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44 2016=403, h e6340n c4(4,403 E)2016 22 = 4032, hence (E). .) The angles in the t\biangle add to 180 2, so2= 180 020 020 = 140, hence (D). T) 30 days is 4 weeks and 2 days. So 30 days f\bom today is the same as 2 days, which is a Satu\bday, hence (E). a) 05+8 = 3, hence (E). g) 25% is 2 0, so 25% of 2 1 is 2 02 2 1= 2 6, hence (A). l) (Also MP12) Each time the pape\b is unfolded, the two pa\bts will be \beflections of each othe\b th\bough the fold line. hence (A). s) 1000(29 + 16 + 8 095) = 100 053095 = 460 05, hence (C). i) In o\bde\b, 0 0009 100019 10008 100109 1004 1 00409 10091, hence (C). t) 12 noon is 10 minutes afte\b sta\bting, 1 pm is 70 minutes, so 1:04 pm is 74 minutes, hence (D). E\b) (Also I4) Estimating, =1423, 1423 1 =14444 1444 = =14 1 = 360. This suggests that 100 1 =1423, 1423 11000. Checking, 201600 1720163 12016000 and so 100 1 =1423, 1423 11000, hence (D). 22 2016 =43 , henc(E )(.eTc(na 2016 AMC – Junior Solutions

4545 220201 6=164 3, h01 h0=11 4en6=14c ,=3( 4(6EE14h h3 E6=)14hc 6=1 .c Tac 6gl si 4en6=1 ngth4\b 201 406l1l =1)t3g 064 6=16 Ta 2. d oc 43 h01 83=ht3g 3, h01 E6=)14h 4en6=1 h06h t4 406l1l t4 TDD 0 2 01 d sy 81=w1ghc 01gw1 kSf\b 210 2016=431,h6 e m, ut6g6 064 2(6=5E14c +6g 064 % 2(6=5E14 6gl %2 2%d2 A %\b M3EPtg) h0t4c s 2 d a 6gl 2d %\b S1hp11g h01(c h01b 06P1 fl2 d Ts (6=5E14c 01gw1 k9f\b 2016=431,h6 n +6g 064 6 =3, h01 (6=5E14 6gl ut6g6 064 4 =\b C,h1= h=6g4,1==tg) h01b 16w0 06P1 4 03, h01 (6=5E14\b M3 h01 % (6=5E14 h=6g4,1==1l (6I1 n8 6 =2 4 0d 4 = 3, h01 (6=5E14\b 201=1,3=1 h01b 06P1 Ts (6=5E14 51hp11g h01(c 01gw1 k9f\b 260 mg 1P1=b hp3 =3p4c h01=1 6=1 o 86P1=4c 3g1 3, p0tw0 ptEE 51 wnh tg 06E,\b M3 h01 gn(51= 3, wnh 86P1=4 t4 ToiD 1o d siDc 01gw1 kCf\b 2=0 7g :3gl6bc m 8E6gh1l TD 688E1 h=114\b 7g 2n14l6bc h01 4(6EE14h gn(51= 3, 3=6g)1 h=114 m w3nEl 06P1 8E6gh1l t4 .c 3g1 51hp11g 16w0 86t= 3, g1t)053n=tg) 688E1 h=114\b 7g ‹1lg14l6bc h01 4(6EE14h gn(51= 3, 816w0 h=114 m w3nEl 06P1 8E6gh1l t4 Tyc 3g1 51hp11g 16w0 86t= 3, g1t)053n=tg) 688E1 6gl 3=6g)1 h=114\b