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30 20+16=03, h e=nn+c (C=).CA l=s=,=03 oU20 + 16 = 36, hence (C). PU (Also UP1) The n\bmbers in order are 555, 556, 565, 566, 655, hence (D). TU The digit 3 is in the tho\bsands position, so it represents 3000, hence (D). \bU I am 12 years old, so o\br combined ages are 6 + 12 = 18, hence (C). mU Here are all possible lines of symmetry: Altho\bgh the Z shape has a point of symmetry, it does not have a line of symmetry, hence (E). bU (Also UP2) One pizza will have 4 q\barters, so two pizzas will have 2 24 = 8 q\barters, hence (D). rU After 30 min\btes it is 5 pm, and after another 15 min\btes it is 5:15 pm, hence (E). iU The n\bmber of wheels is 4 22+22 3+124 = 18, hence (E). dU (Also UP6) 2 0 + 1 6 = 3 , h 2e The opposite chair is both 5 places forward and 5 places back. Five places back from chair 9 is chair 4, hence (D). oaU (Also UP5) 20+16=3+,h1 e In cents, 500 080 = 6r20 so that he b\bys 6 chocolates and has 20 cents left, hence (C). +e 20+1 6=3 , =heenc (Ch).CA lsnoUhsPT 2016 AMC – Middle Primary Solutions

3131 20+16=3+,h1 e 20+16=+3, he nc (C3 nc2).c 2Alc 2sAc 2lcc 2lnc 2o.cU PChT 1\b6,m \b3 b(r (ihCd . b\bhbha +(13, 501 rh1 Dm \b3rb3 gtpU 220 Ih T(y3 1\b3 1h1(+ (, +(C83 (, =h,,65+3m 1\b3 +(C83 d6861, ,\bh0+d \b(H3 =+(b3 H(+03 (, +(C83 (, =h,,65+3U I\b(1 6,m f (rd n :6++ 53 6r 1\b3 13r, =h,616hr, he 1\b3 1:h r0T53C,U I\b3r s (rd A :6++ 53 6r 1\b3 0r61, =h,616hr,U I\b3r 1\b3 ,0T 6, 361\b3C fs Z nA v )Do hC fA Z ns v )Dom 5h1\b :61\b 1h1(+ )Dom \b3rb3 gEpU 2+0 gE+,h O.p z(b\b 16T3 1\b3 =(=3C 6, 0reh+d3dm 1\b3 1:h =(C1, :6++ 53 C3w3b16hr, he 3(b\b h1\b3C 1\bCh08\b 1\b3 eh+d +6r3U \b3rb3 gEpU 210 gE+,h 4qnp k\b3 361\b3C \b(, ( ocb bh6r hC rh1U Fe ,\b3 \b(, ( ocb bh6rm 1\b3r ,\b3 \b(, hr3 h1\b3C )cb bh6r9 oc Z )c v .cU Fe ,\b3 \b(, rh ocb bh6r,m 1\b3r ,\b3 361\b3C \b(, cm )m A hC s Acb bh6r,9 Ac Z Ac Z Ac v .c Ac Z Ac Z )c Z )c v .c AcZ)cZ)cZ)cZ)c v .c )cZ)cZ)cZ)cZ)cZ)c v .c Fr (++m 1\b3C3 (C3 o =h,,656+6163,m \b3rb3 g‘pU 260 k6rb3 1\b3C3 (C3 s bh+h0C,m ’h0 b(r“1 53 ,0C3 1\b(1 1\b3 ”C,1 s 53(r, 6rb+0d3 ( =(6CU •h:3H3Cm :61\b l 53(r,m 1\b3’ b(r“1 (++ 53 d6i3C3r1 bh+h0C,m ,h 1\b3C3 T0,1 53 ( =(6C he 1\b3 ,(T3 bh+h0CU kh l 53(r, g(rd rh ThC3p (C3 r33d3d 1h T(y3 ,0C3 ’h0 \b(H3 ( =(6Cm \b3rb3 g–pU 2=0 gE+,h 4q))p k1(C16r8 