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1、www.linstitute.netTheCENTREforEDUCATIONinMATHEMATICSandCOMPUTINGcemc.uwaterloo.ca2017EuclidContestThursday,April6,2017(inNorthAmericaandSouthAmerica)Friday,April7,2017(outsideofNorthAmericaandSouthAmerica)Solutions©2017UniversityofWaterloowww.linstitute.net1.(a)S
2、ince5(2)+3(3)=19,thenthepairofpositiveintegersthatsatisfies5a+3b=19is
(a,b)=(2,3).(b)Welistthefirstseveralpowersof2inincreasingorder:n12345678910112n24816326412825651210242048Eachpowerof2canbefoundbymultiplyingthepreviouspowerby2.
Fromthetable,thesmallestpowerof2gr
3、eaterthan5is23=8andthelargestpowerof
2lessthan2017is210=1024.Since2nincreasesasnincreases,therecanbenofurther
powersinthisrange.Therefore,thevaluesofnforwhich5<2n<2017aren=3,4,5,6,7,8,9,10.
Thereare8suchvaluesofn.(c)Eachofthe600EurosthatJimmyboughtcost$1.50.Thus,
4、buying600Euroscost600×$1.50=$900.WhenJimmyconverted600Eurosbackintodollars,theratewas$1.00=0.75Euro.
$1.00$600Therefore,Jimmyreceived600Euros×
0.75Euros0.75==$800.Thus,Jimmyhad$900−$800=$100lessthanhehadbeforethesetwotransactions.2.(a)Sincex=0andx=1,wecanmultiply
5、bothsidesofthegivenequationbyx(x−1)to5x(x−1)x(x−1)x(x−1)obtain+or5=(x−1)+x.=x(x−1)xx−1Thus,5=2x−1andso2x=6orx=3.Thismeansthatx=3istheonlysolution.
(Wecansubstitutex=3intotheoriginalequationtoverifythatthisisindeedasolution.)(b)Thesumoftheentriesinthesecondcolumni
6、s20+4+(−12)=12.Thismeansthatthesumoftheentriesineachrow,ineachcolumn,andoneachdiagonal
is12.Inthefirstrow,wehave0+20+a=12andsoa=−8.Onthe“toplefttobottomright”diagonal,wehave0+4+b=12andsob=8.Inthethirdcolumn,wehaveentriesa=−8andb=8whosesumis0.Thus,thethirdentrymust
7、be12.Finally,inthesecondrow,wehavec+4+12=12andsoc=−4.Insummary,a=−8,b=8,andc=−4.Wecancompletethemagicsquaretoobtain:020−8−441216−128www.linstitute.net(c)(i)If1002−n2=9559,thenn2=1002−9559=10000−9559=441.√Sincen>0,thenn=441=21.(ii)From(i),9559=1002−212.Factoringth
8、erightsideasadifferenceofsquares,weseethat9559=(100+21)(100−21)=121·79Therefore,(a,b)=(121,79)satisfiestheconditions.(Inaddition,thepairs(a,b)=(79,121),(869,11),