M3 h01 4(6EE14h gn(51= 3, h=114 h06h m 8E6gh1l 6Eh3)1h01= t4 TD A . A Ty d %oc 01gw1 kŒf\b 240 kCE43 Ž‘s%c mTif C4 h01 gn(51= 3, 83tgh4 81= 1P1gh t4 a 6gl h01 h3h6E gn(51= 3, 83tgh4 )6tg1l t4 y A TT A i d sflc h01=1 (n4h 06P1 511g fl 1P1gh4\b C4 S1hhb 064 TT 83tgh4 401 (n4h 06P1 % ’=4h 8E6w14 6gl 3g1 41w3gl 8E6w1\b Œ6h0b w3nEl g3h 06P1 p3g 6g 1P1ghc 3= 01= 4w3=1 p3nEl 51 a 3= (3=1c 43 401 (n4h 06P1 “n4h 3g1 41w3gl 8E6w1\b M3 Cl=t1gg1 (n4h 06P1 hp3 41w3gl 8E6w14c 01gw1 kŒf\b 230 kCE43 mof m,0 d Tc h01g 00 1 A6 0 = A4 d1A4 A6 0 =\b 7g h01 3h01= 06glc t, p1 4p68 06gl 4c p1 )1h 401 A6 0 = A0 d 401 ATA6 0=c p0tw0 t4 E6=)1=c 4tgw1 53h0 16gl 46=1 s 3= (3=1\b M3 h01 E6=)14h 8344t5E1 P6En1 w6g”h 06P1 0d T\b Mt(tE6=Ebc h01 E6=)14h 8344t5E1 P6En1 w6g”h 06P1 6gb 3, 13 6c 3==1en6E h3 T\b 20n4 4 d T\b 201g p1 3gEb g11l h3 w3g4tl1= h01 ,3EE3ptg) w6414\b 6 s0 %Afl0 iAT d so 6 s0 flA%0 iAT d sfl 6 s0 iA%0 flAT d s% 201=1,3=1c h01 E6=)14h 8344t5E1 P6En1 ,3= h01 1•8=144t3g t4 soc 01gw1 kSf\b 2016 =43 , henc(E )(.eTc(na 20 2016 AMC – Junior Solutions

46 20120 16=43 4,h e66n6c 1(e E411e6, =, h=4)64. T=a g4, le sni,hc tn 1(41 n,e Entt=l=3=10 =t 1(41 1(e n1(e6 1=3e =t T\bad T=a 2016= T==a = T===a 2 6 22 2 2 2 2 2 2 oe ,eeh 1n gn,86. 1(41 1(=t =t 1(e n,30 Entt=l=3=10d D(e )6=h g4, (4ye 41 .nt1 n,e ns =1t w gn6,e6t nggiE=eh l0 1(e n1(e6 1=3ec tn e=1(e6 k n6 w ns 1(e S 1=3et f=33 nggiE0 4 gn6,e6d S4le3 1(e tmi46et ns e4g( S 2c0c1c 6c = 4t t(nf, =, h=4)64. T=ac 1(e, 1(e n,30 tmi46et ns 4, S 1(41 g4, le =, 1(e gn6,e6 ns 1(e u 2 u )6=h 46e 2c64,h =d 5s ==t =, 4 gn6,e6c 4,n1(e6 S .it1 81 =, 4t =, h=4)64. T==ad +n6 1(e 6e.4=,=,) u 2k 6eg14,)3e 1n gn,14=, 1fn St 4,h 3e4ye 1(e 6e.4=,=,) 46e4 =, n,e E=egec 1(e 1fn St .it1 .4%e 4,n1(e6 u 2A 6eg14,)3ec 3e4y=,) 4 u 2M t164=)(1 1=3ed D(=t =t ,n1 4, nE1=n,c tn ,n,e ns 1(e St (4ye tmi46e =n, 4 gn6,e6d Pn n,30 24,h 6g4, le =, 4 gn6,e6d pn,t=he6 1(e b tmi46et .46%eh 2=, h=4)64. T===ad fl, S f=1( 2=, 1(e gn6,e6 f=33 gnye6 A ns 1(etec 4,h 4, S f=1( 6=, 1(e gn6,e6 f=33 gnye6 kd 9e,ge 1(e6e g4,C1 le 1(6ee St f=1( 6=, 1(e gn6,e6c tn 1(e6e .it1 le 41 3e4t1 n,e S f=1( 2=, 1(e gn6,e6d I34ge 4, S f=1( 2=, 4 gn6,e6 4t =, h=4)64. T=yac 1(e, 1(e tmi46et .46%eh 2g4,,n1 le 833eh l0 4,0 ns 1(e 1=3et Tfla7T\bac tn 1(e0 n,30 le 833eh l0 4,n1(e6 S f=1( 2=, 1(e gn6,e6c 4t =, h=4)64. Tyad T=ya 2 2 2 Tya 2 2 2 2 pn,1=,i=,) 3=%e 1(=t 3e4ht 1n 1(e tn3i1=n, 436e4h0 nlte6yehd Pn 1(e6e =t n,30 n,e tn3i1=n,c (e,ge T\bad 261 Tfl3tn :IA‹a Œ 2 Ž Ž Œ Œ Ž Ž +=6t1 160 ,n1 1n ite 4 l3ie gni,1e6 41 433d D(e gni,1e6t g4,C1 le 433 6eh n6 433 )6ee,c tn t1461 f=1( 4 6eh gni,1e6 41 1(e ln11n. ns 1(e g=6g3ed 20 1(e 86t1 6i3ec 1(e gni,1e6t e=1(e6 t=he .it1 le )6ee,d D(e,c l0 1(e tegn,h 6i3ec 1(e gni,1e6t nEEnt=1e 1(ete )6ee, gni,1e6t .it1 le 6ehd D(e 1nE gni,1e6 g4, ,nf le gn3ni6eh )6ee,d D(e t=he gni,1e6t g4,,n1 le 6eh leg4ite 1(e0 46e 4h‘4ge,1 1n 4 6ehd 9nfeye6c n,30 n,e ns 1(e. g4, le )6ee,c tn 1(e n1(e6 .it1 le l3ied D(=t 4664,)e.e,1c 4t t(nf,c t41=t8et 1(e 1(6ee 6i3etd 9e,gec 1(e .=,=.i. ,i.le6 ns l3ie gni,1e6t =t Mc (e,ge T2ad 20 2016 =43 , henc(E )(.eTc(na 2016 AMC – Junior Solutions

4747 201201 6=614 3 ,h1enc, (n hE6 01()(ne. Teag6hl hE616 s(.. i6 t T(6a6, e\bh61 i16ege)6d o\b hE6,6 tl 3 e16 )8e1enh66c h0 i6 eh .6e,h Ee.\b e ,h1encd D0 2 00\b hE6 T(6a6, e16 )8e1enh66c h0 i6 eh .6e,h e, .0n) e, Ee.\b en 8ni10g6n ,h1encl E6na6 ywkd 6=1 yS.,0 fm3uk 50n,(c61 hE6 T1(+6 \beah01(,eh(0n 0\b %uAM P % 223 02pd 2eah01, 0\b %uAM 8nc61 Au e16 Al %l 3l bl Ml pl fl enc 9d on.4 p Ee, T1(+6 \beah01 pl ,0 hE(, +8,h i6 0n6 0\b hE6 e)6,d CE6 3 0(n hE6 T1(+6 \beah01(,eh(0n s(.. 6(hE61 i6 \b10+ 3 2MP% 23 001 \b10+ 9 P 3 0d In hE6 71,h ae,6l hE6 \beah01(,eh(0n (, 3 2M2 p2 AMl sE616 AM (, h00 .e1)6d In hE6 ,6a0nc ae,6l hs0 0\b hE6 e)6, e16 p enc 9d CE6n hE6 16+e(n(n) hs0 e)6, +8.h(T.4 h0 % 2P 3%l ,0 hE64 +8,h i6 b enc fld :6na6 hE6 e)6, e16 bl pl fl enc 9l sE(aE ecc h0 %fll E6na6 y5kd 621 2016=431,h6 e CE6 .e1)6 ‹8) (, i6hs66n tu2 bPA% 10Œ enc tu23PAM 06Œd D0 (h, aeTea(h4 0(, 6(hE61 A3l Abl Atl 01 AM .(h16,d CE6 e+08nh .6\bh (n hE6 ie116. (, 6(hE61 AAl fll t 01 % .