Solutionstothe79thWilliamLowellPutnamMathematicalCompetitionSaturday,December1,2018KiranKedlayaandLennyNgWhilepreparingthesesolutions,welearnedoftheNovem-InMn,performthefollowingoperations,whichdonotber27deathofKentMerryfield,whomoderatedthePut-changethedeterminant:subtractthefinalrowfromnamdiscussionsontheArtofProblemSolvingForumsandrows2n−1through2n−2,andthensubtractthefinaltherebycontributeddirectlyandindirectlytothesesolutionscolumnfromcolumns2n−1through2n−2.Theresultovermanyyears.Hispresencewillbedearlymissed.isthematrixA1Byclearingdenominatorsandregrouping,weseethat0thegivenequationisequivalentto..Mn−1Mn−1.2(3a−2018)(3b−2018)=2018.00···00.()Eachofthefactorsiscongruentto1mod3.There....are6positivefactorsof20182=22·10092thatareM.....n−1congruentto1(mod3):1,22,1009,22·1009,0···0010092,22·10092.Theseleadtothe6possiblepairs:0···00···01(a,b)=(673,1358114),(674,340033),(1009,2018),(2018,1009),(340033,674),and(1358114,673).Wecanremovethefinalrowandcolumnwith-Asfornegativefactors,theonesthatarecongruentto1outchangingthedeterminant.Nowswapthe2(mod3)are−2,−2·1009,−2·1009.However,alloffirst2n−1−1rowswiththefinal2n−1−1rows:theseleadtopairswherea≤0orb≤0.thischangesthedeterminantbyanoverallfactor(2n−1−1)2A2Theansweris1ifn=1and−1ifn>1.WriteMnof(−1)=−1.Theresultistheblock-nn!fora(2−1)×(2−1)matrixofthegivenform,andMn−10notethatdetMdoesnotdependontheorderingofthediagonalmatrix,whosedeterminantisnMn−1Mn−1subsets:transposingtwosubsetshastheeffectoftrans-(detM)2.ThusdetM=−(detM)2asdesired.posingtworowsandthentransposingtwocolumnsinn−1nn−1Mn,andthisdoesnotchangethedeterminant.A3Themaximumvalueis480/49.Sincecos(3xi)=3ClearlydetM1=1.Weclaimthatforn>1,detMn=4cos(xi)−3cos(xi),itisequivalenttomaximize210310−(detMn−1),andthedesiredanswerwillfollowbyin-4∑i=1yifory1,...,y10∈[−1,1]with∑i=1yi=0;noteS0,...,S0thatthisdomainiscompact,sothemaximumvalueduction.Let12n−1−1denotethenonemptysub-setsof{1,...,n−1}inanyorder,withresultingmatrixisguaranteedtoexist.Forconvenience,weestablishn3Mm0(i,j)Msomethingslightlystronger:wemaximize4∑i=1yiforn−1.Letijdenotetheentryofn−1.Nowor-ny1,...,yn∈[−1,1]with∑yi=0,wherenmaybederthenonemptysubsetsS1,...,S2n−1of{1,...,n}asi=1follows:anyevennonnegativeintegerupto10,andshowthatthemaximumisachievedwhenn=10.0n−1Sii≤2−1Wefirststudytheeffectofvaryingyiandyjwhile0n−1nSi=Sn−1∪{n}2≤i≤2−2fixingtheirsum.Ifthatsumiss,thenthefunctioni−2+133{n}i=2n−1.y7→y+(s−y)hasconstantsecondderivative6s,soitiseithereverywhereconvexoreverywherecon-S0,...,S0cave.Consequently,if(y1,...,yn)achievesthemaxi-(Forexample,if12n−1−1areorderedinlex-icographicorderasbinarystrings,thensoaremum,thenforanytwoindicesi−1,wemusthavea1.Incasem>n,setg˜=gh,Nowsincegcd(m,n)=1,thereisanelementφ∈h˜=h,andGL2(Z)sending(n,m)to(1,0),whichthensendsthelinesegmentfrom(0,0)to(n,m)tothesegmentfrom(m−n)k(m−n)(k−1)(0,0)to(1,0).ThenφinducesahomeomorphismofTb=−(k=1,...,n).knnsendingγtog,whichinturninducesanisomorphism−1φ∗:π1(T\{p})→π1(T\{φ(p)}).Bothfundamen-thentalgroupsareequaltohg,hi,andweconcludethatφ∗aaasendsgh1gh2···ghntog.Itfollowsthatφ∗inducesg˜h˜b1···g˜h˜bn=gha1···ghan=e,anisomorphism˜a1a2ansotheinductionhypothesisimpliesthatg˜andhcom-hg,h|ghgh···ghi→hg,h|gi=∼hhi=∼Z.mute;thisimpliesthatgandhcommute.Incasem0forsomex0>0,thenbythemeanvaluetheo-vex.Conversely,iffisultraconvexand0f(x0)remthereexistsx∈(0,x0)forwhichf(x)=<0.liminfx→−∞f(x)≥0,thenfisultraconvex.x0Inparticular,f0(0)=0,sof0∈Salso.–TheclassofultraconvexfunctionsisclosedunderWeshowbyinductionthatforalln≥0,addition,multiplication,andcomposition.Secondsolution.(byZacharyChase)Inthissolution,f(n)(1)f(x)≤xn(f∈S,x∈[0,1]).weuseBernstein’stheoremonmonotonefunctions.Ton!statethisresult,wesaythatafunctionf:[0,∞)→Ristotallymonotoneiffiscontinuous,fisinfinitelydiffer-Weinductwithbasecasen=0,whichholdsbecausen(n)anyf∈Sisnondecreasing.Giventheclaimforn=m,entiableon(0,∞),and(−1)f(x)isnonnegativeforweapplytheinductionhypothesistof0∈Stoseethatallpositiveintegersnandallx>0.Forsuchafunction,Bernstein’stheoremassertsthatthereisanonnegativef(n+1)(1)finiteBorelmeasureµon[0,∞)suchthatf0(t)≤tn(t∈[0,1]),n!Z∞f(x)=e−txdµ(t)(x≥0).thenintegratebothsidesfrom0toxtoconclude.0f(n)(1)Forfasintheproblemstatement,foranyM>0,theNowforf∈S,wehave0≤f(1)≤n!foralln≥0.Ontheotherhand,byTaylor’stheoremwithremainder,restrictionoff(M−x)to[0,∞)istotallymonotone,soBernstein’stheoremprovidesaBorelmeasureµforn(k)R∞−txf(1)whichf(M−x)=0edµ(t)forallx≥0.Takingf(x)≥(x−1)k(x≥1).R∞−Mt∑k!x=M,weseethat0edµ(t)=f(0)=0;sinceµk=0isanonnegativemeasure,itmustbeidenticallyzero.Hencef(x)isidenticallyzeroforx≤M;varyingovernf(k)(1)Applyingthiswithx=2,weobtainf(2)≥∑k=0k!allM,wededucethedesiredresult.f(n)(1)foralln;thisimpliesthatlimn→∞n!=0.SinceThirdsolution.(fromArtofProblemSolvinguser(n)f(1)≤f(1),wemusthavef(1)=0.chronondecay)Inthissolution,weonlyconsiderthen!behavioroffon[0,1].WefirstestablishthefollowingForf∈S,weprovedearlierthatf(x)=0forallx≤0,result.Letf:(0,1)→Rbeafunctionsuchthatforaswellasforx=1.Sincethefunctiong(x)=f(cx)iseachpositiveintegern,f(n)(x)isnonnegativeon(0,1),alsoultraconvexforc>0,wealsohavef(x)=0foralltendsto0asx→0+,andtendstosomelimitasx→1−.x>0;hencefisidenticallyzero.Thenforeachnonnegativeintegern,f(x)x−nisnonde-Tosumup,iff:R→Risinfinitelydifferentiable,creasingon(0,1).f(0)=0,andf(1)=1,thenfcannotbeultraconvex.Toprovetheclaimedresult,weproceedbyinductiononThisimpliesthedesiredresult.n,thecasen=0beingaconsequenceoftheassump-Remark.NoamElkiespointsoutthatonecanrefinetionthatf0(x)isnonnegativeon(0,1).Giventheclaimtheargumenttoshowthatiffisultraconvex