eChT 1\b3 h013C 3rd he 1\b3 ,=6C(+ g1\b3 +hh= hr 1\b3 Ch=3p 1\b3 d(Cy (rd +68\b1 ,3b16hr, (C3 +hr83,1m (rd 1\b3 +68\b1 ,3b16hr, (C3 he ,6T6+(C +3r81\b 1h 1\b3 d(Cy ,3b16hr,U E, ’h0 ThH3 1h:(Cd, 1\b3 h1\b3C 3rd he 1\b3 Ch=3m 5h1\b d(Cy (rd +68\b1 ,3b16hr, 831 ,\bhC13CU —r+’ Ch=3 gEp ,\bh:, 1\b6,m \b3rb3 gEpU 20+1 6=3 , =heenc (Ch).CA lsnoUhsPT 20 2016 AMC – Middle Primary Solutions

32 20+20+1+ 61+ = 3 , 3 h 3 e 3 n c e, (C).+AC(l (s 06oU C0+ Po6(( T( \b (C).+AC(m 20+ , bTC0 s16Ar+ 06C( 61+ TA sA+i06oU sU C0+ Po6((l 6A. (s C0+ sC0+1 Cbs (C).+AC( TA C06C 06oU sU C0+ Po6(( 06d+ ao6P5 06C(m Ds C0+ sAog b6g Cs (toTC C0+ Po6(( TACs Cbs +p)6o r1s)t( T( bTC0 s16Ar+ 6A. ao6P5 TA sA+ r1s)t 6A. 1+.l r1++A 6A. g+oosb TA C0+ sC0+1l 0+AP+ Iy8m 21+ 20+ ()H sU C0+ f1(C fd+ .TrTC( T( hhm 20+1+Us1+ C0+ ()H sU C0+ o6(C Cbs .TrTC( H)(C a+ ehl 6( =n 2hh c ehm 20+1+ 61+ : ts((TaToTCT+( Us1 C0+ o6(C Cbs .TrTC(Z =vl v=l n\bl \bnl E:l :El 6A. ,,l 0+AP+ IO8m 26+ Iyo(s zweh8 20+ o61r+ (p)61+ H)(C 06d+ (T.+ \b PHm 20+A 6oo sU C0+ Ps1A+1( sU C0+ tT+P+( TA C0+ C6Ar16H oT+ sA 6 r1T. sU h PH 0h PH (p)61+(m 20+ (06.+. (p)61+ 06( C0+ (6H+ 61+6 6( h sU C0+ r1T. (p)61+(l s1 h 0Ih PH 0h PH8 c \b PH 2l 0+AP+ I48m 2=+ qs H6CC+1 0sb 6 (TAro+ Uso. T( H6.+l C0+1+ bToo a+ sA+ sU C0+ s1TrTA6o 1Tr0Ci6Aro+ Ps1A+1( C06C T( sA C0+ as)A.61g sU C0+ (06t+m kU C0+ E (06t+(l IF8 .s+( AsC 06d+ 6Ag 1Tr0C 6Aro+(l (s TC P6A9C a+ H6.+ bTC0 6 (TAro+ Uso.m 20+ sC0+1 Us)1 fr)1+( P6A a+ H6.+ bTC0 6 (TAro+ Uso.Z Iy8 IO8 I‘8 I48 0+AP+ IF8m 3,+ 20+16=3+,h1 e 20+1+ 61+ E b6g( sU H65TAr eh U1sH C0+(+ A)Ha+1(Z :3n3e c eh 2:3=3h c eh 2,3E3e c eh 2,3n3h c eh 2E3n3= c eh 20+ A)Ha+1 n T( TA = sU C0+(+ ()H(l 6A. AsC TA e 3 E 3 , 6A. h 3 = 3 :m 20+ A)Ha+1 n Ps)o. a+ +TC0+1 TA 6 Ps1A+1l 6 (T.+ s1 C0+ P+AC1+Z IT8 n ITT8 n ITTT8 n 20 20+1 6=3 , =heenc (Ch).CA lsnoUhsPT 2016 AMC – Middle Primary Solutions

3333 20+161=3 ,h1 henhcenh,1( ceC1) eC 1.Ah A.C 0Ccl s1 o U P U T .C( \b U m U b +heAh (0Cr, h.61 . Cids1= eC A0dd0C3 )0 ae5 .C( aee5 (0Cr, +0=Dg tc.AeCn p eC ,h1 A1C,=13 ,h1=1 .=1 )161=.c +.l) ,0 .==.Cn1 ,h1 0,h1= Cids1=)I p oPT \b mb p oTP m \bb h1CA1 ay5g 20+16=3+,h1 e 8cc )161C Ae=Ac1) .(( ,0 o U \b U m U p U P U T U b H \bfg :h1 ,0Z .C( s0,,0d =0+) ,0n1,h1= .