(h16,d 5e.. hE(, Ž8enh(h4 1P tu 030d CE6 ,+e.. ‹8) Ee, aeTea(h4 6i6hs66n 12b enc 123d CE6 0Th(0n, e16 ,E0sn (n hE6 hei.6‘ 0 1 2 1 2 2 T0,,(i.66 A3 AA %2 1 36 2 3 Ab fl % %6 2 ’ At t A= 1 A6 2 ’ AM % = 6 6 2 ’ D0 hE6 ,+e.. ‹8) Ee, aeTea(h4 3 .(h16,l E6na6 y5kd 2016=431,h6 n D8TT0,6 0n6 .e1)6 ‹8) E0.c, 0.(h16, enc 0n6 ,+e.. ‹8) E0.c, 1.(h16,d I\b 0P Ap 01 +016l hE6n 3 .e1)6 ‹8), aenn0h i6 7..6cl enc (\b 0P A% 01 .6,, hE6n +016 hEen 3 .e1)6 ‹8), s08.c i6 7..6cd D0 0P A3 =Ab= At 01 AMd I\b 0P AMl hE6n % .(h16, 16+e(nl enc 3 ,+e.. ‹8), aen“h i6 7..6cd I\b 0P Atl hE6n t .(h16, 16+e(nl enc hE6 ,+e.. ‹8) +8,h E0.c A .(h16d w8h hE6n t ,+e.. ‹8), aen i6 7..6cd I\b 0P Abl hE6n fl .(h16, 16+e(nl enc hE6 ,+e.. ‹8) +8,h E0.c % .(h16,d w8h hE6n b ‹8), aen i6 7..6cd 2(ne..4l (\b 0P A3l hE6n AA .(h16, 16+e(nd CE6n ,(na6 AA P 3 23 ” %l 6eaE ,+e.. ‹8) +8,h E0.c 3 .(h16,l E6na6 y5kd 2016 =43 , henc(E )(.eTc(na 20 2016 AMC – Junior Solutions

48 2202016=431,h6 e 20 16= 43,,h= ,3e31 3n c( 16=E 16= E)4.=T 4)n1 =316=T 6ag= 16= 0lT4 s c lT c si 2E =at6 tan= 16= \b taEEl1 .= a,dat=E1 1l =316=T 16= o lT 16= 8( nl 31 4)n1 el .=1D==E 16= c aE, 16= s( aE, 16=T= aT= 16=E s Dayn 1l what= 16= o aE, 16= 8i k6=T=0lT= 16=T= aT= \b n)t6 E)4.=Tn D6ln= 43,,h= ,3e31 3n ci Sy ny44=1Ty 16=T= aT= ahnl \b n)t6 E)4.=Tn D6ln= 43,,h= ,3e31 3n 8i 20 16= 43,,h= ,3e31 3n s( 16=E 16= E)4.=T 4)n1 =316=T 6ag= 16= 0lT4 c s o lT o s ci 2E =at6 tan= 16= \b taEEl1 .= a,dat=E1 1l 16= o( nl 31 4)n1 .= .=1D==E 16= c aE, 16= s( aE, 16= wln313lE l0 16= 8 3n 16=E ,=1=T43E=,i k6=T=0lT= 16=T= aT= s n)t6 E)4.=Tn D6ln= 43,,h= ,3e31 3n si Sy ny44=1Ty 16=T= aT= ahnl s n)t6 E)4.=Tn D6ln= 43,,h= ,3e31 3n \bi 20 16= 43,,h= ,3e31 3n o( 16=E 16= E)4.=T 4)n1 =316=T 6ag= 16= 0lT4 s o \b lT \b o si 2E =at6 tan= 16= c 4)n1 .= .=1D==E 16= o aE, 16= \b( aE, 16= wln313lE l0 16= 8 3n 16=E ,=1=T43E=,i k6=T=0lT= 16=T= aT= s n)t6 E)4.=Tn D6ln= 43,,h= ,3e31 3n oi k6= 1l1ah E)4.=T l0 E)4.=Tn D316 16= T=f)3T=, wTlw=T1y 3n \b m \b m s m s m s u c\b( 6=Et= 5S+i 2016=431,h6 n c s o \b 8 c s o \b 8 2E 16= %Tn1 ,3aeTa4( 1Dl ,3e31n aT= dl3E=, 30 16=y taE .= E=3e6.l)T3Ee ,3e31n 3E 16= E)4.=Ti k6= 8A,3e31 E)4.