USA High School Math Contest

HighSchoolMathContest

UniversityofSouthCarolina

February5,2011

Solutions

1.(b)Theleft-handsideisanevennumberunlessmorniszero.Ifm=0,wehave2n≥1and2m 2n≤0.Ifn=0,then2m=64,i.e.m=6.Thus(6,0)istheonlysuchpair.

2.(d)Iftherelationab=1theneithera=±1,orb=0(anda=0).Theequation2x

5=1

2yieldssolutionsx=±2.If2x

5= 1,thenx=±1,butinthiscasex2 2xisodd,so 2 x2 2x2x 5= 1.Fromthelastoptionweobtainx2 2x=0,i.e.x=0orx=2.Thus3

wehavethreesolutions:x=0,±2.

3.(d)Wecanwritex51+51=Q(x)·(x+1)+R,whereQ(x)isthequotientandRistheremainder.Whenx= 1,thisgivesR=( 1)51+51=50.

4.(d)Letxbethetotalnumberofpeople.Thenthetotalnumberofhandshakesis5x,wherewe

havetodivideby2sinceeveryhandshakeiscountedtwice.Thus5x=60,whichleadsto

x=24.

5.(b)Everypowerofsixendsin6.Alsonotethat324=816.Sincethelastdigitof81is1,anypowerofitwillalsoendin1.(Alternatively,forthepowersof3wehave31=3,32=9,33=27,34=81,35=243,...–theunitsdigitsrepeatincyclesoflength4.Since24isdivisibleby4,the naldigitof324isthesameasthe naldigitof34=81,i.e.1.)Thustheunitsdigitof625 324is6 1=5.

6.(c)Obviously,x>y.Ifristheradiusofthecircle,thenx=2πrandy=4r.Hencexπ.2=2<3≈1.14<3.y2.827.(c)Theproducthas2011factors.Inorderforittobenegative,weneedtohaveanoddnumberofnegativefactors(noticethatalltheexponentsareodd)and,hence,anevennumberofpositivefactors.Butthenumberofpositivefactorsissimplyequaltox.Sincethereare1005evenpositiveintegerslessthan2011,theansweris1005(forintegersx≥2011theproductisobviouslypositive).√2