageandtheencryptionkeyindifferentlocations,andwithoutboth"halves",itisimpossibletodecryptthemessage.
Unfortunately,thismethodisimpracticalformostusers,sothemodifiedmethodistouseapasswordasakey.Ifthepasswordisshorterthanthemessage,whichislikely,thekeyisrepeatedcyclicallythroughoutthemessage.Thebalanceforthismethodisusingasufficientlylongpasswordkeyforsecurity,butshortenoughtobememorable.
Yourtaskhasbeenmadeeasy,astheencryptionkeyconsistsofthreelowercasecharacters.Usingcipher.txt(rightclickand'SaveLink/TargetAs...'),afilecontainingtheencryptedASCIIcodes,andtheknowledgethattheplaintextmustcontaincommonEnglishwords,decryptthemessageandfindthesumoftheASCIIvaluesintheoriginaltext.
Problem60:Primepairsets
Theprimes3,7,109,and673,arequiteremarkable.Bytakinganytwoprimesandconcatenatingtheminanyordertheresultwillalwaysbeprime.Forexample,taking7and109,both7109and1097areprime.Thesumofthesefourprimes,792,representsthelowestsumforasetoffourprimeswiththisproperty.
Findthelowestsumforasetoffiveprimesforwhichanytwoprimesconcatenatetoproduceanotherprime.
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Problem61:Cyclicalfiguratenumbers
Triangle,square,pentagonal,hexagonal,heptagonal,andoctagonalnumbersareallfigurate(polygonal)numbersandaregeneratedbythefollowingformulae:
Triangle P3,n=n(n+1)/2 1,3,6,10,15,...Square P4,n=n2 1,4,9,16,25,...Pentagonal P5,n=n(3n1)/2 1,5,12,22,35,...Hexagonal P6,n=n(2n1) 1,6,15,28,45,...Heptagonal P7,n=n(5n3)/2 1,7,18,34,55,...Octagonal P8,n=n(3n2) 1,8,21,40,65,...
Theorderedsetofthree4digitnumbers:8128,2882,8281,hasthreeinterestingproperties.
1. Thesetiscyclic,inthatthelasttwodigitsofeachnumberisthefirsttwodigitsofthenextnumber(includingthelastnumberwiththefirst).
2. Eachpolygonaltype:triangle(P3,127=8128),square(P4,91=8281),andpentagonal(P5,44=2882),isrepresentedbyadifferentnumberintheset.
3. Thisistheonlysetof4digitnumberswiththisproperty.
Findthesumoftheonlyorderedsetofsixcyclic4digitnumbersforwhicheachpolygonaltype:triangle,square,pentagonal,hexagonal,heptagonal,andoctagonal,isrepresentedbyadifferentnumberintheset.
Problem62:Cubicpermutations
Thecube,41063625(3453),canbepermutedtoproducetwoothercubes:56623104(3843)and66430125(4053).Infact,41063625isthesmallestcubewhichhasexactlythreepermutationsofitsdigitswhicharealsocube.
Findthesmallestcubeforwhichexactlyfivepermutationsofitsdigitsarecube.
Problem63:Powerfuldigitcounts
The5digitnumber,16807=75,isalsoafifthpower.Similarly,the9digitnumber,134217728=89,isaninthpower.
Howmanyndigitpositiveintegersexistwhicharealsoannthpower?
Problem64:Oddperiodsquareroots
Allsquarerootsareperiodicwhenwrittenascontinuedfractionsandcanbewrittenintheform:
N=a0+ 1 a1+ 1 a2+ 1 a3+...
Forexample,letusconsider23:
23=4+234=4+ 1 =4+ 1
1234 1+233
7
Ifwecontinuewewouldgetthefollowingexpansion:
23=4+ 1 1+ 1 3+ 1 1+ 1 8+...
Theprocesscanbesummarisedasfollows:
a0=4,1
234=23+4
7 =1+233
77 7(23+3) 233
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a1=1,233= 14 =3+ 2
a2=3,2
233=2(23+3)
14 =1+234
7a3=1,
7234=
7(23+4)7 =8+234
a4=8,1
234=23+4
7 =1+233
7a5=1,
7233=
7(23+3)14 =3+
2332
a6=3,2
233=2(23+3)
14 =1+234
7a7=1,
7234=
7(23+4)7 =8+234
Itcanbeseenthatthesequenceisrepeating.Forconciseness,weusethenotation23=[4(1,3,1,8)],toindicatethattheblock(1,3,1,8)repeatsindefinitely.
Thefirsttencontinuedfractionrepresentationsof(irrational)squarerootsare:
2=[1(2)],period=13=[1(1,2)],period=25=[2(4)],period=16=[2(2,4)],period=27=[2(1,1,1,4)],period=48=[2(1,4)],period=210=[3(6)],period=111=[3(3,6)],period=212=[3(2,6)],period=213=[3(1,1,1,1,6)],period=5
Exactlyfourcontinuedfractions,forN13,haveanoddperiod.
HowmanycontinuedfractionsforN10000haveanoddperiod?
Problem65:Convergentsofe
Thesquarerootof2canbewrittenasaninfinitecontinuedfraction.
2=1+ 1 2+ 1 2+ 1 2+ 1 2+...
Theinfinitecontinuedfractioncanbewritten,2=[1(2)],(2)indicatesthat2repeatsadinfinitum.Inasimilarway,23=[4(1,3,1,8)].
Itturnsoutthatthesequenceofpartialvaluesofcontinuedfractionsforsquarerootsprovidethebestrationalapproximations.Letusconsidertheconvergentsfor2.
1+1=3/2 21+ 1 =7/5 2+1 21+ 1 =17/12 2+ 1 2+1 21+ 1 =41/29 2+ 1 2+ 1 2+1 2
Hencethesequenceofthefirsttenconvergentsfor2are:
1,3/2,7/5,17/12,41/29,99/70,239/169,577/408,1393/985,3363/2378,...
Whatismostsurprisingisthattheimportantmathematicalconstant,e=[21,2,1,1,4,1,1,6,1,...,1,2k,1,...].
Thefirsttentermsinthesequenceofconvergentsforeare:
2,3,8/3,11/4,19/7,87/32,106/39,193/71,1264/465,1457/536,...
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Thesumofdigitsinthenumeratorofthe10thconvergentis1+4+5+7=17.
Findthesumofdigitsinthenumeratorofthe100thconvergentofthecontinuedfractionfore.
Problem66:Diophantineequation
ConsiderquadraticDiophantineequationsoftheform:
x2Dy2=1
Forexample,whenD=13,theminimalsolutioninxis6492131802=1.
ItcanbeassumedthattherearenosolutionsinpositiveintegerswhenDissquare.
ByfindingminimalsolutionsinxforD={2,3,5,6,7},weobtainthefollowing:
32222=122312=192542=152622=182732=1
Hence,byconsideringminimalsolutionsinxforD7,thelargestxisobtainedwhenD=5.
FindthevalueofD1000inminimalsolutionsofxforwhichthelargestvalueofxisobtained.
Problem67:MaximumpathsumII
Bystartingatthetopofthetrianglebelowandmovingtoadjacentnumbersontherowbelow,themaximumtotalfromtoptobottomis23.
374
2468593
Thatis,3+7+4+9=23.
Findthemaximumtotalfromtoptobottomintriangle.txt(rightclickand'SaveLink/TargetAs...'),a15Ktextfilecontainingatrianglewithonehundredrows.
NOTE:ThisisamuchmoredifficultversionofProblem18.Itisnotpossibletotryeveryroutetosolvethisproblem,asthereare299altogether!Ifyoucouldcheckonetrillion(1012)routeseveryseconditwouldtakeovertwentybillionyearstocheckthemall.Thereisanefficientalgorithmtosolveit.o)
Problem68:Magic5gonring
Considerthefollowing"magic"3gonring,filledwiththenumbers1to6,andeachlineaddingtonine.
Workingclockwise,andstartingfromthegroupofthreewiththenumericallylowestexternalnode(4,3,2inthisexample),eachsolutioncanbedescribeduniquely.Forexample,theabovesolutioncanbedescribedbytheset:4,3,26,2,15,1,3.
Itispossibletocompletetheringwithfourdifferenttotals:9,10,11,and12.Thereareeightsolutionsintotal.
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Total SolutionSet9 4,2,35,3,16,1,29 4,3,26,2,15,1,310 2,3,54,5,16,1,310 2,5,36,3,14,1,511 1,4,63,6,25,2,411 1,6,45,4,23,2,612 1,5,62,6,43,4,512 1,6,53,5,42,4,6
Byconcatenatingeachgroupitispossibletoform9digitstringsthemaximumstringfora3gonringis432621513.
Usingthenumbers1to10,anddependingonarrangements,itispossibletoform16and17digitstrings.Whatisthemaximum16digitstringfora"magic"5gonring?