,thenitisforsomen≥0,notethatsincef0alsosatisfiesthehy-analytic(i.e.,itisrepresentedbyanentireTaylorseriespothesesoftheproblem,f0(x)x−nisalsonondecreasingaboutanypoint,asopposedtoafunctionlikef(x)=on(0,1).Choosec∈(0,1)andconsiderthefunction2e−1/xwhoseTaylorseriesat0isidenticallyzero);hef0(c)attributesthefollowingargumenttoPeterShalen.Letg(x)=xn(x∈[0,1)).n1(kkcngn(x)=∑k=0k!f)(0)xbethen-thorderTaylorpoly-nomialoff.ByTaylor’stheoremwithremainder(a/k/aForx∈(0,c),f0(x)x−n≤f0(c)c−n,sof0(x)≤g(x);Lagrange’stheorem),f(x)−g(x)iseverywherenon-nsimilarly,forx∈(c,1),f0(x)≥g(x).Itfollowsthatnegative;consequently,forallx≥0,theTaylorseries0∞1(n)niff(c)>0,then∑n=0n!f(0)xconvergesandisboundedabovebyf.(n+1)Butsincef(x)isnondecreasing,Lagrange’sthe-R1f0(x)dxR1g(x)dxRcf0(x)dxRcg(x)dx1(n+1)cc00oremalsoimpliesthatf(x)−gn(x)≤f(x);Rc≥Rc⇒≤(n+1)!f0(x)dxg(x)dxR1f0(x)dxR1g(x)dxforfixedx≥0,therightsidetendsto0asn→∞.Hence0000fisrepresentedbyitsTaylorseriesforx≥0,andsoisandsof(c)/f(1)≤cn+1.(Hereforconvenience,weanalyticforx>0;byreplacingf(x)withf(x−c),weextendfcontinuouslyto[0,1].)Thatis,f(c)/cn+1≤mayconcludethatfiseverywhereanalytic.f(1)forallc∈(0,1).Foranyb∈(0,1),wemayapplyRemark.Werecordsomepropertiesoftheclassofthesamelogictothefunctionf(bx)todeducethatifultraconvexfunctions.f0(c)>0,thenf(bc)/cn+1≤f(b),orequivalently–Anynonnegativeconstantfunctionisultraconvex.f(bc)f(b)≤.Theexponentialfunctionisultraconvex.(bc)n+1bn+14Thisyieldstheclaimunlessf0isidentically0on(0,1),thenarebutinthatcasetheclaimisobviousanyway.√22(x1a)Wenowapplytheclaimtoshowthatforfasinthex1=(n)aproblemstatement,itcannotbethecasethatf(x)is√2nonnegativeon(0,1)foralln.Supposethecontrary;2(x2a)x2=thenforanyfixedx∈(0,1),wemayapplytheprevious√a√claimwitharbitrarilylargentodeducethatf(x)=0.(x1a)(x2a)xx=Bycontinuity,wealsothenhavef(1)=0,acontradic-12ation.2222y1=(x1+y1)−x1Fourthsolution.(byAlexanderKarabegov)Asinthey2=(x2+y2)−x2.firstsolution,wemayseethatf(n)(0)=0foralln.Con-2222sequently,forallnwehaveNownotethatthequantityZx1n−1(n)(x−x)2+(y−y)2=x2−2xx+x2+y2−2yy+y2f(x)=(x−t)f(t)dt(x∈R)121211221122(n−1)!0isknowntoberational,asiseverysummandontheandhence2rightexcept−2y1y2;thusy1y2isalsorational.Sincey12Z1Z1isalsorational,sothenisy1/y2=(y1y2)/(y);since1n(n)1f(x)dx=(1−t)f(t)dt.√√0n!0area(4ABC)=ay1,area(4ABD)=ay2,Supposenowthatfisinfinitelydifferentiable,f(1)=thisyieldsthedesiredresult.1,andf(n)(x)≥0forallnandallx∈[0,1].ThenSecondsolution.(byManjulBhargava)Letb,c,dbeZ111Z1thevectorsAB,AC,ADviewedascolumnvectors.Thef(x)dx=·(1−t)nf(n)(t)dt0n(n−1)!0desiredratioisgivenbyZ111n−1(n)det(b,c)det(b,c)Tdet(b,c)≤·(1−t)f(t)dt=n(n−1)!0det(b,d)det(b,c)Tdet(b,d)11!!