Solutions to the 74th William Lowell Putnam Mathematical Competition

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1、Solutionstothe74thWilliamLowellPutnamMathematicalCompetitionSaturday,December7,2013KiranKedlayaandLennyNgA1Supposeotherwise.Theneachvertexvisavertexforw0j(i.e.,thelastdigitofw0jcomesrightbeforethefirstfivefaces,allofwhichhavedifferentlabels,andsothedigi

2、tofw0j+1).SincewjhaslengthZ(wj)+N(wj),thesumofthelabelsofthefivefacesincidenttovisatleastsumofthelengthsofw1;:::;wkisk(Z+N),andsothe0+1+2+3+4=10.Addingthissumoverallverticesconcatenationofw01;:::;w0isastringofk(Z+N)con-kvgives339=117,sinceeachface’sla

3、beliscountedsecutivedigitsaroundthecircle.(Thisstringmaywrapthreetimes.Sincethereare12vertices,weconcludearoundthecircle,inwhichcasesomeofthesedigitsthat1012117,contradiction.mayappearmorethanonceinthestring.)BreakthisRemark:Onecanalsoobtainthedesir

4、edresultbystringintokarcsw001;:::;w00eachoflengthZ+N,eachkshowingthatanycollectionoffivefacesmustcontainadjacenttothepreviousone.(Notethatifthenum-twofacesthatshareavertex;itthenfollowsthateachberofdigitsaroundthecircleism,thenZ+Nmlabelcanappearatmost

5、4times,andsothesumofallsinceZ(wj)+N(wj)mforallj,andthuseachoflabelsisatleast4(0+1+2+3+4)=40>39,contra-w00;:::;w00isindeedanarc.)1kdiction.Weclaimthatforsomej=1;:::;k,Z(w00j)=ZandN(w00j)=N(wherethesecondequationfollowsfromA2Supposetothecontrarythatf(n

6、)=f(m)withn

7、0j)Z+1forallj.perfectsquare.NoweliminateanyfactorinthisproductIneithercase,jkZåjZ(w0j)j=jkZåjZ(w00j)jthatappearstwice(i.e.,ifai=bjforsomei;j,thenk.Butsincew1=w01,wehavejkZåjZ(w0j)j=deleteaiandbjfromthisproduct).Theproductofwhatk0k0remainsmustalso

8、beaperfectsquare,butthisisnowajåj=1(Z(wj)Z(wj))j=jåj=2(Z(wj)Z(wj))jåkjZ(w)Z(w0)jk1,contradiction.productofdistinctintegers,thesmallestofwhichisnj=2jjandthelargestofwhichisstrictlysmallerthanar=bs.3Thiscontradictstheminimalityofa.A5LetA1;