x3
(A)?2
(B)?
12
x2
(C)
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Wehave0 64 16 64 4 64 hence(C). 2. 2014S10 ppIf=3thenequalsp?2qq(A)?3 (B)3 (C) 1 3 (D) p =3,q23 (E)2 Wehavep=3(p?2q),so6q=2pandp=3q.Then hence(B). 3. 2014S15 Inthediagram,PS=5,PQ=3,??PQSisright-angledatQ,∠QSR=30?andQR=RS.ThelengthofRSis √ √3 (A)(B)3(C)2 2 √43(D)(E)4 3 R ||||Q3P 5 30?S Copyright ? 2014, 2019 Australian Mathematics Trust AMTT Limited ACN 083 950 341 (Set1) — SeniorDuetotheright-angledtriangle??PQS,Pythagoras’theoremgivesQS=4.Then??QRSisisosceles,soitsaltitudeRTbisectsQS. Q3P 2T5 R ||60?||xS √3so 230?Now,??SRTisstandard30?,60?√,90?trianglewithRT:RS:ST=1:2: 2443thatx=RS=√ST=√=, 333 hence(D). 42 √=. cos30?3Comment Thisproblemcanalsobesolvedusingtrigonometry:x=4. 2014S20 Giventhatf1(x)=(A) x2014x+1 x andfn+1(x)=f1(fn(x)),thenf2014(x)equalsx+12014x2014x+1 (C) xx+2014 (D) 2014xx+1 (E) x 2014(x+1) (B) Alternative1 ????x =f2(x)=f x+1 xx = 1x+x+12x+1 x xx2x+1==f3(x)=x+1x+2x+13x+12x+1andingeneral,byinduction x xxxnx+1fn(x)==?fn+1(x)=x==, nx+1+1x+nx+1(n+1)x+1nx+1x ,sof2014(x)= 2014x+1 = xx+1x+x+1 hence(A). Alternative2Consider 1 .fn(x)1111 =1+=?=f1(fn(x))=1+f1(x)xfn+1(x)fn(x) 111 =1+=2+=···=? f2014(x)f2013(x)f2012(x) 12014x+11 =2014+=···=2013+ f1(x)xx Hencef2014(x)= x , 2014x+1 hence(A). AMC Practice Questions and Solutions — Senior5. 2014S25Thesequence 2,2,2 2 22 ,2 22 2 ,... isde?nedbya1=2andan+1=2anforalln≥1.Whatisthe?rstterminthesequencegreaterthan10001000?(A)a4=22 22 (B)a5=22 222 (C)a6=22 2222 (D)a7=22 222 22 (E)a8=22 222222 Wewantan>10001000=103000.Weknowthata1=2,a2=22=4,a3=24=16 anda4=216=65536,alllessthan103000.Also210=1024>103,sothatwecanestimatea5, a5=265536=(210)655326>(103)655326=64×1019659 Thisisgreaterthan103000, hence(B). 6. 2014S26 Whatisthelargestthree-digitnumberwiththepropertythatthenumberisequaltothesumofitshundredsdigit,thesquareofitstensdigitandthecubeofitsunitsdigit? Alternative1 Letthenumberbeabc.Then 100a+10b+c=a+b2+c399a+10b?b2=c(c2?1) 99a+b(10?b)=(c?1)c(c+1) Considerthepossibilities: 99a99×1=9999×2=19899×3=29799×4=39699×5=49599×6=59499×7=69399×8=79299×9=891b(10?b)1×9=92×8=163×7=214×6=245×5=256×4=247×3=218×2=169×1=9(c?1)c(c+1)1×2×3=62×3×4=243×4×5=604×5×6=1205×6×7=2106×7×8=3367×8×9=5048×9×10=720Lookingatthepossibilitiesfor99a+b(10?b)=(c?1)c(c+1),wehavetwo:99+21=120=?a=1,b=3or7,c=5=?n=135orn=175.495+9=504=?a=5,b=1or9,c=8=?n=518orn=598. So,therearefour3-digitnumberswhichsatisfytherequirementsandthelargestofthesefournumbersis598, hence(598). AMC Practice Questions and Solutions — SeniorAlternative2 Thenumberabcisequaltoa+b2+c3,andthesearethepossiblevaluesofb2andc3: DigitSquareCube 000111248349162764 525125 636216 749343 864512 981729 Wetrythesenumbersinanadditiongrid,tryingthelargevaluesofc?rst,then?llinginpossiblevaluesforaandb.Thistrial-and-errorsearchispresentedhereasatree. 87298b9a729ab9a:b2:c3:a+abca512ab877297b965126b85+1+2=87+3+9=196+0+2=88+2+9=1982(nob2)7298b983(nob2)7298b9600512608??85159??051515128512855125b8a343ab7abc<598Likewiseforc=6,5,...Thelargestsolutionfoundis598,andanysolutionsonbranchesc=7,c=6,...,c=1mustbelessthanthis,hence(598).
澳大利亚2014高中奥数竞赛试题及答案