Problem69:Totientmaximum
Euler'sTotientfunction,(n)[sometimescalledthephifunction],isusedtodeterminethenumberofnumberslessthannwhicharerelativelyprimeton.Forexample,as1,2,4,5,7,and8,arealllessthannineandrelativelyprimetonine,(9)=6.
n RelativelyPrime (n) n/(n)2 1 1 23 1,2 2 1.54 1,3 2 25 1,2,3,4 4 1.256 1,5 2 37 1,2,3,4,5,6 6 1.1666...8 1,3,5,7 4 29 1,2,4,5,7,8 6 1.510 1,3,7,9 4 2.5
Itcanbeseenthatn=6producesamaximumn/(n)forn10.
Findthevalueofn1,000,000forwhichn/(n)isamaximum.
Problem70:Totientpermutation
Euler'sTotientfunction,(n)[sometimescalledthephifunction],isusedtodeterminethenumberofpositivenumberslessthanorequaltonwhicharerelativelyprimeton.Forexample,as1,2,4,5,7,and8,arealllessthannineandrelativelyprimetonine,(9)=6.Thenumber1isconsideredtoberelativelyprimetoeverypositivenumber,so(1)=1.
Interestingly,(87109)=79180,anditcanbeseenthat87109isapermutationof79180.
Findthevalueofn,1
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Considerthefraction,n/d,wherenanddarepositiveintegers.Ifn
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30cm:(5,12,13)36cm:(9,12,15)40cm:(8,15,17)48cm:(12,16,20)
Incontrast,somelengthsofwire,like20cm,cannotbebenttoformanintegersidedrightangletriangle,andotherlengthsallowmorethanonesolutiontobefoundforexample,using120cmitispossibletoformexactlythreedifferentintegersidedrightangletriangles.
120cm:(30,40,50),(20,48,52),(24,45,51)
GiventhatListhelengthofthewire,forhowmanyvaluesofL1,500,000canexactlyoneintegersidedrightangletrianglebeformed?
Problem76:Countingsummations
Itispossibletowritefiveasasuminexactlysixdifferentways:
4+13+23+1+12+2+12+1+1+11+1+1+1+1
Howmanydifferentwayscanonehundredbewrittenasasumofatleasttwopositiveintegers?
Problem77:Primesummations
Itispossibletowritetenasthesumofprimesinexactlyfivedifferentways:
7+35+55+3+23+3+2+22+2+2+2+2
Whatisthefirstvaluewhichcanbewrittenasthesumofprimesinoverfivethousanddifferentways?
Problem78:Coinpartitions
Letp(n)representthenumberofdifferentwaysinwhichncoinscanbeseparatedintopiles.Forexample,fivecoinscanbeseparatedintopilesinexactlysevendifferentways,sop(5)=7.
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
Findtheleastvalueofnforwhichp(n)isdivisiblebyonemillion.
Problem79:Passcodederivation
Acommonsecuritymethodusedforonlinebankingistoasktheuserforthreerandomcharactersfromapasscode.Forexample,ifthepasscodewas531278,theymayaskforthe2nd,3rd,and5thcharacterstheexpectedreplywouldbe:317.
Thetextfile,keylog.txt,containsfiftysuccessfulloginattempts.
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Giventhatthethreecharactersarealwaysaskedforinorder,analysethefilesoastodeterminetheshortestpossiblesecretpasscodeofunknownlength.
Problem80:Squarerootdigitalexpansion
Itiswellknownthatifthesquarerootofanaturalnumberisnotaninteger,thenitisirrational.Thedecimalexpansionofsuchsquarerootsisinfinitewithoutanyrepeatingpatternatall.
Thesquarerootoftwois1.41421356237309504880...,andthedigitalsumofthefirstonehundreddecimaldigitsis475.
Forthefirstonehundrednaturalnumbers,findthetotalofthedigitalsumsofthefirstonehundreddecimaldigitsforalltheirrationalsquareroots.
Problem81:Pathsum:twoways
Inthe5by5matrixbelow,theminimalpathsumfromthetoplefttothebottomright,byonlymovingtotherightanddown,isindicatedinboldredandisequalto2427.
Findtheminimalpathsum,inmatrix.txt(rightclickand"SaveLink/TargetAs..."),a31Ktextfilecontaininga80by80matrix,fromthetoplefttothebottomrightbyonlymovingrightanddown.
Problem82:Pathsum:threeways
NOTE:ThisproblemisamorechallengingversionofProblem81.
Theminimalpathsuminthe5by5matrixbelow,bystartinginanycellintheleftcolumnandfinishinginanycellintherightcolumn,andonlymovingup,down,andright,isindicatedinredandboldthesumisequalto994.
Findtheminimalpathsum,inmatrix.txt(rightclickand"SaveLink/TargetAs..."),a31Ktextfilecontaininga80by80matrix,fromtheleftcolumntotherightcolumn.
Problem83:Pathsum:fourways
NOTE:ThisproblemisasignificantlymorechallengingversionofProblem81.
Inthe5by5matrixbelow,theminimalpathsumfromthetoplefttothebottomright,bymovingleft,right,up,anddown,isindicatedinboldredandisequalto2297.
Findtheminimalpathsum,inmatrix.txt(rightclickand"SaveLink/TargetAs..."),a31Ktextfilecontaininga80by80matrix,fromthetoplefttothebottomrightbymovingleft,right,up,anddown.
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Problem84:Monopolyodds
Inthegame,Monopoly,thestandardboardissetupinthefollowingway:
GO A1 CC1 A2 T1 R1 B1 CH1 B2 B3 JAIL
H2 C1
T2 U1
H1 C2
CH3 C3
R4 R2
G3 D1
CC3 CC2
G2 D2
G1 D3
G2J F3 U2 F2 F1 R3 E3 E2 CH2 E1 FP
AplayerstartsontheGOsquareandaddsthescoresontwo6sideddicetodeterminethenumberofsquarestheyadvanceinaclockwisedirection.Withoutanyfurtherruleswewouldexpecttovisiteachsquarewithequalprobability:2.5%.However,landingonG2J(GoToJail),CC(communitychest),andCH(chance)changesthisdistribution.
InadditiontoG2J,andonecardfromeachofCCandCH,thatorderstheplayertogodirectlytojail,ifaplayerrollsthreeconsecutivedoubles,theydonotadvancetheresultoftheir3rdroll.Insteadtheyproceeddirectlytojail.
Atthebeginningofthegame,theCCandCHcardsareshuffled.WhenaplayerlandsonCCorCHtheytakeacardfromthetopoftherespectivepileand,afterfollowingtheinstructions,itisreturnedtothebottomofthepile.Therearesixteencardsineachpile,butforthepurposeofthisproblemweareonlyconcernedwithcardsthatorderamovementanyinstructionnotconcernedwithmovementwillbeignoredandtheplayerwillremainontheCC/CHsquare.
CommunityChest(2/16cards):1. AdvancetoGO2. GotoJAIL
Chance(10/16cards):1. AdvancetoGO2. GotoJAIL3. GotoC14. GotoE35. GotoH26. GotoR17. GotonextR(railwaycompany)8. GotonextR9. GotonextU(utilitycompany)
10. Goback3squares.
Theheartofthisproblemconcernsthelikelihoodofvisitingaparticularsquare.Thatis,theprobabilityoffinishingatthatsquareafteraroll.Forthisreasonitshouldbeclearthat,withtheexceptionofG2Jforwhichtheprobabilityoffinishingonitiszero,theCHsquareswillhavethelowestprobabilities,as5/8requestamovementtoanothersquare,anditisthefinalsquarethattheplayerfinishesatoneachrollthatweareinterestedin.Weshallmakenodistinctionbetween"JustVisiting"andbeingsenttoJAIL,andweshallalsoignoretheruleaboutrequiringadoubleto"getoutofjail",assumingthattheypaytogetoutontheirnextturn.
BystartingatGOandnumberingthesquaressequentiallyfrom00to39wecanconcatenatethesetwodigitnumberstoproducestringsthatcorrespondwithsetsofsquares.
Statisticallyitcanbeshownthatthethreemostpopularsquares,inorder,areJAIL(6.24%)=Square10,E3(3.18%)=Square24,andGO(3.09%)=Square00.Sothesethreemostpopularsquarescanbelistedwiththesixdigitmodalstring:102400.
If,insteadofusingtwo6sideddice,two4sideddiceareused,findthesixdigitmodalstring.
Problem85:Countingrectangles
Bycountingcarefullyitcanbeseenthatarectangulargridmeasuring3by2containseighteenrectangles:
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Althoughthereexistsnorectangulargridthatcontainsexactlytwomillionrectangles,findtheareaofthegridwiththenearestsolution.