−1=f(1)=.b·bb·cb·bb·dnn=detdet.c·bc·cc·bc·dR1Sincethisholdsforalln,wehave0f(x)dx=0,andsof(x)=0forx∈[0,1];thisyieldsthedesiredcontra-ThesquareofthelengthofABisb·b,sothisquantityisdiction.rational.ThesquareofthelengthsofACandBCarec·cA6Firstsolution.ChooseaCartesiancoordinatesys-and(c−b)·(c−b)=b·b+c·c−2b·c,sob·c=c·btemwithoriginatthemidpointofABandpositivex-isrational.Similarly,usingADandBD,wededucethatd·dandb·disrational;thenusingCD,wededucethataxiscontaining√A.By√theconditiononAB,wehaveA=(a,0),B=(−a,0)forsomepositiverationalc·disrational.numbera.Let(x1,y1)and(x2,y2)betherespectiveco-Thirdsolution.(byDavidRusin)RecallthatHeron’sordinatesofCandD;bycomputingthelengthsoftheformula(fortheareaofatriangleintermsofitssidesegmentsAC,BC,AD,BD,CD,weseethatthequanti-length)admitsthefollowingthree-dimensionalana-tieslogueduetoPierodellaFrancesca:ifVdenotesthevolumeofatetrahedronwithverticesA,B,C,D∈3,√22√22R(x1−a)+y1,(x1+a)+y1,then√22√22(x2−a)+y2,(x2+a)+y2,0AB2AC2AD21(x−x)2+(y−y)21212AB20BC2BD212222areallrationalnumbers.Byaddingandsubtractingthe288V=detACBC0CD1firsttwoquantities,andsimilarlyforthenexttwo,weAD2BD2CD201seethatthequantities1111022√22√x1+y1,x1a,x2+y2,x2aInparticular,thedeterminantvanishesifandonlyifA,B,C,Darecoplanar.Fromtheidentityarerationalnumbers.Sinceaisarationalnumber,so64(4Area(4ABC)2Area(4ABD)2−9AB2V2)=(AB4−AB2(AC2+AD2+BC2+BD2−2CD2)+(AC2−BC2)(AD2−BD2))25weseethatArea(4ABC)Area(4ABD)isrational;pairs,assignhalfofthemtoS1andhalftoS2,andsinceeachoftheareashasrationalsquare,wededucethencompletethepartitionofP\{v}byassign-theclaim.ing(0,100−b),(2,25+b/2),and(1,50)toS1andRemark.DerekSmithobservesthatthisresultis(1,100−b),(2,b),and(0,75−b/2)toS2.(NotethatProposition1of:M.Knopf,J.Milzman,D.Smith,D.thethreevectorsassignedtoeachofS1andS2havetheZhuandD.Zirlin,Latticeembeddingsofplanarpointsamesum(3,175−b/2).)Byconstruction,S1andS2sets,DiscreteandComputationalGeometry56(2016),havethesamenumberofelements,andΣ(S1)=Σ(S2).693–710.Forb=50,thisconstructiondoesnotworkbe-Remark.Itisworthpointingoutthatitisindeedpossi-cause(1,b)=(100−b),butaslightvariationcanbletochoosepointsA,B,C,Dsatisfyingtheconditionsbemade.Inthiscase,threeofthepairsinP\oftheproblem;onecanevenensurethatthelengths{(1,50)}are{(2,50),(0,50)},{(1,51),(1,49)},andofallfoursegmentsarethemselvesrational.Forex-{(0,49),(2,51)}.Assignhalfoftheother148pairsample,itwasoriginallyobservedbyEulerthatonecantoS1andhalftoS2,andcompletethepartitionoffindaninfinitesetofpointsontheunitcirclewhoseP\{(1,50)}byassigning(2,50),(1,51),and(0,49)pairwisedistancesareallrationalnumbers.OnewaytotoS1and(0,50),(1,49),and(2,51)toS2.seethisistoapplythelinearfractionaltransformationB2Firstsolution.Notefirstthatfn(1)>0,so1isnotaf(z)=z+itotheRiemannspheretocarrytherealaxisz−irootoffn.Next,notethat(plus∞)totheunitcircle,thencomputethat(z−1)f(z)=zn+···+z−n;2|z−z||n|f(z)−f(z)|=12.12n|(z1−i)(z2−i)|however,for|z|≤1