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1.Answer: 121 o3wc1u3c01Let the integers be w ,v2 w ,v4 4l4l 1lw =v100. Then 2w =v aerv ,v2w ,v 8erv ,v1l4l4l ,v2w ,v100)2 -v300850 Let cv-vwv ,v51 and regrouping the terms, we obtain 72c 2 8derv ,v2cv,v 8der Fv,v l2c 247)2 ,v2cv,v47)2 ] ,vxxxv,v 72c 2nerv ,v2cv,v1)2 ] -v300850 which simplifies to 6icrv ,va2nrv ,v5rv =6r ,v9rv ,vxxxv,492 ) -v300850 . 4 1 . Usmg the fact that 12 ,v32 +52 +· ·· + (2n-1)rv -v 3i. - 3n, we obtam cv-v72. Hence w-v21, so that the required number is 121. 2) 2.Answer: 7986 o3wc1u3c01Note that 2k+1 - 2k -v2k , and that 2k -;.l>.Fel2k+av if and only if llog 2 >.F1n-k. Then the requires sum (denoted by D:-;can be obtained by cl-v 3. Answer: 4.)a - r. .H .H.(=a gF>. 0F,vllog 2 3j- BHllog 2 >. 0 k= 2 r.dE r..mv o6JaaJ) 2(.)0(a - r. F1k2k ,v1 - orJ)9 8192 ,v1 -23(9) -v7986 l) o3wc1u3c01Let g(x) -vxf(x)-2, hence g(x) is a polynomial of degree 2013. Since g(l) - g(2) -vg(3) -v 4l4l4l-vg(2013) -v 0, we must have g(x) -v >.(x-l)(x -2)(x- 3) 4l4l 4l(x-2012)( x-2013) 2 for some >.. Also, g(O) --2 -v- >. 4l2013! , we thus have >. -v 20131 Hence, g(2014) -v 20a3! (2013!) -v2014. f(2014) - 2 concluding that 2014 4lf(2014) -v4 --l2)

AdFAnswer: 3333 )202 d1202(201) . . . h37c1u3c01First note that there are 2 = 2 = 20301 positive mteger sets (1(v c(vz) which satisfy the given equation. These solution sets include those where two of the three values are equal. If 1v= 3xlthen 2x +z = 203. By enumerating, z = 1, 3, 5, m. m. me-;201. There are thus 101 solutions of the form 31(v1(vz). Similarly, there are 101 solutions of the form 21( c(1evand 21( c(ce4vSince 1vel cvelz, the required answer is 2) 5.Answer: 62 h37c1u3c01Since .o n1 = (n-1)(n2 + n + 1), we know that n2 + n + 1 divides .o n1. Also, since .E 5 8 o n1 = t.o c7l 8n n1, we also know that n2 + n + 1 divides .E 5 8 o n1. As .E 5 8 on + 61 = .E 5 8 on l. 1 + 62, we must have that n2 + n + 1 divides .E 5 9 on + 61 if and only if n2 + n + 1 divides 62. Case (i): If n2 + n + 1 = 1, then n = 0, -1. Case (ii): If n2 + n + 1 = 2, there is no integer solution for n. Case (iii): If n2 + n + 1 = 31, then n = 6, -5. Case (iv): If n2 + n + 1 = 62, there is no integer solution for n. Thus, all the integer values of n are 0, -1, 6, -5. Hence the sum of squares is 1 + 36 + 25 = 62. 6.Answer: 3015 h3wc1u3c01The recurrence relation can be written as DgDn n f0 Dn= )1n AL 1 n !-10 )1!1n n61 1d1 m. Summing for n = 2 toN, we obtain showing that fc 2DnF.3 )11 1 -1 25D-gN- N +1. Summing this up for N = 2 toN= 2011, we obtain E 5 88n S ='"""' f Jbn = � (2010) -3-: b-1-E1= 3014.5 -1-0Hf Jn2 2 2012 + 2012 showing that the integer closest to S is 3015. -dl2) l)