(( ,0 \b 2o\b H \bpg v0 ,h1 de((c1 Ae=Ac1 di), s1 \bf 0\bp H p3 h1CA1 ay5g 20+ veCA1 1.Ah ,1.d Zc.l1( ,h=11 n.d1)3 +1 A.C A0dZc1,1 ,h1 =1A0=( E0= ,h1 O.nc1)3 z.cA0C) .C( w0C(0=)g tc.l1( 4eC y=.+ q0)) t0eC,) O.nc1) m m k k F 2.+D) m z.cA0C) m k o \b o w0C(0=) m k o \b o :h1 O.nc1) +0C .cc ,h1e= n.d1)3 )0 ,h1 2.+D) c0), ,0 ,h1 O.nc1)g :h1 z.cA0C) .C( ,h1 w0C(0=) h.( C0 +eC)3 )0 ,h1 n.d1 s1,+11C ,h1d di), h.61 s11C (=.+Cg :h1=1E0=1 s0,h c0), ,0 ,h1 2.+D)g :h1 2.+D)r ,h=11 n.d1) +1=1 \b +eC) .C( o c0))3 E0= T Z0eC,)3 h1CA1 aw5g 22+ q.s1c ,h1 Z0eC,) .) )h0+C .C( +0=D (0+C ,h1 (e.n=.dg 2 0 + 1 6 = 3 , :h1=1 e) o =0i,1 E=0d 2,0 +3 .C( .c)0 o =0i,1 E=0d 2,0 =g 90i,1) ,0 1+ecc A0d1 ,h=0inh 1e,h1= +0= =3 +heAh ne61) 0Ccl \b =0i,1)g qeD1+e)13 ,h1=1 .=1 \b =0i,1) ,0 3g 90i,1) ,0 6+ecc A0d1 ,h=0inh 1e,h1= 10= 3g :h1=1 .=1 \b =0i,1) ,0 1.C( \b =0i,1) ,0 33 )0 ,h1=1 .=1 p =0i,1) ,0 6g qeD1+e)13 ,h1=1 .=1 p =0i,1) ,0 ,g 90i,1) ,0 0.=1 ,h1 p =0i,1) ,h., A0d1 ,h=0inh 6Zci) ,h1 p =0i,1) ,h., A0d1 ,h=0inh ,3 d.DeCn f =0i,1)3 h1CA1 ay5g 21+ a8c)0 ‘t\bo5 :h1 (e)A) .=1 s.AD eC ,h1e= 0=eneC.c Z0)e,e0C) .E,1= P d061)g = s n l 0 o 0+ n l 0 = s \b 0+ 0 = s n l m 0+ s n l 0 = p 0+ l 0 = s n P 0+ = s n l 0 20+1 6=3 , =heenc (Ch).CA lsnoUhsPT 22 2016 AMC – Middle Primary Solutions

34 20+1 6=33 ,+ =h e0+=n cn=(=hC3 )c.=e=ch. C(C=h CAe+n lso lU ChP Ts \bcm+.b rAe+n l \bcn+ \bcm+o ,3i+ 6=33 ,+ ch e0+ ,ceec\bo0+hd+ a5Db 20+ ar3.c gtTTD 20+16=3+,h1 e p+)3Cd=h( e0+ 3+c)CnP ,1 Chce0+n 3=ch acA e0+ .C\b+ 6+=(0e C. e0+ 3=chD 6ci3P CPP Is y(o ChP n+)3Cd=h( e0+ e=(+n ,1 Chce0+n 3=ch 6ci3P CPP Us y(b 20+h 8 3=ch. 6+=(0 8ls H Is H Us f :Us y( ChP l 3=ch 6+=(0. :Us 28 f lUs y(o 0+hd+ a5Db 20+16=3+,h1 n ZA e0+ 3=ch 6+=(0. lss y(o e0+h e0+ 3+c)CnP 6+=(0. ls y( ChP e0+ e=(+n Us y( Acn C eceC3 cA lvs y(b 20=. =. lUs y( ecc 3=(0eb rPP=h( lUs 28 f Us y( ec +Cd0 6+=(0e y++). e0+ P=E+n+hd+. =h 6+=(0e e0+ .C\b+b Oc e0+ 3=ch 6+=(0. lUs y(o 0+hd+ a5Db 21+ zCh+ \bi.e 0Cm+ (=m+h C6C1 0+n wT dc=ho ce0+n6=.+ 2c\b 6ci3P 0Cm+ w: cn \bcn+b 2c\b \bi.e +hP i) 6=e0 Ch +m+h hi\b,+n cA d+he.o .c 0+ \bi.e 0Cm+ (=m+h C6C1 0=. Ud dc=hb 4=e0 qi.e e0+.