,wehave|z+···+z|≤nbythetriangleinequality;equalitycanonlyoccurifz,...,zn2LetSbethesetofrationalnumberszforwhich2(z+1)havenorm1andthesameargument,whichonlyhap-isaperfectsquare;thesetf(S)hasthedesiredpropertypensforz=1.Thustherecanbenorootoffnwithprovidedthatitisinfinite.Thatcanbecheckedinvar-|z|≤1.iousways;forinstance,theequation2(x2+1)=(2y)2Remark.Arelatedapproachistowriteequatestox2−2y2=−1(amodifiedBrahmagupta-Pellequation),whichhasinfinitelymanysolutionseven(z−1)2f(z)=zn+1−(n+1)z+novertheintegers:n√√andagainusethetriangleinequalitytoseethatfor|z|≤2n+1x+y2=(1+2).1,zn+1+n≤|(n+1)z|withequalityonlywhenzn+1isapositiverealnumber,necessarily1.However,ifB1Theansweristhecollectionofvectors(1,b)where0≤zn+1=1andz6=1,thenn+16=(n+1)zandsozisnotb≤100andbiseven.(Foreaseoftypography,wearootoffn.writetuplesinsteadofcolumnvectors.)Secondsolution.(byKarlMahlburg)Definethepoly-FirstweshowthatifP\{v}canbepartitionedintonomialsubsetsS1andS2ofequalsizeandequalsum,thenvn−1mustbeoftheform(1,b)wherebiseven.Forafinitegn(z)=nz+···+2z+1nonemptysetSofvectorsin2,letΣ(S)denotetheZn−1−1sumofthevectorsinS.Sincetheaveragex-andy-andnotethatzgn(z)=fn(z).Sincefn(0)6=0,tocoordinatesinPare1and50,respectively,andthereprovetheclaimitisequivalenttoshowthatgnhasnoare3·101elementsinP,wehaverootsintheregion|z|≥1.0Nownotethatgn(z)=hn(z)forΣ(P)=303·(1,50)=(303,15150).nhn(z)=z+···+z+1,Ontheotherhand,apolynomialwithrootse2πij/(n+1)forj=0,...,n.ByΣ(P)=v+Σ(S1)+Σ(S2)=v+2Σ(S1).theGauss-Lucastheorem,therootsofgnlieinthecon-vexhulloftherootsofh,andmoreovercannotbeByparityconsiderations,theentriesofvmustbeoddnverticesoftheconvexhullbecausehhasnorepeatedandeven,respectively,andthusvisoftheclaimednroots.Thisimpliestheclaim.form.Remark.YetanotherapproachistousetheEnestrom-Nextsupposev=(1,b)wherebiseven.NotethatKakeyatheorem:ifP(z)=a+···+aznisapolyno-P\{(1,50)}canbepartitionedinto151pairsof(dis-n0nmialwithrealcoefficientssatisfying|a|≥···≥|a|>tinct)vectors(x,y)and(2−x,100−y),eachsum-n00,thentherootsofP(z)allsatisfy|z|≤1.Namely,ap-mingto(2,100).Ifb6=50thenthreeofthesenplyingthistothepolynomialg(z/c)forc=n/(n−1)pairsare{(1,b),(1,100−b)},{(2,b),(0,100−b)},andnshowsthattherootsofgallsatisfy|z|≤1/c.{(2,25+b/2),(0,75−b/2)}.Oftheremaining148nRemark.Forarelatedproblem,seeproblemA5fromthe2014Putnamcompetition.62`B3Thevaluesofnwiththispropertyare2for`=andsoxm+1=2xmxm−1−xm−2=cos(Fm+1b).This1,2,4,8.First,notethatndivides2nifandonlyifncompletestheinduction.misitselfapowerof2;wemaythuswriten=2andSincex=cos(Fb),ifx=0forsomenthenFb=100nnnnnotethatifn<10,thenk2πforsomeoddintegerk.Inparticular,wecanwriteb=c(2π)wherec=kandd=4Fareintegers.2m=n<10100<(103)34<(210)34=2340.dnLetxndenotethepair(Fn,Fn+1),whereeachentryinMoreover,thecasem=0doesnotleadtoasolutionthispairisviewedasanelementofZ/dZ.Sincetherenbecauseforn=1,n−1=0doesnotdivide2−1=1;areonlyfinitelymanypossibilitiesforxn,theremustwemaythusassume1≤m≤340.besomen2>n1suchthatxn1=xn2.NowxnuniquelyNext,notethatmodulon−1=2m−1,thepowersofdeterminesbothxn+1andxn−1,anditfollowsthatthe2cyclewithperiodm(theterms20,...,2m−1remainsequence{xn}isperiodic:for`=n2−n1,xn+`=xnthesameuponreduction,andthenthenexttermrepeatsforalln≥0.Inparticular,Fn+`≡Fn(modd)foralln.nFn+`cFnctheinitial1);consequently,n−1divides2−1ifandButthend−disaninteger,andsoonlyifmdividesn,whichhappensifandonlyifmisapowerof2.Writem=2`andnotethat2`<340<512,Fn+`cxn+`=cos(2π)so`<9.Thecase`=0doesnotleadtoasolutiondnbecauseforn=2,n−2=0doesnotdivide2−2=2;Fc=cosn(2π)=xwemaythusassume1≤`≤8.dnFinally,notethatn−2=2m−2divides2n−2ifandonlyif2m−1−1divides2n−1−1.Bythesamelogicforalln.Thusthesequence{xn}isperiodic,asdesired.asthepreviousparagraph,thishappensifandonlyifRemark.KarlMahlburgpointsoutthatonecanmoti-`mm−1dividesn−1,thatis,if2−1divides2−1.vatetheprevioussolutionbycomputingthetermsThisinturnhappensifandonlyif`dividesm=2`,2353whichhappensifandonlyif`isapowerof2.Thex2=2a−1,x3=4a−3a,x4=16a−20a+5avaluesallowedbythebound`<9are`=1,2,4,8;forthesevalues,m≤28=256andandrecognizingtheseastheChebyshevpolynomialsT2,T3,T5.(NotethatT3wasusedinthesolutionofprob-n=2m≤2256≤(23)86<1086<10100,lemA3.)Remark.Itisnotnecessarytohandlethecase|a|>1sothesolutionslisteddosatisfytheoriginalinequality.separately;thecosinefunctionextendstoasurjectiveB4Wefirstruleoutthecase|a|>1.Inthiscase,weproveanalyticfunctiononCandcontinuestosatisfytheaddi-that|xn+1|≥|xn|foralln,meaningthatwecannothavetionformula,soonecanwritea=cosbforsomeb∈Cxn=0.Weproceedbyinduction;theclaimistrueforandthenproceedasabove.n=0,1byhypothesis.Toprovetheclaimforn≥2,002B5Let(a1,a2)and(a1,a2)bedistinctpointsinR;write00wewanttoshowthatf(a1,a2)6=f(a1,a2).Write(v,v)=(a0,a0)−(a,a),andletγ(t)=(a,a)+|xn+1|=|2xnxn−1−xn−2|12121212t(v1,v2),t∈[0,1],bethepathbetween(a1,a2)and≥2|xn||xn−1|−|xn−2|00(a1,a2).Defineareal-valuedfunctiongbyg(t)=≥|xn|(2|xn−1|−1)≥|xn|,(v1,v2)·f(γ(t)).BytheChainRule,wherethelaststepfollowsfrom|xn−1|≥|xn−2|≥···≥!!0∂f1/∂x1∂f1/∂x2v1|x0|=1.f(γ(t))=.∂f2/∂x1∂f2/∂x2v2Wemaythusassumehereafterthat|a|≤1.Wecanthenwritea=cosbforsomeb∈[0,π].Let{Fn}betheAbbreviate∂fi/∂xjbyfij;thenFibonaccisequence,definedasusualbyF1=F2=1andFn+1=Fn+Fn−1.Weshowbyinductionthat!!0f11f12v1g(t)=v1v2xn=cos(Fnb)(n≥0).f21f22v2=fv2+(f+f)vv+fv2Indeed,thisistrueforn=0,1,2;giventhatitistruefor11112211222222n≤m,thenf12+f214f11f22−(f12+f21)2=f11v1+v2+v22f114f112xmxm−1=2cos(Fmb)cos(Fm−1b)≥0=cos((Fm−Fm−1)b)+cos((Fm+Fm−1)b)2=cos(Fm−2b)+cos(Fm+1b)sincef11andf11f22−(f12+f21)/4arepositivebyas-sumption.Sincetheonlywaythatequalitycould
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