=Tn 3E 16= f)=n13lE tlTT=nwlE, 1l wa16n 16a1 g3n31 =g=Ty ,3e31 =Mat1hy lEt=i k6= n=tlE, ,3aeTa4 6an 16= na4= =,e=n( .)1 3n T=aTTaEe=, 0lT thaT31yi 20 =,e= cP8 3n El1 )n=,( 16=T= aT= cp wlnn3.3h313=n( n3Et= 16=T= aT= 8 t6l3t=n l0 n1aT13Ee ,3e31( 16=E s t6l3t=n l0 n=tlE, ,3e31( aE, 16=E ahh l16=T ,3e31n 0lhhlDi 20 =,e= cP8 3n )n=,( 16=E 16=T= aT= \b wlnn3.3h313=ni k6=n= taE .= tl)E1=, .y t6lln3Ee =316=T =,e= cPo lT =,e= 8Po( aE, 16=E ,=t3,3Ee D6=16=T 16= wa16 D3hh n1aT1 lT =E, a1 o( n3Et= 16= T=n1 l0 16= wa16 3n ,=1=T43E=, .y 163ni k63n e3g=n 0l)T wlnn3.3h313=nb oc8s\b( \bs8co( o8c\bs aE, s\bc8oi 2E ahh 16=T= aT= cp m \b u c\b wlnn3.3h313=n( 6=Et= 5S+i 210 fl)4.=T 16= w=lwh= c2 s2o 2000 i 9=TnlE c =316=T 53+ ,l=n D=aT a 6a1 lT 533+ ,l=n El1i 2E =316=T tan=( 0lT w=TnlEn s( o( \b( i i i ( D= ,=t3,= D6=16=T 16=y 6ag= a 6a1 .y 4aC3Ee n)T= 16a1 w=TnlEn c( s( o( i i i 6ag= =Mat1hy lE= E=3e6.l)T D316 a 6a1b 53+ 2 0 1 6 = 4 3 , h 2e 22 20 21 26 2= 24 23 222 222 222 533+ Iat6 l0 16=n= wa11=TEn T=w=a1n =g=Ty \b w=lwh=( n3Et= 30 13n 6a1h=nn 16=E 1m s 3n 6a11=,( aE, g3t=Ag=Tnai 7Tl4 16=n= wa11=TEn( w=TnlEn c 2o 28 2: 2000 5l,, E)4.=Tn+ 4ay lT 4ay El1 6ag= 6a1n( w=TnlEn \b2 ‹2cs2 cŒ2000 54)h13wh=n l0 \b+ E=g=T 6ag= 6a1n aE, w=TnlEn s 2Œ 2cp2 c\b2000 5=g=E( El1 4)h13wh=n l0 \b+ ahDayn 6ag= 6a1ni Ž)n1 an 16= n=tlE, w=TnlE 4)n1 6ag= a 6a1( nl 4)n1 16= n=tlE,Ahan1i ‘lD=g=T( w=TnlE cpp taE’1 6ag= a 6a1( nl 16=T= taE’1 .= cpc w=lwh=i k6= l16=T aEnD=Tn “‹( ““( cpp aE, cps aT= ahh wlnn3.h=( n3Et= ElE= l0 “:2 “‹2““ aE, cpc 3n a 4)h13wh= l0 \b( nl =at6 20 2016 =43 , henc(E )(.eTc(na 2016 AMC – Junior Solutions

4949 2011 6=43 = 6=, 0h 30,63e n0c (e n00c =E(43)63h.3 nTca 201 2016=431,h6 e g, 3=.6 l,3s) ,63 s3el(h e3.3040hi 1(1103l t(\bE13l ,630e s013a d(eo0hi E=.o2=etl) 8(l6 D\b,6 y=w kS kS kS nf(1103l =, 3htc mu mu u5 n8(l6 =ht D\b,6 =E(\b, ,( t(\bE13 ,( kSc+ %u kS n8(l6 =E(\b, ,( t(\bE13 ,( mu =ht y=w ,( u5c %S S+ mu nD\b,6 =E(\b, ,( t(\bE13 ,( %u =ht y=w ,( kSc 63h.3 nAca 2016=431,h6 n y\bss(l3 8(l6 l,=e,l 20,6 21(1103l (\b, (M ,63 Pu ,(,=1a D\b,6 =ht y=w ,(i3,63e 6=43 