Problem86:Cuboidroute
Aspider,S,sitsinonecornerofacuboidroom,measuring6by5by3,andafly,F,sitsintheoppositecorner.Bytravellingonthesurfacesoftheroomtheshortest"straightline"distancefromStoFis10andthepathisshownonthediagram.
However,thereareuptothree"shortest"pathcandidatesforanygivencuboidandtheshortestroutedoesn'talwayshaveintegerlength.
Itcanbeshownthatthereareexactly2060distinctcuboids,ignoringrotations,withintegerdimensions,uptoamaximumsizeofMbyMbyM,forwhichtheshortestroutehasintegerlengthwhenM=100.ThisistheleastvalueofMforwhichthenumberofsolutionsfirstexceedstwothousandthenumberofsolutionswhenM=99is1975.
FindtheleastvalueofMsuchthatthenumberofsolutionsfirstexceedsonemillion.
Problem87:Primepowertriples
Thesmallestnumberexpressibleasthesumofaprimesquare,primecube,andprimefourthpoweris28.Infact,thereareexactlyfournumbersbelowfiftythatcanbeexpressedinsuchaway:
28=22+23+24
33=32+23+24
49=52+23+24
47=22+33+24
Howmanynumbersbelowfiftymillioncanbeexpressedasthesumofaprimesquare,primecube,andprimefourthpower?
Problem88:Productsumnumbers
Anaturalnumber,N,thatcanbewrittenasthesumandproductofagivensetofatleasttwonaturalnumbers,{a1,a2,...,ak}iscalledaproductsumnumber:N=a1+a2+...+ak=a1a2...ak.
Forexample,6=1+2+3=123.
Foragivensetofsize,k,weshallcallthesmallestNwiththispropertyaminimalproductsumnumber.Theminimalproductsumnumbersforsetsofsize,k=2,3,4,5,and6areasfollows.
k=2:4=22=2+2k=3:6=123=1+2+3k=4:8=1124=1+1+2+4k=5:8=11222=1+1+2+2+2k=6:12=111126=1+1+1+1+2+6
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Hencefor2k6,thesumofalltheminimalproductsumnumbersis4+6+8+12=30notethat8isonlycountedonceinthesum.
Infact,asthecompletesetofminimalproductsumnumbersfor2k12is{4,6,8,12,15,16},thesumis61.
Whatisthesumofalltheminimalproductsumnumbersfor2k12000?
Problem89:Romannumerals
ForanumberwritteninRomannumeralstobeconsideredvalidtherearebasicruleswhichmustbefollowed.Eventhoughtherulesallowsomenumberstobeexpressedinmorethanonewaythereisalwaysa"best"wayofwritingaparticularnumber.
Forexample,itwouldappearthatthereareatleastsixwaysofwritingthenumbersixteen:
IIIIIIIIIIIIIIIIVIIIIIIIIIIIVVIIIIIIXIIIIIIVVVIXVI
However,accordingtotherulesonlyXIIIIIIandXVIarevalid,andthelastexampleisconsideredtobethemostefficient,asitusestheleastnumberofnumerals.
The11Ktextfile,roman.txt(rightclickand'SaveLink/TargetAs...'),containsonethousandnumberswritteninvalid,butnotnecessarilyminimal,RomannumeralsseeAbout...RomanNumeralsforthedefinitiverulesforthisproblem.
Findthenumberofcharacterssavedbywritingeachoftheseintheirminimalform.
Note:YoucanassumethatalltheRomannumeralsinthefilecontainnomorethanfourconsecutiveidenticalunits.
Problem90:Cubedigitpairs
Eachofthesixfacesonacubehasadifferentdigit(0to9)writtenonitthesameisdonetoasecondcube.Byplacingthetwocubessidebysideindifferentpositionswecanformavarietyof2digitnumbers.
Forexample,thesquarenumber64couldbeformed:
Infact,bycarefullychoosingthedigitsonbothcubesitispossibletodisplayallofthesquarenumbersbelowonehundred:01,04,09,16,25,36,49,64,and81.
Forexample,onewaythiscanbeachievedisbyplacing{0,5,6,7,8,9}ononecubeand{1,2,3,4,8,9}ontheothercube.
However,forthisproblemweshallallowthe6or9tobeturnedupsidedownsothatanarrangementlike{0,5,6,7,8,9}and{1,2,3,4,6,7}allowsforallninesquarenumberstobedisplayedotherwiseitwouldbeimpossibletoobtain09.
Indeterminingadistinctarrangementweareinterestedinthedigitsoneachcube,nottheorder.
{1,2,3,4,5,6}isequivalentto{3,6,4,1,2,5}{1,2,3,4,5,6}isdistinctfrom{1,2,3,4,5,9}
Butbecauseweareallowing6and9tobereversed,thetwodistinctsetsinthelastexamplebothrepresenttheextendedset{1,2,3,4,5,6,9}forthepurposeofforming2digitnumbers.
Howmanydistinctarrangementsofthetwocubesallowforallofthesquarenumberstobedisplayed?
Problem91:Righttriangleswithintegercoordinates
ThepointsP(x1,y1)andQ(x2,y2)areplottedatintegercoordinatesandarejoinedtotheorigin,O(0,0),to
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formOPQ.
Thereareexactlyfourteentrianglescontainingarightanglethatcanbeformedwheneachcoordinateliesbetween0and2inclusivethatis,0x1,y1,x2,y22.
Giventhat0x1,y1,x2,y250,howmanyrighttrianglescanbeformed?
Problem92:Squaredigitchains
Anumberchainiscreatedbycontinuouslyaddingthesquareofthedigitsinanumbertoformanewnumberuntilithasbeenseenbefore.
Forexample,
443213101185891454220416375889
Thereforeanychainthatarrivesat1or89willbecomestuckinanendlessloop.WhatismostamazingisthatEVERYstartingnumberwilleventuallyarriveat1or89.
Howmanystartingnumbersbelowtenmillionwillarriveat89?
Problem93:Arithmeticexpressions
Byusingeachofthedigitsfromtheset,{1,2,3,4},exactlyonce,andmakinguseofthefourarithmeticoperations(+,,*,/)andbrackets/parentheses,itispossibletoformdifferentpositiveintegertargets.
Forexample,
8=(4*(1+3))/214=4*(3+1/2)19=4*(2+3)136=3*4*(2+1)
Notethatconcatenationsofthedigits,like12+34,arenotallowed.
Usingtheset,{1,2,3,4},itispossibletoobtainthirtyonedifferenttargetnumbersofwhich36isthemaximum,andeachofthenumbers1to28canbeobtainedbeforeencounteringthefirstnonexpressiblenumber.
Findthesetoffourdistinctdigits,a
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Problem94:Almostequilateraltriangles
Itiseasilyprovedthatnoequilateraltriangleexistswithintegrallengthsidesandintegralarea.However,thealmostequilateraltriangle556hasanareaof12squareunits.
Weshalldefineanalmostequilateraltriangletobeatriangleforwhichtwosidesareequalandthethirddiffersbynomorethanoneunit.
Findthesumoftheperimetersofallalmostequilateraltriangleswithintegralsidelengthsandareaandwhoseperimetersdonotexceedonebillion(1,000,000,000).
Problem95:Amicablechains
Theproperdivisorsofanumberareallthedivisorsexcludingthenumberitself.Forexample,theproperdivisorsof28are1,2,4,7,and14.Asthesumofthesedivisorsisequalto28,wecallitaperfectnumber.
Interestinglythesumoftheproperdivisorsof220is284andthesumoftheproperdivisorsof284is220,formingachainoftwonumbers.Forthisreason,220and284arecalledanamicablepair.
Perhapslesswellknownarelongerchains.Forexample,startingwith12496,weformachainoffivenumbers:
1249614288154721453614264(12496...)
Sincethischainreturnstoitsstartingpoint,itiscalledanamicablechain.
Findthesmallestmemberofthelongestamicablechainwithnoelementexceedingonemillion.
Problem96:SuDoku
SuDoku(Japanesemeaningnumberplace)isthenamegiventoapopularpuzzleconcept.Itsoriginisunclear,butcreditmustbeattributedtoLeonhardEulerwhoinventedasimilar,andmuchmoredifficult,puzzleideacalledLatinSquares.TheobjectiveofSuDokupuzzles,however,istoreplacetheblanks(orzeros)ina9by9gridinsuchthateachrow,column,and3by3boxcontainseachofthedigits1to9.Belowisanexampleofatypicalstartingpuzzlegridanditssolutiongrid.