7.Answer: 38 h3w01u3ce1Let n be an even positive integer. Then each of the following expresses n as the sum of two odd integers: n = (n-15) + 15 (n-25) + 25 or (n-35) + 35. Note that at least one of n -15 n -25 n -35 is divisible by 3 so that it is composite, and hence n can be expressed as the sum of two composite odd numbers if n 1)38. Indeed, it can be verified that 38 cannot be expressed as the sum of two composite odd positive integers. yl -]lAnswer: 76 h3w01u3ce1Let the lengths of the sides of the triangle in centimetres be 16, 16r and 16r2 1 -2r + 3r2 19 3 (where r 1)1). Then 2 - so that r = - 2. Hence, the perimeter of the r -2r + oF9 triangle= 16 (1 + ;L + L d1= 76cmlI 9.Answer: 4 h3w01u3ce1Since 2002 nB4(mod9) 43 nB1(mod9) and 2002 = 667 ")3 + 1, it follows that 20022002 nB4667x3+1 nB4(mod9). Observe that for positive integers x, the possible residues modulo 9 for x3 are 0 ±1 . Therefore, none of the following X --;X --;+ G� G--;+ X � + G� can have a residue of 4 modulo 9. However, since 2002 = 103 + 103 + 13 + 13 , it follows that 20022002 2002 . (2002667 )3 (1 03 + 103 + 13 + 13 ) (200 2667 )3 (10 . 2002667 )3 + (10 . 2002667 )3 + (2002667 )3 + (2002667 )3This shows that x� + x � + x � + x! = 20022002 is indeed solvable. Hence the least integral value of n is 4. yl 10.Answer: 2015 Let f( x) = nx3 + 2x -n. It is easy to see that f is a strictly increasing function for n = 2 3 4 1. m.m-.Further, 1 ( n. 191= n ( n.1 93 +2 ( n. 191-n = (n1 1) 3 ( -n2 +n +1 )

dda d+en1=FFii h3w01u3ce1-121+3Fb)F2=6Fe--tF2t02Fb)4F)6 .F , = =.0) +)0)+=. 0)+) 0)+=. 0)+) 0)+=. 0)+) 0) 0) 0) 0) 0) 0) l) 0) o) 0) 0) o) 0) 0) 0) 0) o) 0) Oacmox!') c 72F,eF-(10.F2t02Fcb)vF0+3Fc6)vF0.1F-=+a.-1+2lFI1+-1F6vF4FbvF4F60+3F4v6)FB. 4vb) .F1t-ebF4v6)F4F4vb)F4F4vC)F,sh(,1eF2t02F 2t1Fh=,+2eFvbF)bFCF0.1F6F 0.1F -=+-8-(,- .F9,r1+F2t02F'v)bF4L; +bFt1+-1F46)vF4L;+lFv)C6F,eF0F-8-(,-Fc-03.,(021.0(F n,2tF6vF0eF0F3,--fl:s121. .F1t-ebF46CvF4L;+ .F7+F2t1F=,at2X0+a(13F2.,0+a(1F6CvbFn1Ft0r1F 3,

13. d+en1cTF86 h37c1u3ce1J12Fa, b, c, d n1F2t1Fx=-cF.==2eF=xFx4 -18x3 + kx2 + 200x -1984 = 0 e-=tF2t02 ab = -32. kt1+F 12a + b + c + d = 18, ab + ac + ad + be+ bd + cd = k, abc+ abd + acd +bed= -200, abed = -1984. 3,+=1Fab = -32 0+3Fabed = -1984, n1Ft0r1Fed = 62. kt1+bFxc=sabc + abd + acd +bed= -200 n1Ft0r1F -200 = -32c-32d + 62a nl62b = -32(e +d) + 62(a +b). 3= (r,+aF2t,eF1c202,=+F2=a12t1.Fn,2tF2t1F1c202,=+Fa + b + c + d = 18 a,r1eF2t02 a + b = 4, c +d = 14. o=sFab + ac + ad+ be + bd + ed = k, n1Ft0A1 = 30 +(a+ b)(c +d)= 86. 14. d+en1cTF337 h37c1u3ce1dee-s1F2t02F 0+3Fe=F (n nl1)(2n + 1) =6m2 . kt2eF6 (n + 1)(2n + 1), ,sh(8,+aF2t02Fn nB1 =.Fu=s=3F6). =dgk1m-l il= 6k + u.Fl) kt1+Fm2 = (k + 1)(12k + 11). 3,+=1F(k + 1) 0+3F(12k + 11) 0.1Fc1(02,r1(8Fh.,s1bFn=2tFs-e2 n1Fec20c1e .F3=F2t1c1F0.1Fh=e,2,r1F,+21a1ceFa 0+3Fb e--tF2t02Fk + 1 = a2 0+3F12k + 11 = b2 . kt2eF12a2 = b2 .l1. +-2bF0eF12a2 nB!=s=3F4) 0+3Fb2 + 1 nB1 =cFi=s=3F4), 2t1c1F0c::;F+= ,+21a1ceFa 0+3Fb e-=tF2t02F12a2 = b2 + 1. I1+=1Fy0e1F1 -0++=2Ft0hh1+ .F =dgk1o-ln = 6k + 1. kt1+Fm2 = (3k + 1)(4k + 1). 3,+-1F3k + 1 0+3F4k Xl1 0c1Fc1(02,r1(8Fhc,s1bFn=2tFs2e2Fn1 ec20c1e .F3=F2t1.1F0c1Fh=e,2,r1F,+21a1ceFa 0+3Fb e2=tF2t02F3k + 1 = a2 0+3F4k + 1 = b2 . kt1+F3b2 = (2a -1)(2a + 1). 7ne1cr1F2t02F,+F2t1F(1x2Xt0+3Fe,31bF1r1c8Fhc,s1Fx0=2=cF11=1h2 3t0eF0+F1r1+Fh=n1c .F3=F+1,2t1cF2a- 1 +=.F2a + 1 -0+Fn1F0Fhc,s1F=2t1cF2t0+F3. 6=nFn1F-=+e,31.Fh=e,2,r1F,+21a1ceFa e--tF 2t02F+1,2t1cF2a - 1 +=cF2a + 1 =0+Fn1F0Fh.,s1 =2t1cF2t0+F3. 7xFa= 1, 2t1+Fb = 1 0+3Fn = 1. 3=Fn1F ==+e,31cFa ((l2. kt1F+112F es0((-e2 e2,20n(1Fr0(21Fx=.Fa ,eF13. Lt1+Fa = 13, n1Ft0r1F 3b2 = iuFF)27 0+3Fsob= 9ubF,sh(8,+aF2t02Fk =56 0+3Fe=1= 6k + 1 = 337. l) -.l