nPu 22c) l( 8(l6 i043l =2=p nPu 22c 1(1103l) 13=40hi 2 2 nPu 22cbS 22 Pu b Sn2 25+ca D\b,6 ,63h i043l 8(l6 Sn2 25+c w(e3 1(1103l l( ,6=, 8(l6 6=l 5n2 25+ca y=w ,63h i043l 8(l6 5n2 25+c w(e3 1(1103l l( ,6=, 8(l6 6=l +n2 25+c 0h ,63 3hta y(140hi) +n2 25+c b kS) ,63h 22 5+ b 5 =ht 2b %S) 63h.3 nAca 261 2016=431,h6 e fl(e .(h43h03h.3) e3se3l3h, ,63 10h3l Ep lpwE(1l 0)1)6(e =l( ,6=, 10h3l \bl3 ,63 l=w3 lpwE(1 0M =ht (h1p 0M ,63p e6pw3a 9(,3 ,6=, l,e\b.,\be3l l\b.6 =l 0106=ht6=60 =e3 h(, .(hl0t3e3t t0C3e3h, l0h.3 30,63e (h3 0ht0.=,3l ,6=, ,63 Iel, =ht ,60et 10h3l e6pw3 20,6 3=.6 (,63e) 26013 h30,63e (M ,63 l3.(ht (e M(\be,6 10h3l e6pw3l 20,6 =hp (,63ela 763 M(11(20hi e\b13l 2011 i3h3e=,3 = 10l, (M t0C3e3h, e6pw0hi l,e\b.,\be3l: ma763 Iel, 13,,3e w\bl, E3 0a Sa g 13,,3e .=h E3 \bl3t (h1p 0M =11 (M 0,l se3t3.3ll(el 0h ,63 =1s6=E3, 6=43 =1e3=tp E33h \bl3ta 763 M\b11 10l,) 0h =1s6=E3,0.=1 (et3e) 0l 00004 00014 00104 00114 00164 01004 01014 01064 01104 01114 01164 01604 01614 01664 016= l( ,63e3 =e3 m% t0C3e3h, e6pw0hi l,e\b.,\be3l 0h ,(,=1a 7( .(\bh, ,63w w(e3 lpl,3w=,‹ 0.=11p) .(hl0t3e ,63 I43 ,6e33‹10h3 s(3wl: 0004 0014 0104 0114 0163 g11 M(\be‹10h3 s(3wl =e3 .(hl,e\b.,3t Ep =ss3ht0hi = l0hi13 13,,3e ,( ,63l3) l\bE Œ3., ,( e\b13 S =E(43a 763 m% s(ll0E010,03l =e3 000Ž0 (e 1 b S s(ll0E010,03l 001 010 011 2 0 1 Ž 04 1 (e6 bk0k b P s(ll0E010,03l 016 Ž04 14 6 (e= b 5 s(ll0E010,03l 63h.3 n‘ca 2016 =43 , henc(E )(.eTc(na 20 2016 AMC – Junior Solutions

50 2016=431,h6 e 20166=43 ,he nh3c=(E )1,,en(6 .3 ,he (Tc.en a4 0=(e6 ,h1, nh3cegl=s i=,h 100 t 0=(e6 nh3c=(E\b ,hene =6 a(03 a(e )1,,en( d d d d d d d d d d d d d d d d d d d d d d d o l==s i=,h 8 0=(e6 nh3c=(E 1(D a(e 0=(e (a,\b ,he T(nh3ceD 0=(e =6 0=(e o\b y\b 8 an td wa ,hene 1ne t )1,,en(6 ddddddddddddddddddddddddddddddddddddddddddddd t l===s i=,h y )1=n6 a4 y 0=(e6 nh3c=(E\b ,he 0=(e ,h1, nh3ce6 k=,h 0=(e o =6 0=(e y\b 8 an td wa ,hene 1ne 8 )1,,en(6 ddddddddddddddddddddddddddddddddddddddddd 8 l=Ss i=,h y 0=(e6 nh3c=(E 1(D ,he a,hen ,ka (a, nh3c=(E\b ,hene 1ne y )1,,en(6 4an eSen3 )1,,en( =( lfsd wa ,hene 