003900001
020305806
600001400
008700006
102000708
900008200
002800005
609203010
500009300
483967251
921345876
657821493
548729136
132564798
976138245
372814695
689253417
514769382
AwellconstructedSuDokupuzzlehasauniquesolutionandcanbesolvedbylogic,althoughitmaybenecessarytoemploy"guessandtest"methodsinordertoeliminateoptions(thereismuchcontestedopinionoverthis).Thecomplexityofthesearchdeterminesthedifficultyofthepuzzletheexampleaboveisconsideredeasybecauseitcanbesolvedbystraightforwarddirectdeduction.
The6Ktextfile,sudoku.txt(rightclickand'SaveLink/TargetAs...'),containsfiftydifferentSuDokupuzzlesrangingindifficulty,butallwithuniquesolutions(thefirstpuzzleinthefileistheexampleabove).
Bysolvingallfiftypuzzlesfindthesumofthe3digitnumbersfoundinthetopleftcornerofeachsolutiongridforexample,483isthe3digitnumberfoundinthetopleftcornerofthesolutiongridabove.
Problem97:LargenonMersenneprime
Thefirstknownprimefoundtoexceedonemilliondigitswasdiscoveredin1999,andisaMersenneprimeoftheform269725931itcontainsexactly2,098,960digits.SubsequentlyotherMersenneprimes,oftheform2p1,havebeenfoundwhichcontainmoredigits.
However,in2004therewasfoundamassivenonMersenneprimewhichcontains2,357,207digits:2843327830457+1.
Findthelasttendigitsofthisprimenumber.
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Problem98:Anagramicsquares
ByreplacingeachofthelettersinthewordCAREwith1,2,9,and6respectively,weformasquarenumber:1296=362.Whatisremarkableisthat,byusingthesamedigitalsubstitutions,theanagram,RACE,alsoformsasquarenumber:9216=962.WeshallcallCARE(andRACE)asquareanagramwordpairandspecifyfurtherthatleadingzeroesarenotpermitted,neithermayadifferentletterhavethesamedigitalvalueasanotherletter.
Usingwords.txt(rightclickand'SaveLink/TargetAs...'),a16KtextfilecontainingnearlytwothousandcommonEnglishwords,findallthesquareanagramwordpairs(apalindromicwordisNOTconsideredtobeananagramofitself).
Whatisthelargestsquarenumberformedbyanymemberofsuchapair?
Problem99:Largestexponential
Comparingtwonumberswritteninindexformlike211and37isnotdifficult,asanycalculatorwouldconfirmthat211=2048519432525806wouldbemuchmoredifficult,asbothnumberscontainoverthreemilliondigits.
Usingbase_exp.txt(rightclickand'SaveLink/TargetAs...'),a22Ktextfilecontainingonethousandlineswithabase/exponentpaironeachline,determinewhichlinenumberhasthegreatestnumericalvalue.
Problem100:Arrangedprobability
Ifaboxcontainstwentyonecoloureddiscs,composedoffifteenbluediscsandsixreddiscs,andtwodiscsweretakenatrandom,itcanbeseenthattheprobabilityoftakingtwobluediscs,P(BB)=(15/21)(14/20)=1/2.
Thenextsucharrangement,forwhichthereisexactly50%chanceoftakingtwobluediscsatrandom,isaboxcontainingeightyfivebluediscsandthirtyfivereddiscs.
Byfindingthefirstarrangementtocontainover1012=1,000,000,000,000discsintotal,determinethenumberofbluediscsthattheboxwouldcontain.
Problem101:Optimumpolynomial
Ifwearepresentedwiththefirstktermsofasequenceitisimpossibletosaywithcertaintythevalueofthenextterm,asthereareinfinitelymanypolynomialfunctionsthatcanmodelthesequence.
Asanexample,letusconsiderthesequenceofcubenumbers.Thisisdefinedbythegeneratingfunction,un=n3:1,8,27,64,125,216,...
Supposewewereonlygiventhefirsttwotermsofthissequence.Workingontheprinciplethat"simpleisbest"weshouldassumealinearrelationshipandpredictthenexttermtobe15(commondifference7).Evenifwewerepresentedwiththefirstthreeterms,bythesameprincipleofsimplicity,aquadraticrelationshipshouldbeassumed.
WeshalldefineOP(k,n)tobethenthtermoftheoptimumpolynomialgeneratingfunctionforthefirstktermsofasequence.ItshouldbeclearthatOP(k,n)willaccuratelygeneratethetermsofthesequencefornk,andpotentiallythefirstincorrectterm(FIT)willbeOP(k,k+1)inwhichcaseweshallcallitabadOP(BOP).
Asabasis,ifwewereonlygiventhefirsttermofsequence,itwouldbemostsensibletoassumeconstancythatis,forn2,OP(1,n)=u1.
HenceweobtainthefollowingOPsforthecubicsequence:
OP(1,n)=1 1,1,1,1,...
NOTE:Allanagramsformedmustbecontainedinthegiventextfile.
NOTE:Thefirsttwolinesinthefilerepresentthenumbersintheexamplegivenabove.
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OP(2,n)=7n6 1,8,15,...OP(3,n)=6n211n+6 1,8,27,58,...OP(4,n)=n3 1,8,27,64,125,...
ClearlynoBOPsexistfork4.
ByconsideringthesumofFITsgeneratedbytheBOPs(indicatedinredabove),weobtain1+15+58=74.
Considerthefollowingtenthdegreepolynomialgeneratingfunction:
un=1n+n2n3+n4n5+n6n7+n8n9+n10
FindthesumofFITsfortheBOPs.
Problem102:Trianglecontainment
ThreedistinctpointsareplottedatrandomonaCartesianplane,forwhich1000x,y1000,suchthatatriangleisformed.
Considerthefollowingtwotriangles:
A(340,495),B(153,910),C(835,947)
X(175,41),Y(421,714),Z(574,645)
ItcanbeverifiedthattriangleABCcontainstheorigin,whereastriangleXYZdoesnot.
Usingtriangles.txt(rightclickand'SaveLink/TargetAs...'),a27Ktextfilecontainingthecoordinatesofonethousand"random"triangles,findthenumberoftrianglesforwhichtheinteriorcontainstheorigin.
Problem103:Specialsubsetsums:optimum
LetS(A)representthesumofelementsinsetAofsizen.Weshallcallitaspecialsumsetifforanytwononemptydisjointsubsets,BandC,thefollowingpropertiesaretrue:
i. S(B)S(C)thatis,sumsofsubsetscannotbeequal.ii. IfBcontainsmoreelementsthanCthenS(B)>S(C).
IfS(A)isminimisedforagivenn,weshallcallitanoptimumspecialsumset.Thefirstfiveoptimumspecialsumsetsaregivenbelow.
n=1:{1}n=2:{1,2}n=3:{2,3,4}n=4:{3,5,6,7}n=5:{6,9,11,12,13}
Itseemsthatforagivenoptimumset,A={a1,a2,...,an},thenextoptimumsetisoftheformB={b,a1+b,a2+b,...,an+b},wherebisthe"middle"elementonthepreviousrow.
Byapplyingthis"rule"wewouldexpecttheoptimumsetforn=6tobeA={11,17,20,22,23,24},withS(A)=117.However,thisisnottheoptimumset,aswehavemerelyappliedanalgorithmtoprovideanearoptimumset.Theoptimumsetforn=6isA={11,18,19,20,22,25},withS(A)=115andcorrespondingsetstring:111819202225.
GiventhatAisanoptimumspecialsumsetforn=7,finditssetstring.
NOTE:ThisproblemisrelatedtoProblem105andProblem106.
Problem104:PandigitalFibonacciends
TheFibonaccisequenceisdefinedbytherecurrencerelation:
Fn=Fn1+Fn2,whereF1=1andF2=1.
ItturnsoutthatF541,whichcontains113digits,isthefirstFibonaccinumberforwhichthelastninedigitsare19pandigital(containallthedigits1to9,butnotnecessarilyinorder).AndF2749,whichcontains575
NOTE:Thefirsttwoexamplesinthefilerepresentthetrianglesintheexamplegivenabove.
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digits,isthefirstFibonaccinumberforwhichthefirstninedigitsare19pandigital.
GiventhatFkisthefirstFibonaccinumberforwhichthefirstninedigitsANDthelastninedigitsare19pandigital,findk.
Problem105:Specialsubsetsums:testing
LetS(A)representthesumofelementsinsetAofsizen.Weshallcallitaspecialsumsetifforanytwononemptydisjointsubsets,BandC,thefollowingpropertiesaretrue:
i. S(B)S(C)thatis,sumsofsubsetscannotbeequal.ii. IfBcontainsmoreelementsthanCthenS(B)>S(C).
Forexample,{81,88,75,42,87,84,86,65}isnotaspecialsumsetbecause65+87+88=75+81+84,whereas{157,150,164,119,79,159,161,139,158}satisfiesbothrulesforallpossiblesubsetpaircombinationsandS(A)=1286.
