jmc

First, for simplicity, let's make all the amounts of money into integers by considering them all in cents. For example, \5.43$becomes543.Letthepurchasepricebe$A=A_1A_2A_3$andtheamountofchangebe$B_1B_2B_3$where$A_1$representsthefirstdigitof$A$,$B_1$representsthefirstdigitof$B$,$A_2$representstheseconddigitof$A$,etc.<br/><br/>Weknowthat$A+B=1000$,andwecanconcludethat$A_1+B_1=9$becauseif$A_1+B_1<9$then$A+B<1000$andif$A_1+B_1=10$then$A_2=B_2=A_3=B_3=0$,butthentheonlywaythatBcanbearearrangementofthedigitsofAisif$A_1=B_1=5$,whichmeans$A=B=500$,buttheproblemstatesthatthepriceandtheamountofchangearedifferent.<br/><br/>Since9isodd,wecanalsoconcludethat$A_1$and$B_1$aredistinct,which,usingthefactthat$A${}^{\prime }sdigitscanberearrangedtoget{B}^{\prime }sdigits,impliesthat$A_1=B_2$or$A_1=B_3$and$B_1=A_2$or$B_1=A_3$.WecanalsoobservethatAandBhavethesameremainderwhendividedby9becausetheremainderwhen$n$isdividedby9isequaltotheremainderwhenthesumofthedigitsof$n$isdividedby9forall$n$andthesumofthedigitsofAisobviouslyequaltothesumofthedigitsofB.<br/><br/>Sincetheremainderwhen1000isdividedby9is1,wecaninfactconcludethattheremainderwhenAandBaredividedby9(andwhenthesumoftheirdigitsisdividedby9)is5.Keepinginmindthattwoofthedigitsof$A$are$A_1$and$B_1$andthat$A_1+B_1=9$,wecanconcludethattheotherdigitis5,whichistheonlydigitthatwouldresultinthesumhavingaremainderof5whendividedby9.Bysimilarlogicwecanconcludethat5isalsooneofthedigitsof$B$.Alittlethoughtmakesitclearthatatleastoneofthese{5}^{\prime }sappearsasthelastdigitinitsnumber(thatis,$A_3=5$or$B_3=5$)sinceifneitherofthemappearsasthelastdigitinanumber,then$A_1=B_3$and$B_1=A_3$and$A_3+B_3=9\Rightarrow A+B$endsina9,whichisacontradiction.Butif$A_3=5$thentheonlywayforthesumof$A$and$B$toendina0isfor$B_3=5$,sowecanconcludethat$A_3=B_3=5$,$A_1=B_2$,and$A_2=B_1$.Sooncewehavepickedavaluefor$A_1$,theother5digitsarealldetermined.Sincebothamountsaregreaterthanadollar,weknowthat$A_1$canbeanynumberbetween1and8foratotalof8possibleprices(andthus8possibleamountsofchange).Todoublecheck,wecanworkout$A$and$B$foreachvalueof$A_1$andreconvertthemtodollarstomakesurethatthepriceandtheamountofchangesatisfythegivenconditions:<br/><br/>$A_1=1\Rightarrow A=\1.85, B=\8.15$;<br/><br/>$A_1=2\Rightarrow A=\2.75, B=\7.25$;<br/><br/>$A_1=3\Rightarrow A=\3.65, B=\6.35$;<br/><br/>$A_1=4\Rightarrow A=\4.55, B=\5.45$;<br/><br/>$A_1=5\Rightarrow A=\5.45, B=\4.55$;<br/><br/>$A_1=6\Rightarrow A=\6.35, B=\3.65$;<br/><br/>$A_1=7\Rightarrow A=\7.25, B=\2.75$;andfinally<br/><br/>$A_1=8\Rightarrow A=\8.15, B=\1.85$.<br/><br/>Thisconfirmsthatthereare$\boxed{8}$ possible amounts of change.