1ne m )1,,en(6 d d d d d d d d d d d d d d d d d d d d d d d d d d m lSs u4 (a 0=(e6 nh3ce\b ,hene =6 a(03 o )1,,en( ddddddddddddddddddddddddddddddddd o 5he( o+t+8+m+o % oA\b he(fe lMsd 201 Pe,2.e ,he (Tc.en 016d u( ,he T(=,6 fa0Tc(\b t6 h16 T(=,6 D=E=, y\b 6a6% 8 an pd u4 6% p\b ,he( opbbb =yt2= oflbbbd 9nac ,he cT0,=)0e6 a4 yt 0=6,eD .e0ak\b ke h1Se yt 2Cbb % ompbb 1(D yt 2pbb % oflybb 6a ,h1, 2cT6, .e =( ,he Cbb6 1(D 0% Cd u( ,he ,e(6 fa0Tc(\b ke cT6, h1Se 8 + p + m % oCg C1p 2 yt 8y II yp om II ot op 1Cy 4 yt44 yt4 o yt m ott y tp C omp 8 Cy p ofly t flm fl yom A oyb 5he( t 1e(D6 =( p\b 6a 1% y an 1% C\b .T, (e=,hen a4 ,he6e kan7d u4 6% 8\b ,he( o8bbb =yt2= otbbbd 9nac ,he cT0,=)0e6 a4 yt\b Abb =2 =mbb\b 6a 0 % Ad 5he( =( ,he ,e(6 fa0Tc(\b o + p + m % oAg A18 2 yt oy II yb m II ob o8 1Ay :E1=(\b t 1e(D6 =( p 6a ,h1, 1% y an 1% Cd 20e1n03 1% y =6 ,aa 6c100\b .T, 1% C E=Se6 1 6a0T,=a(g AC8 2yt % o8CAyd wa 2% AC8 =6 ,he a(03 6a0T,=a(\b he(fe lAC8sd 20 2016 =43 , henc(E )(.eTc(na 2016 AMC – Junior Solutions

5151 2012016= 43,he n,c( E).T. aT. g)T.. la6.6 s=T )=i g). gTta\bd1.6 aT. l=1=oT.8D 2t( 011 g)T.. 6t8.6 aT. g). 6ay. l=1=oTe itg) w k=66tSt1tgt.6f 2tt( Ei= 6t8.6 aT. g). 6ay. a\b8 =\b. 6t8. t6 8tm.T.\bge itg) w 2c u ,5 k=66tSt1tgt.6f 2ttt( 011 g)T.. 6t8.6 aT. 8tm.T.\bge itg) 22 02 1 6 u +5 k=66tSt1tgt.6f %= g).T. aT. w A ,5 A +5 u Mw k=66tSt1tgt.6 t\b a11e ).\bl. 2Mw(f 2016= E). la1lo1agt=\b t\b 2ttt( T.1t.6 =\b g). =S6.TPagt=\b g)ag ts i. aT. l)==6t\bd g). 6t8.6 t\b =T8.Te g).T. aT. w k=66tSt1tgt.6 s=T g). pT6g 6t8.e g).\b c s=T g). 6.l=\b8e g).\b M s=T g). g)tT8f b=i.P.Te t\b g).6. w 2c2 M u fl5 k=66tSt1tgt.6e .al) 6.1.lgt=\b 201 it11 akk.aT fl gty.6D 2016 2106 0216 0126 1206 102 f E)t6 t8.a akk.aT6 t\b g). d.\b.Ta1 s=Tyo1a s=T 2 2 00 e g). \boyS.T =s ia96 =s l)==6t\bd ==S C.lg6 sT=y 4=S C.lg6f 261 I.g3,hS. g). g)T..78tdtg \boyS.Tf E).\b +55 3A +5 ,A hu M:3 A M:,A M: hf E)t6 dtP.6 flM3 u ,:,A Mflh a\b8 8tPt8t\bd S9 he : 3uM ,Ac hf E)t6 dtP.6 ya\b9 6=1ogt=\b6 i).T. 3u ,u he Sog tg t6 a T.‹otT.y.\bg g)ag 3e,a\b8 haT. a11 8tm.T.\bgf b.