Usingsets.txt(rightclickand"SaveLink/TargetAs..."),a4Ktextfilewithonehundredsetscontainingseventotwelveelements(thetwoexamplesgivenabovearethefirsttwosetsinthefile),identifyallthespecialsumsets,A1,A2,...,Ak,andfindthevalueofS(A1)+S(A2)+...+S(Ak).
NOTE:ThisproblemisrelatedtoProblem103andProblem106.
Problem106:Specialsubsetsums:metatesting
LetS(A)representthesumofelementsinsetAofsizen.Weshallcallitaspecialsumsetifforanytwononemptydisjointsubsets,BandC,thefollowingpropertiesaretrue:
i. S(B)S(C)thatis,sumsofsubsetscannotbeequal.ii. IfBcontainsmoreelementsthanCthenS(B)>S(C).
Forthisproblemweshallassumethatagivensetcontainsnstrictlyincreasingelementsanditalreadysatisfiesthesecondrule.
Surprisingly,outofthe25possiblesubsetpairsthatcanbeobtainedfromasetforwhichn=4,only1ofthesepairsneedtobetestedforequality(firstrule).Similarly,whenn=7,only70outofthe966subsetpairsneedtobetested.
Forn=12,howmanyofthe261625subsetpairsthatcanbeobtainedneedtobetestedforequality?
NOTE:ThisproblemisrelatedtoProblem103andProblem105.
Problem107:Minimalnetwork
Thefollowingundirectednetworkconsistsofsevenverticesandtwelveedgeswithatotalweightof243.
Thesamenetworkcanberepresentedbythematrixbelow.
A B C D E F G
A 16 12 21
B 16 17 20
C 12 28 31
D 21 17 28 18 19 23
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E 20 18 11
F 31 19 27
G 23 11 27
However,itispossibletooptimisethenetworkbyremovingsomeedgesandstillensurethatallpointsonthenetworkremainconnected.Thenetworkwhichachievesthemaximumsavingisshownbelow.Ithasaweightof93,representingasavingof24393=150fromtheoriginalnetwork.
Usingnetwork.txt(rightclickand'SaveLink/TargetAs...'),a6Ktextfilecontaininganetworkwithfortyvertices,andgiveninmatrixform,findthemaximumsavingwhichcanbeachievedbyremovingredundantedgeswhilstensuringthatthenetworkremainsconnected.
Problem108:DiophantinereciprocalsI
Inthefollowingequationx,y,andnarepositiveintegers.
1x +
1y =
1n
Forn=4thereareexactlythreedistinctsolutions:
15 +
120 =
14
16 +
112 =
14
18 +
18 =
14
Whatistheleastvalueofnforwhichthenumberofdistinctsolutionsexceedsonethousand?
NOTE:ThisproblemisaneasierversionofProblem110itisstronglyadvisedthatyousolvethisonefirst.
Problem109:Darts
Inthegameofdartsaplayerthrowsthreedartsatatargetboardwhichissplitintotwentyequalsizedsectionsnumberedonetotwenty.
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Thescoreofadartisdeterminedbythenumberoftheregionthatthedartlandsin.Adartlandingoutsidethered/greenouterringscoreszero.Theblackandcreamregionsinsidethisringrepresentsinglescores.However,thered/greenouterringandmiddleringscoredoubleandtreblescoresrespectively.
Atthecentreoftheboardaretwoconcentriccirclescalledthebullregion,orbullseye.Theouterbullisworth25pointsandtheinnerbullisadouble,worth50points.
Therearemanyvariationsofrulesbutinthemostpopulargametheplayerswillbeginwithascore301or501andthefirstplayertoreducetheirrunningtotaltozeroisawinner.However,itisnormaltoplaya"doublesout"system,whichmeansthattheplayermustlandadouble(includingthedoublebullseyeatthecentreoftheboard)ontheirfinaldarttowinanyotherdartthatwouldreducetheirrunningtotaltooneorlowermeansthescoreforthatsetofthreedartsis"bust".
Whenaplayerisabletofinishontheircurrentscoreitiscalleda"checkout"andthehighestcheckoutis170:T20T20D25(twotreble20sanddoublebull).
Thereareexactlyelevendistinctwaystocheckoutonascoreof6:
D3 D1 D2 S2 D2 D2 D1 S4 D1 S1 S1 D2S1 T1 D1S1 S3 D1D1 D1 D1D1 S2 D1S2 S2 D1
NotethatD1D2isconsidereddifferenttoD2D1astheyfinishondifferentdoubles.However,thecombinationS1T1D1isconsideredthesameasT1S1D1.
Inadditionweshallnotincludemissesinconsideringcombinationsforexample,D3isthesameas0D3and00D3.
Incrediblythereare42336distinctwaysofcheckingoutintotal.
Howmanydistinctwayscanaplayercheckoutwithascorelessthan100?
Problem110:DiophantinereciprocalsII
Inthefollowingequationx,y,andnarepositiveintegers.
1x +
1y =
1n
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Itcanbeverifiedthatwhenn=1260thereare113distinctsolutionsandthisistheleastvalueofnforwhichthetotalnumberofdistinctsolutionsexceedsonehundred.
Whatistheleastvalueofnforwhichthenumberofdistinctsolutionsexceedsfourmillion?
NOTE:ThisproblemisamuchmoredifficultversionofProblem108andasitiswellbeyondthelimitationsofabruteforceapproachitrequiresacleverimplementation.
Problem111:Primeswithruns
Considering4digitprimescontainingrepeateddigitsitisclearthattheycannotallbethesame:1111isdivisibleby11,2222isdivisibleby22,andsoon.Buttherearenine4digitprimescontainingthreeones:
1117,1151,1171,1181,1511,1811,2111,4111,8111
WeshallsaythatM(n,d)representsthemaximumnumberofrepeateddigitsforanndigitprimewheredistherepeateddigit,N(n,d)representsthenumberofsuchprimes,andS(n,d)representsthesumoftheseprimes.
SoM(4,1)=3isthemaximumnumberofrepeateddigitsfora4digitprimewhereoneistherepeateddigit,thereareN(4,1)=9suchprimes,andthesumoftheseprimesisS(4,1)=22275.Itturnsoutthatford=0,itisonlypossibletohaveM(4,0)=2repeateddigits,butthereareN(4,0)=13suchcases.
Inthesamewayweobtainthefollowingresultsfor4digitprimes.
Digit,d M(4,d) N(4,d) S(4,d)
0 2 13 67061
1 3 9 22275
2 3 1 2221
3 3 12 46214
4 3 2 8888
5 3 1 5557
6 3 1 6661
7 3 9 57863
8 3 1 8887
9 3 7 48073
Ford=0to9,thesumofallS(4,d)is273700.
FindthesumofallS(10,d).
Problem112:Bouncynumbers
Workingfromlefttorightifnodigitisexceededbythedigittoitsleftitiscalledanincreasingnumberforexample,134468.
Similarlyifnodigitisexceededbythedigittoitsrightitiscalledadecreasingnumberforexample,66420.
Weshallcallapositiveintegerthatisneitherincreasingnordecreasinga"bouncy"numberforexample,155349.
Clearlytherecannotbeanybouncynumbersbelowonehundred,butjustoverhalfofthenumbersbelowonethousand(525)arebouncy.Infact,theleastnumberforwhichtheproportionofbouncynumbersfirstreaches50%is538.
Surprisingly,bouncynumbersbecomemoreandmorecommonandbythetimewereach21780theproportionofbouncynumbersisequalto90%.
Findtheleastnumberforwhichtheproportionofbouncynumbersisexactly99%.
Problem113:Nonbouncynumbers
Workingfromlefttorightifnodigitisexceededbythedigittoitsleftitiscalledanincreasingnumberforexample,134468.
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Similarlyifnodigitisexceededbythedigittoitsrightitiscalledadecreasingnumberforexample,66420.
Weshallcallapositiveintegerthatisneitherincreasingnordecreasinga"bouncy"numberforexample,155349.
Asnincreases,theproportionofbouncynumbersbelownincreasessuchthatthereareonly12951numbersbelowonemillionthatarenotbouncyandonly277032nonbouncynumbersbelow1010.
Howmanynumbersbelowagoogol(10100)arenotbouncy?
Problem114:CountingblockcombinationsI
Arowmeasuringsevenunitsinlengthhasredblockswithaminimumlengthofthreeunitsplacedonit,suchthatanytworedblocks(whichareallowedtobedifferentlengths)areseparatedbyatleastoneblacksquare.Thereareexactlyseventeenwaysofdoingthis.
Howmanywayscanarowmeasuringfiftyunitsinlengthbefilled?
NOTE:Althoughtheexampleabovedoesnotlenditselftothepossibility,ingeneralitispermittedtomixblocksizes.Forexample,onarowmeasuringeightunitsinlengthyoucouldusered(3),black(1),andred(4).