9u .d+en1.FFia99 o=sF2t1F.1-2..1+-1F.1(02E=+bFf(f(1)f(b)) 4Fxo9B)F0+3FxoxoDBxo9BBF4Ff(b) m.9dI1+-1bf(b) 4 xo9BD .F+8F(122E+aFb 4FGo9BbFn1F=n20E+Fxoxo9BBF4Foxo9BBg .Fd(e=bFx.=sF2t1F aEr1+Fx2+-2E=+0(F 1c202E=+bFn1Ft0r1Fxoxo9BBF4F9bF t1+--FoLo9BBgF qF 9bFx=((=nE+aF2t02FGo9BFEeF1E2t1.F9F=.Fz9dF I1+-1Fxoia99BF4Fia99dFlI 9:dd+en1.FF9y h3w01u3c016=21F2t02Fx-9FelLxJ 6lx . 6=21F2t02FExFx zCl3, 2t1.1FnE((Fn1F+=Fe=(22E=+F0e x3 -4LxJ enlx3 -4x 4F x(x2 -4) zCl3(5) 4F9ud d(e=bFExFx 6lzibF2t1.1FnE((Fn1F+=Fe=(22E=+F0eFx3 - 4 L x J elx3 -4(x z9BF 4Fx(x2 -4) + 4 64 . I1+-1F 2t1Fe=(22E=+Fs2e2Fn1FE+F2t1FE+21.r0(FoFzibF3) . If L x J 4FzibF2t1+Fx3 4F-3, aErE+aFx 4F:--;ntE-tFEeF0Fe=(22E=+d If LxJ 4Fz9bF2t1+Fx3 4F9bF aErE+aFx 4F 9FntE-)F-=+2.03E-2eFnE2tFLxJ 4Fz9 . =LlL x J 4FabF2t1+Fx3 4F5, t1+-1F 2t1.1FEeF+=Fe=(22E=+ .F If L x J 4F9bF2t1+Fx3 4F y .F3E+-1FiFelL'-;el3, 2t1.1FEeF+=Fe=(22E=+d If LxJ 4FibF2t1+Fx3 4F9odF3E+-1FiFel�-;el3, X4F�-;EeF0Fe=(22E=+ . 1t2ebF2t1F.1c2E.13F0+en1.FEeF-3 + 9oF 4F9a .F 9.dd+en1.F584 h3w01u3c0172FEeF-(10.F2t02F T FiL+ gCiiL+ TgiiL+ ... + g 9± liLlI EeF0Fs=+=2=+1FE+-.10eE+aFx2+-2E=+F=xFx, 0+3Fnt1+Fx 4F:lbF2t1F0n=r1F11h.1eeE=+Ft0eF 0Fr0(21 (0.a1.F2t0+F9aa9dF1t2eF10-tFe=(22E=+F=xF2t1F1c202E=+FT Si iL,v T CiiL+ Tga iL+ . .. + T 9± liL4F 9aa9FEeF(1eeF2t0+F:lF0+3Fe=F ExFxEeF0Fe=(22E=+bF2t1+F T SiiL+ gCiiL+ TgiiL+ ... + T 9± liL4F T SiiL+ T Ca iL+ ggi pL+ TliL+ T1iiLBL deFxel:lbFE2Ft0eF0F 2+Ec21F11h.1eeE=+F=xF2t1F x=.sF XFqF CFF)ulF + b F)AlF + k) F)olF + d F)ilF + :: () nt1.1F±3b, c, d, ::) 0.1F+=+l+1a02Er1FE+21a1.FnE2tF CF6l5, b 6l4, c 6l3, d hlibF::) 6l9dF6=21F2t02 ExF XF4F CFF)ulF + b F)AlF + k) F)olF + d F)ilF + :: () 2t1+Fg Si .L+ T CiiL+ Tg i iL+ Tc iiL+ To iL4F i a: CF + 41 b + 9a 'F +3 d + ::B)3E+-1F41b + 9a'F + 3d + ::) hlia9]Fn1Ft0r1FvaaF6lia:CF6l9aa9F0+3Fe=FCF 4F4 . 1t2e 41b+ lOc+ 3d + ::) 4F9.. b -ml