\bl. i. aT. 1==Œt\bd s=T M, Achg= S. a yo1gtk1. =s : i).T. ,0u hf 06 i. aT. 1==Œt\bd s=T g). 1aTd.6ge i. gT9 3u hf E).\b M, Achu flMe i)tl) )a6 6=1ogt=\b6 2+6 +w(e 2w6+,(e 2h6h(e 2+M6fl(e 2+:6M( a\b8 2,+ 65(f Ž=\b. =s g).6. i=TŒf ET9t\bd 3u ‘e g).\b M, Achu wfle dtPt\bd 256 +c(e 2c6++(e 2‘6‘(e 2+,6w( a\b8 2+fl6 ,(f Ž=\b. =s g).6. i=TŒf ET9t\bd 3u :e g).\b M, Achu ch dtPt\bd 2M6 +5(e 2:6:(e 2++ 6c( a\b8 2+w 6+(f Ž=\b. =s g).6. i=TŒf ET9t\bd 3u fle g).\b M, Achu c, dtPt\bd 2,6 h(e 2fl6fl(e 2+56 M( a\b8 2+c 65(f %= g). 1aTd.6g 6=1ogt=\b t6 fl,he ).\bl. 2fl,h(f Ž=g.D ns 9=o Œ\b=i =T =S6.TP. g)ag M 2M: u +++e g)t6 dtP.6 a\b .’lt.\bg ia9 g= 6.aTl) s=T 6=1ogt=\b6 itg)=og g). aS=P. a1d.STaf 2=1 %t\bl. .al) g.Ty t6 g). 6oy =s g). kT.Pt=o6 gi=e g). ay=o\bg =s .TT=T s=11=i6 g). 6ay. kagg.T\bf %gaTgt\bd sT=y g). ‘hg) g.Ty i)tl) t6 l=TT.lg a\b8 g). h5g) g.Ty i)tl) )a6 a\b .TT=T =s +e i. )aP. g). s=11=it\bdD E.Ty ‘h h5 h+ h, hM hc hw hfl h: h‘ hh +55 “TT=T 5 + + , M w ‘ +M ,+ Mc ww ‘h Ž=g. g)ag i. 8=\b”g algoa119 \b..8 g= Œ\b=i i).g).T g). pT6g .TT=T ia6 aS=P. =T S.1=i g). l=TT.lg Pa1o.e ).\bl. 2‘h(f 2016 =43 , henc(E )(.eTc(na 20 2016 AMC – Junior Solutions

52 2012011 6=4 14341, h en034c 2( En 6) . eT)1Eag 0(l si0tc0a ei( ,=)\b, 6=4 n),,id14 4ot4, \b=4c4 8 T0g ,64n Enl ei( 2 1 0 eii( 2 1 0 eiii( 2 1 0 D=4c4 0c4 yw 4ot4, d46\b44g 14341 y 0go 14341 kl S)c 6=4 6)n 4ot4 )f T)1Eag 1m oi0tc0a eii( ,=)\b, 6=4 )g1u n06= fc)a 26=06 5gi,=4, du Tc),,igt 6=06 4ot4l si0tc0a eiii( ,=)\b, 6=06 f)c 6=4 )6=4c y+ 4ot4, 6=4c4 0c4 6\b) n06=,l %) 6=4c4 0c4 .A n06=, 6) 14341 k 6=06 4go du ,64nnigt 0Tc),, )g4 )f 6=4,4 4ot4,l M)\b if 6i, 6=4 14341Pk T)1Eag 5c,6 ,64nn4o )gm 6=4g )f 6=4 p 4ot4, )g T)1Eag 0m )g4 =0, )g4 n06= fc)a 66) 0oic4T61u Tc),,igt 6=06 4ot4m 0go 6=4 )6=4c A =034 6\b) n06=,m 0, ,=)\bg ig oi0tc0a, ei3( 0go e3( c4,n4T6i341ul %) 6=4c4 0c4 yy n06=, fc)a 6 6) 0l ei3( 0 6 e3( 06 D=4c4 0c4 g) c4,6ciT6i)g, )g T)adigigt 6=4 n06= fc)a 26) 6\bi6= 0 n06= fc)a 66) 0 m ,) 6=4 6)601 gEad4c )f n06=, i, .A 2yy b .wAm =4gT4 e.wA(l 20 2016 =43 , henc(E )(.eTc(na 2016 AMC – Junior Solutions