Problem115:CountingblockcombinationsII
NOTE:ThisisamoredifficultversionofProblem114.
Arowmeasuringnunitsinlengthhasredblockswithaminimumlengthofmunitsplacedonit,suchthatanytworedblocks(whichareallowedtobedifferentlengths)areseparatedbyatleastoneblacksquare.
Letthefillcountfunction,F(m,n),representthenumberofwaysthatarowcanbefilled.
Forexample,F(3,29)=673135andF(3,30)=1089155.
Thatis,form=3,itcanbeseenthatn=30isthesmallestvalueforwhichthefillcountfunctionfirstexceedsonemillion.
Inthesameway,form=10,itcanbeverifiedthatF(10,56)=880711andF(10,57)=1148904,son=57istheleastvalueforwhichthefillcountfunctionfirstexceedsonemillion.
Form=50,findtheleastvalueofnforwhichthefillcountfunctionfirstexceedsonemillion.
Problem116:Red,greenorbluetiles
Arowoffiveblacksquaretilesistohaveanumberofitstilesreplacedwithcolouredoblongtileschosenfromred(lengthtwo),green(lengththree),orblue(lengthfour).
Ifredtilesarechosenthereareexactlysevenwaysthiscanbedone.
Ifgreentilesarechosentherearethreeways.
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Andifbluetilesarechosentherearetwoways.
Assumingthatcolourscannotbemixedthereare7+3+2=12waysofreplacingtheblacktilesinarowmeasuringfiveunitsinlength.
Howmanydifferentwayscantheblacktilesinarowmeasuringfiftyunitsinlengthbereplacedifcolourscannotbemixedandatleastonecolouredtilemustbeused?
NOTE:ThisisrelatedtoProblem117.
Problem117:Red,green,andbluetiles
Usingacombinationofblacksquaretilesandoblongtileschosenfrom:redtilesmeasuringtwounits,greentilesmeasuringthreeunits,andbluetilesmeasuringfourunits,itispossibletotilearowmeasuringfiveunitsinlengthinexactlyfifteendifferentways.
Howmanywayscanarowmeasuringfiftyunitsinlengthbetiled?
NOTE:ThisisrelatedtoProblem116.
Problem118:Pandigitalprimesets
Usingallofthedigits1through9andconcatenatingthemfreelytoformdecimalintegers,differentsetscanbeformed.Interestinglywiththeset{2,5,47,89,631},alloftheelementsbelongingtoitareprime.
Howmanydistinctsetscontainingeachofthedigitsonethroughnineexactlyoncecontainonlyprimeelements?
Problem119:Digitpowersum
Thenumber512isinterestingbecauseitisequaltothesumofitsdigitsraisedtosomepower:5+1+2=8,and83=512.Anotherexampleofanumberwiththispropertyis614656=284.
Weshalldefineantobethenthtermofthissequenceandinsistthatanumbermustcontainatleasttwodigitstohaveasum.
Youaregiventhata2=512anda10=614656.
Finda30.
Problem120:Squareremainders
Letrbetheremainderwhen(a1)n+(a+1)nisdividedbya2.
Forexample,ifa=7andn=3,thenr=42:63+83=72842mod49.Andasnvaries,sotoowillr,butfora=7itturnsoutthatrmax=42.
For3a1000,findrmax.
Problem121:Discgameprizefund
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Abagcontainsonereddiscandonebluedisc.Inagameofchanceaplayertakesadiscatrandomanditscolourisnoted.Aftereachturnthediscisreturnedtothebag,anextrareddiscisadded,andanotherdiscistakenatrandom.
Theplayerpays1toplayandwinsiftheyhavetakenmorebluediscsthanreddiscsattheendofthegame.
Ifthegameisplayedforfourturns,theprobabilityofaplayerwinningisexactly11/120,andsothemaximumprizefundthebankershouldallocateforwinninginthisgamewouldbe10beforetheywouldexpecttoincuraloss.Notethatanypayoutwillbeawholenumberofpoundsandalsoincludestheoriginal1paidtoplaythegame,sointheexamplegiventheplayeractuallywins9.
Findthemaximumprizefundthatshouldbeallocatedtoasinglegameinwhichfifteenturnsareplayed.
Problem122:Efficientexponentiation
Themostnaivewayofcomputingn15requiresfourteenmultiplications:
nn...n=n15
Butusinga"binary"methodyoucancomputeitinsixmultiplications:
nn=n2
n2n2=n4
n4n4=n8
n8n4=n12
n12n2=n14
n14n=n15
Howeveritisyetpossibletocomputeitinonlyfivemultiplications:
nn=n2
n2n=n3
n3n3=n6
n6n6=n12
n12n3=n15
Weshalldefinem(k)tobetheminimumnumberofmultiplicationstocomputenkforexamplem(15)=5.
For1k200,findm(k).
Problem123:Primesquareremainders
Letpnbethenthprime:2,3,5,7,11,...,andletrbetheremainderwhen(pn1)n+(pn+1)nisdividedbypn2.
Forexample,whenn=3,p3=5,and43+63=2805mod25.
Theleastvalueofnforwhichtheremainderfirstexceeds109is7037.
Findtheleastvalueofnforwhichtheremainderfirstexceeds1010.
Problem124:Orderedradicals
Theradicalofn,rad(n),istheproductofthedistinctprimefactorsofn.Forexample,504=23327,sorad(504)=237=42.
Ifwecalculaterad(n)for1n10,thensortthemonrad(n),andsortingonniftheradicalvaluesareequal,weget:
Unsorted Sorted
n rad(n) n rad(n) k1 1 1 1 12 2 2 2 23 3 4 2 3
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4 2 8 2 45 5 3 3 56 6 9 3 67 7 5 5 78 2 6 6 89 3 7 7 910 10 10 10 10
LetE(k)bethekthelementinthesortedncolumnforexample,E(4)=8andE(6)=9.
Ifrad(n)issortedfor1n100000,findE(10000).
Problem125:Palindromicsums
Thepalindromicnumber595isinterestingbecauseitcanbewrittenasthesumofconsecutivesquares:62
+72+82+92+102+112+122.
Thereareexactlyelevenpalindromesbelowonethousandthatcanbewrittenasconsecutivesquaresums,andthesumofthesepalindromesis4164.Notethat1=02+12hasnotbeenincludedasthisproblemisconcernedwiththesquaresofpositiveintegers.
Findthesumofallthenumberslessthan108thatarebothpalindromicandcanbewrittenasthesumofconsecutivesquares.
Problem126:Cuboidlayers
Theminimumnumberofcubestocovereveryvisiblefaceonacuboidmeasuring3x2x1istwentytwo.
Ifwethenaddasecondlayertothissoliditwouldrequirefortysixcubestocovereveryvisibleface,thethirdlayerwouldrequireseventyeightcubes,andthefourthlayerwouldrequireonehundredandeighteencubestocovereveryvisibleface.
However,thefirstlayeronacuboidmeasuring5x1x1alsorequirestwentytwocubessimilarlythefirstlayeroncuboidsmeasuring5x3x1,7x2x1,and11x1x1allcontainfortysixcubes.
WeshalldefineC(n)torepresentthenumberofcuboidsthatcontainncubesinoneofitslayers.SoC(22)=2,C(46)=4,C(78)=5,andC(118)=8.
Itturnsoutthat154istheleastvalueofnforwhichC(n)=10.
FindtheleastvalueofnforwhichC(n)=1000.
Problem127:abchits
Theradicalofn,rad(n),istheproductofdistinctprimefactorsofn.Forexample,504=23327,sorad(504)=237=42.
Weshalldefinethetripletofpositiveintegers(a,b,c)tobeanabchitif:
1. GCD(a,b)=GCD(a,c)=GCD(b,c)=12. a
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1. GCD(5,27)=GCD(5,32)=GCD(27,32)=12. 5
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GiventhatnisapositiveintegerandGCD(n,10)=1,itcanbeshownthattherealwaysexistsavalue,k,forwhichR(k)isdivisiblebyn,andletA(n)betheleastsuchvalueofkforexample,A(7)=6andA(41)=5.
Youaregiventhatforallprimes,p>5,thatp1isdivisiblebyA(p).Forexample,whenp=41,A(41)=5,and40isdivisibleby5.
However,therearerarecompositevaluesforwhichthisisalsotruethefirstfiveexamplesbeing91,259,451,481,and703.
FindthesumofthefirsttwentyfivecompositevaluesofnforwhichGCD(n,10)=1andn1isdivisiblebyA(n).
Problem131:Primecubepartnership
Therearesomeprimevalues,p,forwhichthereexistsapositiveinteger,n,suchthattheexpressionn3+n2pisaperfectcube.