nt,-tF,sh(,1.F2t02Fb4F AF0+3F.=F=+bFa,r,+aF2t02F-)4Fd4F pF0+3F::) 4F alF 1t-.F .l4F AFF)ulF5FAFF)AlF 5F pFF)olF JF pFF)ilF5FaF 4FumAlF d.FumAF,.F2t1F=+(8F,+21a1.F.=(-2,=+bF2t1F0+.n1.F,.FumAlF FI pmld+.n1.OFiu h3wc1u3ce1!:1F.t0((F.t=nF2t02F0 i o±F.,+oCF5F.,+ovF5F.,+obo±FoDJil �..-s1F2t02FCF o±F vF o±FblF1t1+FCF o±F:allF1t-.F.,+FoCFenlalF72F,.F-(10.F2t02F.,+FovFenl0 pF0+3 .,+obFzCl09lF1t-.F.,+oCF5F.,+ovF5F.,+obF47l0ilFJ12FvF4blF1t1+FvF 4F bF4L; 4zCJil 7xFCF,.Fr1.8F.s0((bF vF0+3FbF0.1F-(=.1F2=FL; 4bF0+3F2t-.F.,+FoCFOl.,+FovF5F.,+FobF,.F-(=.1F2= 0il 6=nFn1F.t=nF2t02F.,+FoCF5.,+FovF5.,+FobFo±FoDJilFi,..2F2t1F-hh1.Fn=-+3F-0+Fn1F.10-t13 nt1+FC4FvF 4FialF0+3FbF 4FpAall J12Fe4FoCbFsl4F ovF0+3FDF 4FoobF0pialBlF1t1+FeJFsl5FDF 4FpmalF0+3 .,+oCF5F.,+ovF5F.,+obF4.,+Fe5F.,+Fsl5F.,+DlF 3-hh=.1F2t02FebFsllDF.02,.-F2t1F-=+3,2,=+F2t02Fe5Fsl5DF4FpmalF.--tF2t02F.,+FeF5.,+Fsl 5F .,+DFt0.F2t1Fs01,s-sFr0(-1lF !:1F-0+F2t1+F.t=nF2t02Fe4Fsl4FDlF d..-s1F2t02FF --)l s ---:lDlF7xFeoFDbF2t1+F ,shL8,+aF2t02FlF lF lFe5De0 DF lFe5D .FFeF5F.FFDF 4F iF.FF iF-=.F iFoF iF.FF iFbF lF eFLl -llF illFe5D Ll -llFe5D !FJ !F5 !Fo !F iF5!F5!F iF nt,-tF-=+2.03,-2.F2t1F0..-sh2,=+F2t02F .,+F(F5.,+Fsl5.,+FDFt0.F2t1Fs01,s-sFr0(-1lFI1+-1F e4Fsl4F DF 4F:albF,sh(8,+aF2t02FC4FialbF vF 4FialF0+3FbF 4FpAalF0+3F .,+FoCF5.,+FovF5.,+FobF4FoA.JilF 3,+-1F--;oBpl.oibF2t1F0+.n1.F,.F2t1+F =n20,+13llI pLlFd+.n1.OF pF h3wc1u3ce16=21F2t02F=l4FabF3l·F aF0+3F8l4F aF,.F0F.=(-2,=+F=xF2t,.F1c-02,=+lF!:1F .t0((F .t=nF 2t02F2t,.F,.F,2.F=+(8F,+21a1.F.=(-2,=+Fn8Fh.=r,+aF2t02F,xF=xl3xl8l,.F0F.=(-2,=+F=xF2t,.F1:-02,=+F 0+3Fnt1+1r1.F=xl3xl8l0.1F3,r,.,n(1Fn8Fi1bF2t18F0.1F0(.=F3,r,.,n(1Fn8Fi1L_-F x=.F0+8FnF47lal J12F=l4FrT =9 xl 3l4Fi1fNF0+3F8l4Fi1F89 4l1t1+F=ol 5F3ol 5F8ol 4F=o 3ol ,.F-t0+a13F2=F 72F,.F10.8F2=Fr1.,-F2t02F=+(8Fnt1+F=9 xl 39 xl 89l 0.1F0((F1r1+bF=9ol 5F39ol 5F8Aol 0+3Fro T =Ao 39ol t0r1F 2t1F.0s1F.1s0,+31.Fnt1+F3,r,313Fn8FAlF1t-.F=xl3xl8l0.1F3,r,.,n(1Fn8FiTL_Fs (FI dvl