+ e6c dc=h. (=m+h ec rh(i.o zCh+ 0C. wlbkU ChP 2c\b 0C. w8bksb Oc =e dChFe ,+ Pch+ 6=e0 qi.e T dc=h. (=m+h ec rh(i.b Ze dCh ,+ Pch+ 6=e0 8 dc=h.9 =A 2c\b (=m+. C6C1 0=. Ud ChP 0=. lsd 0+ 0C. w82 ‘so ChP =A zCh+ (=m+. C6C1 0+n wTo .0+ 0C. wlbkUo 0+hd+ a5Db 26+ 20=. eC,3+ 3=.e. e0+ hi\b,+n. CddcnP=h( ec e0+=n ’n.e e6c P=(=e.b “=n.e O+dchP P=(=e P=(=e slT8:Uv‘k ”cihe l lsl llT lT8 l8: l:U lUv lv‘ l‘k lkI I T TsT Tl8 TT: T8U T:v TU‘ Tvk T‘I k 8 8s8 8l: 8TU 88v 8:‘ 8Uk 8vI ‘ : :s: :lU :Tv :8‘ ::k :UI v U UsU Ulv UT‘ U8k U:I U v vsv vl‘ vTk v8I : ‘ ‘s‘ ‘lk ‘TI 8 k ksk klI T I IsI l :U 20+ eceC3 hi\b,+n cA )c..=,=3=e=+. =. I H k H ‘ H v H U H : H 8 H T H l f :Uo 0+hd+ a:UDb 2=+ •hd+ e0+ C(+ cA e0+ 1cih(+n PCi(0e+n =. yhc6ho e0+ ce0+n Acin C(+. ChP e0+ eceC3 dCh ,+ dC3di3Ce+Pb –+n+ Cn+ e0+ ’n.e A+6 )c..=,=3=e=+.9 —cih(+n PCi(0e+n •3P+n PCi(0e+n —cih(+n .ch •3P+n .ch 2ceC3 s T :‘ l8 l 8 vI lI T :k ll TU 20 20+1 6=3 , =heenc (Ch).CA lsnoUhsPT 2016 AMC – Middle Primary Solutions

3535 20+1 6=33,he nce3+e(,1C )+30 30, 3c3=. Ac+eA (6 +e l1C 1+en, sch ,=n0 oU ce 30, Pc(eA,h T=(A03,hC 30,h, +1 oU ce 30, c.T,h T=(A03,h =eT o\b ce mc30 1ce1b r 3c3=. cs ii h,d(+h,1 a5 Dch, 30=e +e 30, .=13 .+e, =mcg,C )0+n0 )+.. 0=66,e i hc)1 .=3,hb 20, te=. .+e, 10c)1pIc(eA,h T=(A03,h y.T,h T=(A03,h Ic(eA,h 1ce y.T,h 1ce 2c3=. 8 HUf \bU ii 0,en, :8Zb 20+ vs r =eT E =h, mc30 n0c1,eC 30,e 30, 30+hT 13=D6 +1 ,+30,h O ch zb :\b 6c11+m+.+3+,1Z vs r m(3 ec3 E +1 n0c1,eC 30,e O D(13 m, n0c1,eC =eT ,+30,h w ch zb :\b 6c11+m+.+3+,1Z vs E m(3 ec3 r +1 n0c1,eC 30,e z D(13 m, n0c1,eC =eT ,+30,h O ch 4b :\b 6c11+m+.+3+,1Z vs e,+30,h r ech E +1 n0c1,eC 30,e 30, a 13=D61 D(13 m, +e = hc) cs 30h,,p wOz ch Oz4b :\b 6c11+m+.+3+,1Z ve =.. 30,h, =h, f 6c11+m+.+3+,1C 0,en, :fZb 21+ :r.1c qk\bfZ 20, th13 n(m, (1,1 U\b D=3n0,1C 30,e ,=n0 1(m1,d(,e3 n(m, (1,1 f D=3n0,1b F+en, \b5Ul 2U\b 9 \b55‘ =eT \b55‘ 0f 9 \bi5h‘C 30,h, =h, U o \bi5 9 \biU n(m,1 D=T,C )+30 ‘ D=3n0,1 .,s3 cg,hC 0,en, :\biUZb 6=+ :r.1c ’\b5Z wce1+T,h 30, 6h+D, s=n3ch+1=3+ce cs \b5Ul 9 \b 2+a 0+8b 20, s=n3ch1 cs \b5Ul (eT,h U5 =h, UC \bC aC ‘C lC 8C f =eT Hb ye.P 8 0=1 6h+D, s=n3ch 8C 1c 30+1 D(13 m, ce, cs 30, =A,1b 20, a 0+e 30, 6h+D, s=n3ch+1=3+ce )+.. ,+30,h m, shcD a +l9\b +a 0ch shcD H 9 a 0b ve 30, th13 n=1,C 30, s=n3ch+1=3+ce +1 a +l+ 8+ UlC )0,h, Ul +1 3cc .=hA,b ve 30, 1,nceT n=1,C 3)c cs 30, =A,1 =h, 8 =eT Hb 20,e 30, h,D=+e+eA 3)c =A,1 D(.3+6.P 3c \b 29 a\bC 1c 30,P D(13 m, ‘ =eT fb “,en, 30, =A,1 =h, ‘C 8C f =eT HC )0+n0 =TT 3c \bfC 0,en, :\bfZb 20+1 6=3 , =heenc (Ch).CA lsnoUhsPT 20 2016 AMC – Middle Primary Solutions