Forexample,whenp=19,83+8219=123.
Whatisperhapsmostsurprisingisthatforeachprimewiththispropertythevalueofnisunique,andthereareonlyfoursuchprimesbelowonehundred.
Howmanyprimesbelowonemillionhavethisremarkableproperty?
Problem132:Largerepunitfactors
Anumberconsistingentirelyofonesiscalledarepunit.WeshalldefineR(k)tobearepunitoflengthk.
Forexample,R(10)=1111111111=11412719091,andthesumoftheseprimefactorsis9414.
FindthesumofthefirstfortyprimefactorsofR(109).
Problem133:Repunitnonfactors
Anumberconsistingentirelyofonesiscalledarepunit.WeshalldefineR(k)tobearepunitoflengthkforexample,R(6)=111111.
LetusconsiderrepunitsoftheformR(10n).
AlthoughR(10),R(100),orR(1000)arenotdivisibleby17,R(10000)isdivisibleby17.YetthereisnovalueofnforwhichR(10n)willdivideby19.Infact,itisremarkablethat11,17,41,and73aretheonlyfourprimesbelowonehundredthatcanbeafactorofR(10n).
FindthesumofalltheprimesbelowonehundredthousandthatwillneverbeafactorofR(10n).
Problem134:Primepairconnection
Considertheconsecutiveprimesp1=19andp2=23.Itcanbeverifiedthat1219isthesmallestnumbersuchthatthelastdigitsareformedbyp1whilstalsobeingdivisiblebyp2.
Infact,withtheexceptionofp1=3andp2=5,foreverypairofconsecutiveprimes,p2>p1,thereexistvaluesofnforwhichthelastdigitsareformedbyp1andnisdivisiblebyp2.LetSbethesmallestofthesevaluesofn.
FindSforeverypairofconsecutiveprimeswith5p11000000.
Problem135:Samedifferences
Giventhepositiveintegers,x,y,andz,areconsecutivetermsofanarithmeticprogression,theleastvalueofthepositiveinteger,n,forwhichtheequation,x2y2z2=n,hasexactlytwosolutionsisn=27:
342272202=1229262=27
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Itturnsoutthatn=1155istheleastvaluewhichhasexactlytensolutions.
Howmanyvaluesofnlessthanonemillionhaveexactlytendistinctsolutions?
Problem136:Singletondifference
Thepositiveintegers,x,y,andz,areconsecutivetermsofanarithmeticprogression.Giventhatnisapositiveinteger,theequation,x2y2z2=n,hasexactlyonesolutionwhenn=20:
13210272=20
Infacttherearetwentyfivevaluesofnbelowonehundredforwhichtheequationhasauniquesolution.
Howmanyvaluesofnlessthanfiftymillionhaveexactlyonesolution?
Problem137:Fibonaccigoldennuggets
ConsidertheinfinitepolynomialseriesAF(x)=xF1+x2F2+x3F3+...,whereFkisthekthtermintheFibonaccisequence:1,1,2,3,5,8,...thatis,Fk=Fk1+Fk2,F1=1andF2=1.
ForthisproblemweshallbeinterestedinvaluesofxforwhichAF(x)isapositiveinteger.
SurprisinglyAF(1/2)=(1/2).1+(1/2)2.1+(1/2)3.2+(1/2)4.3+(1/2)5.5+... =1/2+1/4+2/8+3/16+5/32+... =2
Thecorrespondingvaluesofxforthefirstfivenaturalnumbersareshownbelow.
x AF(x)21 11/2 2
(132)/3 3(895)/8 4(343)/5 5
WeshallcallAF(x)agoldennuggetifxisrational,becausetheybecomeincreasinglyrarerforexample,the10thgoldennuggetis74049690.
Findthe15thgoldennugget.
Problem138:Specialisoscelestriangles
Considertheisoscelestrianglewithbaselength,b=16,andlegs,L=17.
ByusingthePythagoreantheoremitcanbeseenthattheheightofthetriangle,h=(17282)=15,whichisonelessthanthebaselength.
Withb=272andL=305,wegeth=273,whichisonemorethanthebaselength,andthisisthesecondsmallestisoscelestrianglewiththepropertythath=b1.
FindLforthetwelvesmallestisoscelestrianglesforwhichh=b1andb,Larepositiveintegers.
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Problem139:Pythagoreantiles
Let(a,b,c)representthethreesidesofarightangletrianglewithintegrallengthsides.Itispossibletoplacefoursuchtrianglestogethertoformasquarewithlengthc.
Forexample,(3,4,5)trianglescanbeplacedtogethertoforma5by5squarewitha1by1holeinthemiddleanditcanbeseenthatthe5by5squarecanbetiledwithtwentyfive1by1squares.
However,if(5,12,13)triangleswereusedthentheholewouldmeasure7by7andthesecouldnotbeusedtotilethe13by13square.
Giventhattheperimeteroftherighttriangleislessthanonehundredmillion,howmanyPythagoreantriangleswouldallowsuchatilingtotakeplace?
Problem140:ModifiedFibonaccigoldennuggets
ConsidertheinfinitepolynomialseriesAG(x)=xG1+x2G2+x3G3+...,whereGkisthekthtermofthesecondorderrecurrencerelationGk=Gk1+Gk2,G1=1andG2=4thatis,1,4,5,9,14,23,....
ForthisproblemweshallbeconcernedwithvaluesofxforwhichAG(x)isapositiveinteger.
Thecorrespondingvaluesofxforthefirstfivenaturalnumbersareshownbelow.
x AG(x)(51)/4 1
2/5 2(222)/6 3
(1375)/14 41/2 5
WeshallcallAG(x)agoldennuggetifxisrational,becausetheybecomeincreasinglyrarerforexample,the20thgoldennuggetis211345365.
Findthesumofthefirstthirtygoldennuggets.
Problem141:Investigatingprogressivenumbers,n,whicharealsosquare
Apositiveinteger,n,isdividedbydandthequotientandremainderareqandrrespectively.Inadditiond,q,andrareconsecutivepositiveintegertermsinageometricsequence,butnotnecessarilyinthatorder.
Forexample,58dividedby6hasquotient9andremainder4.Itcanalsobeseenthat4,6,9areconsecutivetermsinageometricsequence(commonratio3/2).Wewillcallsuchnumbers,n,progressive.
Someprogressivenumbers,suchas9and10404=1022,happentoalsobeperfectsquares.Thesumofallprogressiveperfectsquaresbelowonehundredthousandis124657.
Findthesumofallprogressiveperfectsquaresbelowonetrillion(1012).
Problem142:PerfectSquareCollection
Findthesmallestx+y+zwithintegersx>y>z>0suchthatx+y,xy,x+z,xz,y+z,yzareallperfectsquares.
Problem143:InvestigatingtheTorricellipointofatriangle
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LetABCbeatrianglewithallinterioranglesbeinglessthan120degrees.LetXbeanypointinsidethetriangleandletXA=p,XC=q,andXB=r.
FermatchallengedTorricellitofindthepositionofXsuchthatp+q+rwasminimised.
TorricelliwasabletoprovethatifequilateraltrianglesAOB,BNCandAMCareconstructedoneachsideoftriangleABC,thecircumscribedcirclesofAOB,BNC,andAMCwillintersectatasinglepoint,T,insidethetriangle.MoreoverheprovedthatT,calledtheTorricelli/Fermatpoint,minimisesp+q+r.Evenmoreremarkable,itcanbeshownthatwhenthesumisminimised,AN=BM=CO=p+q+randthatAN,BMandCOalsointersectatT.
Ifthesumisminimisedanda,b,c,p,qandrareallpositiveintegersweshallcalltriangleABCaTorricellitriangle.Forexample,a=399,b=455,c=511isanexampleofaTorricellitriangle,withp+q+r=784.
Findthesumofalldistinctvaluesofp+q+r120000forTorricellitriangles.
Problem144:Investigatingmultiplereflectionsofalaserbeam
Inlaserphysics,a"whitecell"isamirrorsystemthatactsasadelaylineforthelaserbeam.Thebeamentersthecell,bouncesaroundonthemirrors,andeventuallyworksitswaybackout.
Thespecificwhitecellwewillbeconsideringisanellipsewiththeequation4x2+y2=100
Thesectioncorrespondingto0.01x+0.01atthetopismissing,allowingthelighttoenterandexitthroughthehole.
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Thelightbeaminthisproblemstartsatthepoint(0.0,10.1)justoutsidethewhitecell,andthebeamfirstimpactsthemirrorat(1.4,9.6).
Eachtimethelaserbeamhitsthesurfaceoftheellipse,itfollowstheusuallawofreflection"angleofincidenceequalsangleofreflection."Thatis,boththeincidentandreflectedbeamsmakethesameanglewiththenormallineatthepointofincidence.