20. u-

i o dd+en1gFF a h3wc1u3c01J12 1 pn4F20+Fv pan 3,+-1FiF n;-;4F20+F0-; -7-;bF,2Fx=((=neF2t02 20 +F = p)n 20 + 0 EYF 2L l 3L 1 p2mn4F20+ :,. .. L_xF4 0-;a f-;420+Fv pn)n wiF w0 20+F= pn20+F mEn 3=bF 1 p2DEn 420+Fe = p)nlw-;420+F= pB. 1 psn,sh(8,+aF2t02F2t,eFe1c-1+-1Ft0eF0Fh1g,=3F=-Fwid 7ne1gr1F2t02FwaawFnBuos=3F9iBF0+3FAa9FnBuos=3F9iBdFy=+e1c=1+2(8bF 1 D m 0mnn1u . 8n4F 1 fn n1 fn4F a lI iAdd+en1gFFiawo h3wc1u3c01m,ge2F,5F,eF-(10gF2t02F2t1F0+en1gF,eF0+F,+21a1gFn12n11+FaF0+3FiawAdF+-2F,2F-0++=2 n1FiawAOF0em. m. m. 1 mn1En n1 onn1unm. m. m (1E .gun0+3 1 m)n 1En )m. m. m.)n 1E .gun4F wF dliF)m. m. m.)niawAF4F9aa.FF)iawuF t0r1F2t1Fe0s1Fh0g,28 .F 6=nFn1F---e2F+113F 2=Fet=nF2t02FiawoF-0+Fn1F0-t,1r13 .Fm=gF0+8F,+21a1gFnOF 1t-eF iudd+en1gFFwoFAnF)niBF 0 oFAnF)nABF 0 oFAnF)nuBF 0 oFAnF)noBF 4F a iF 0 AF nuF noF ndddF noFAnF)niBF 0 oFAnF)nABF0 oFAnF)nuBF noFAnF)noB 0 iawaF0iawiF0iawoF 0iawwF 0ia9AF0 9F B.a/3iawAF/3 wFB.iawoF lI o3wc1u3c01y=+e,31gF2t1F-=sh(11F+-sn1gFq)4F -=eFa-;)n lne,+Fa±-;7+F=+1Ft0+3bF-e,+aF2t1 n,+=s,0(F2t1=g1sF=+1Ft0eF 2Lb)b3LE 5gE L 1F -=eF50)nlne,+F.0 1(e+(E .mE L 1F±1 )lrn .± 81E 5mE1.l; i9i..± 81E .m 0n ,c E.n )nia AwiF=o 51E .. Fn t/i 61)naaan )t/Lp.) 61c9Fg wF0, iawi.1En, iawi .1 dFananBm 55 7n,iawi.I -E .mEnBn idlBn AF)n )n ia 9iF d)l

7+F2t1Fr2t1.Ft0+3bF-e,+aF 2t1Ft1F-r,r.1-eF2t1r.13Fr+1Ft0eF 1t-e bFL 1oB g1 - g BF 4FL 1F0 -re:=C : l1 dl-: e,+F:=C: l o14-reF:=C : l1 4-reF: l1 4Fl -: oFoFoF oF :1LCX1G Cc oF e:=C : o1ogF e:=f: o1mmm1 0o-11 lF e:=C : o1o-11IFe:=C : oy1 4Fk-:ig1 - gF :1dl 4 L_l :=C=1dl :=C:1 :1erF2t02F --;T f11 oF e:=C: o1ogF e:=C : o1mmm1 0 o-11 lF e:=C : o1o-11IF e:=C : o11 4 ig1 --F :1dl 4 dl :=C=1dl :=C:1C1 c(llI