Inthefigureontheleft,theredlineshowsthefirsttwopointsofcontactbetweenthelaserbeamandthewallofthewhitecellthebluelineshowsthelinetangenttotheellipseatthepointofincidenceofthefirstbounce.
Theslopemofthetangentlineatanypoint(x,y)ofthegivenellipseis:m=4x/y
Thenormallineisperpendiculartothistangentlineatthepointofincidence.
Theanimationontherightshowsthefirst10reflectionsofthebeam.
Howmanytimesdoesthebeamhittheinternalsurfaceofthewhitecellbeforeexiting?
Problem145:Howmanyreversiblenumbersaretherebelowonebillion?
Somepositiveintegersnhavethepropertythatthesum[n+reverse(n)]consistsentirelyofodd(decimal)digits.Forinstance,36+63=99and409+904=1313.Wewillcallsuchnumbersreversibleso36,63,409,and904arereversible.Leadingzeroesarenotallowedineithernorreverse(n).
Thereare120reversiblenumbersbelowonethousand.
Howmanyreversiblenumbersaretherebelowonebillion(109)?
Problem146:InvestigatingaPrimePattern
Thesmallestpositiveintegernforwhichthenumbersn2+1,n2+3,n2+7,n2+9,n2+13,andn2+27areconsecutiveprimesis10.Thesumofallsuchintegersnbelowonemillionis1242490.
Whatisthesumofallsuchintegersnbelow150million?
Problem147:Rectanglesincrosshatchedgrids
Ina3x2crosshatchedgrid,atotalof37differentrectanglescouldbesituatedwithinthatgridasindicatedinthesketch.
Thereare5gridssmallerthan3x2,verticalandhorizontaldimensionsbeingimportant,i.e.1x1,2x1,3x1,1x2and2x2.Ifeachofthemiscrosshatched,thefollowingnumberofdifferentrectanglescouldbesituatedwithinthosesmallergrids:
1x1:12x1:43x1:81x2:42x2:18
Addingthosetothe37ofthe3x2grid,atotalof72differentrectanglescouldbesituatedwithin3x2andsmallergrids.
Howmanydifferentrectanglescouldbesituatedwithin47x43andsmallergrids?
Problem148:ExploringPascal'striangle
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WecaneasilyverifythatnoneoftheentriesinthefirstsevenrowsofPascal'strianglearedivisibleby7:
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 11 6 15 20 15 6 1
However,ifwecheckthefirstonehundredrows,wewillfindthatonly2361ofthe5050entriesarenotdivisibleby7.
Findthenumberofentrieswhicharenotdivisibleby7inthefirstonebillion(109)rowsofPascal'striangle.
Problem149:Searchingforamaximumsumsubsequence
Lookingatthetablebelow,itiseasytoverifythatthemaximumpossiblesumofadjacentnumbersinanydirection(horizontal,vertical,diagonalorantidiagonal)is16(=8+7+1).
2 5 3 2
9 6 5 1
3 2 7 3
1 8 4 8
Now,letusrepeatthesearch,butonamuchlargerscale:
First,generatefourmillionpseudorandomnumbersusingaspecificformofwhatisknownasa"LaggedFibonacciGenerator":
For1k55,sk=[100003200003k+300007k3](modulo1000000)500000.For56k4000000,sk=[sk24+sk55+1000000](modulo1000000)500000.
Thus,s10=393027ands100=86613.
Thetermsofsarethenarrangedina20002000table,usingthefirst2000numberstofillthefirstrow(sequentially),thenext2000numberstofillthesecondrow,andsoon.
Finally,findthegreatestsumof(anynumberof)adjacententriesinanydirection(horizontal,vertical,diagonalorantidiagonal).
Problem150:Searchingatriangulararrayforasubtrianglehavingminimumsum
Inatriangulararrayofpositiveandnegativeintegers,wewishtofindasubtrianglesuchthatthesumofthenumbersitcontainsisthesmallestpossible.
Intheexamplebelow,itcanbeeasilyverifiedthatthemarkedtrianglesatisfiesthisconditionhavingasumof42.
Wewishtomakesuchatriangulararraywithonethousandrows,sowegenerate500500pseudorandomnumbersskintherange219,usingatypeofrandomnumbergenerator(knownasaLinearCongruentialGenerator)asfollows:
t:=0fork=1uptok=500500:t:=(615949*t+797807)modulo220
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sk:=t219
Thus:s1=273519,s2=153582,s3=450905etc
Ourtriangulararrayisthenformedusingthepseudorandomnumbersthus:
s1s2s3
s4s5s6s7s8s9s10
...
Subtrianglescanstartatanyelementofthearrayandextenddownasfaraswelike(takinginthetwoelementsdirectlybelowitfromthenextrow,thethreeelementsdirectlybelowfromtherowafterthat,andsoon).The"sumofasubtriangle"isdefinedasthesumofalltheelementsitcontains.Findthesmallestpossiblesubtrianglesum.
Problem151:Papersheetsofstandardsizes:anexpectedvalueproblem
Aprintingshopruns16batches(jobs)everyweekandeachbatchrequiresasheetofspecialcolourproofingpaperofsizeA5.
EveryMondaymorning,theforemanopensanewenvelope,containingalargesheetofthespecialpaperwithsizeA1.
Heproceedstocutitinhalf,thusgettingtwosheetsofsizeA2.ThenhecutsoneoftheminhalftogettwosheetsofsizeA3andsoonuntilheobtainstheA5sizesheetneededforthefirstbatchoftheweek.
Alltheunusedsheetsareplacedbackintheenvelope.
Atthebeginningofeachsubsequentbatch,hetakesfromtheenvelopeonesheetofpaperatrandom.IfitisofsizeA5,heusesit.Ifitislarger,herepeatsthe'cutinhalf'procedureuntilhehaswhatheneedsandanyremainingsheetsarealwaysplacedbackintheenvelope.
Excludingthefirstandlastbatchoftheweek,findtheexpectednumberoftimes(duringeachweek)thattheforemanfindsasinglesheetofpaperintheenvelope.
Giveyouranswerroundedtosixdecimalplacesusingtheformatx.xxxxxx.
Problem152:Writing1/2asasumofinversesquares
Thereareseveralwaystowritethenumber1/2asasumofinversesquaresusingdistinctintegers.
Forinstance,thenumbers{2,3,4,5,7,12,15,20,28,35}canbeused:
Infact,onlyusingintegersbetween2and45inclusive,thereareexactlythreewaystodoit,theremainingtwobeing:{2,3,4,6,7,9,10,20,28,35,36,45}and{2,3,4,6,7,9,12,15,28,30,35,36,45}.
Howmanywaysaretheretowritethenumber1/2asasumofinversesquaresusingdistinctintegersbetween2and80inclusive?
Problem153:InvestigatingGaussianIntegers
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Asweallknowtheequationx2=1hasnosolutionsforrealx.Ifwehoweverintroducetheimaginarynumberithisequationhastwosolutions:x=iandx=i.Ifwegoastepfurthertheequation(x3)2=4hastwocomplexsolutions:x=3+2iandx=32i.x=3+2iandx=32iarecalledeachothers'complexconjugate.Numbersoftheforma+biarecalledcomplexnumbers.Ingenerala+biandabiareeachother'scomplexconjugate.
AGaussianIntegerisacomplexnumbera+bisuchthatbothaandbareintegers.TheregularintegersarealsoGaussianintegers(withb=0).TodistinguishthemfromGaussianintegerswithb0wecallsuchintegers"rationalintegers."AGaussianintegeriscalledadivisorofarationalintegerniftheresultisalsoaGaussianinteger.Ifforexamplewedivide5by1+2iwecansimplify inthefollowingmanner:
Multiplynumeratoranddenominatorbythecomplexconjugateof1+2i:12i.
Theresultis .
So1+2iisadivisorof5.Notethat1+iisnotadivisorof5because .
NotealsothatiftheGaussianInteger(a+bi)isadivisorofarationalintegern,thenitscomplexconjugate(abi)isalsoadivisorofn.
Infact,5hassixdivisorssuchthattherealpartispositive:{1,1+2i,12i,2+i,2i,5}.Thefollowingisatableofallofthedivisorsforthefirstfivepositiverationalintegers:
n GaussianintegerdivisorswithpositiverealpartSums(n)ofthesedivisors
1 1 12 1,1+i,1i,2 53 1,3 44 1,1+i,1i,2,2+2i,22i,4 135 1,1+2i,12i,2+i,2i,5 12
Fordivisorswithpositiverealparts,then,wehave: .
For1n105,s(n)=17924657155.
Whatiss(n)for1n108?
Problem154:ExploringPascal'spyramid
Atriangularpyramidisconstructedusingsphericalballssothateachballrestsonexactlythreeballsofthenextlowerlevel.