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5353 2016=30,h e n,=c(Ac)3.=c +343h30, 5920216 = 320, hence (A). s9 1 2+ 1 4+ 1 5= 10+5+4 20 = 19 20 so that 2 01 \bs shaded, hence (A). o9 Between 11:15 am and 2:09 pm, there are 45 + 120 + 9 = 174 m\bnutes, hence (B). t9 (Also J10) Est\bmat\bng, 6012=3 012= 0 601111 0111 = 601 0 = 360. Th\bs suggests that 100 2 6012=3 012= 21000. Check\bng, 201600 2720163 22016000 and so 100 2 6012=3 012= 21000, hence (D). a9 2 01 3 ,= 2 02 , 3= , == 0 3, hence (A). \b9 2016=3,1he6 n Let the number be 0. Then 3 ,2 2 211 20= 6, so 2 ,11 = 2 and 0= 800, hence (A). 2016=3,1he6 c 0.25% of the number \bs 2 (one-th\brd of the g\bven amount), so 1% of the number \bs 8 and 100% of the number \bs 800, hence (A). d9 (Also J16) If1 = 1, then 12 6 += 2 3 +, =6+, += 2 3. On the other hand, \bf we swap 1and ,, we get ,26 += 2 3 +1 = ,26 +1+= 23, wh\bch \bs larger, s\bnce both 6and ,are 2 or more. So the largest poss\bble value can’t have 1= 1. S\bm\blarly, the largest poss\bble value can’t have any of 6h =, or3equal to 1. Thus , = 1. Then we only need to cons\bder the follow\bng cases. 6 22 3+42 5+1 = 27 6 22 4+32 5+1 = 24 6 22 5+32 4+1 = 23 Therefore, the largest poss\bble value for the express\bon \bs 27, hence (B). 2016 =3, h enc(A)(.+4c( 59soc+9nt 20 2016 AMC – Intermediate Solutions

54 202016= 3,h enc (cA).c+cA6 =4 59 s9 o ta\b 3 tAc d 2 B9 w9 B2 B ta\b d Ac6(c:+)mc1p9 nca:c 2rh7 10 2016=3,1he6 n enc cul)1t+cAt1 +A)taJ1c E)+n :=AacA6 gB9 d ta\b w nt6 +nAcc taJ1c6 =4 TC 29 6= +nc 6)\bc 4A=. gB += w )6 TC 24A=. n=A)k=a+t17 enc 1)ac 4A=. D += , )6 t16= n=A)k=a+t19 6= +nc taJ1c Lc+Ecca +nc +E= 1)ac6 )6 TC 29 nca:c 2rh7 2016=3,1he6 c enc 1)ac gb8 )6 (tAt11c1 += gBbw9 ta\b 6= .t%c6 +nc 6t.c taJ1c E)+n +nc 1)ac ,bD9 \blc += +nc :=AAc6(=a\b)aJ taJ1c Al1c7 3)a:c L=+n gb8 ta\b ,bD (t66 +nA=lJn +nc :ca+Ac =4 +nc :1=:%9 +nc taJ1c Lc+Ecca +nc. )6 2 02 0 ,TC 2f TC 29 nca:c 2rh7 6=0 enc .t-).l. al.LcA =4 (ca6 v :ta +t%c E)+n=l+ ntm)aJ t+ 1ct6+ =ac (ca =4 ct:n :=1=lA )6 d I O f D7 en)6 =::lA6 )4 v +t%c +nc d Ac\b (ca6 ta\b +nc O pc11=E (ca67 3= v acc\b += +t%c gC (ca6 += Lc :cA+t)a +nt+ v ntmc t+ 1ct6+ =ac (ca =4 ct:n :=1=lA9 nca:c 2Sh7 660 2016= 3Th 2016=3,1he6 n enc 6l. =4 +nc c-+cA)=A taJ1c6 =4 +nc (ca+tJ=a )6 ,TC f DC I d 02gwC 12h9 6= +nt+ gwC 12f B8C 0d f T81 O ta\b2f gwC 1T81 OfggB 1O9 nca:c 2’h7 2016=3,1he6 c enc 6l. =4 +nc )a+cA)=A taJ1c6 =4 +nc (ca+tJ=a )6 , 0gwC f DC I d 29 6= +nt+2f dOC 6d f ggB1 O9 nca:c 2’h7 630 2016= 38h v46 )6 +nc (=)a+ \b)Ac:+1p Lc1=E =ta\b += +nc 1c4+ =4 39 +nca +nc A)Jn+ +A)taJ1c =3 6 nt6 6)\bc6 gT ta\b gB9 ta\b np(=+cal6c 27 enca2 2f gT 2I gB 2f dCC ta\b 2f BC9 nca:c 20h7 6,0 g B enc tAct =4 +nc 6c.):)A:1c )6 0 2, B 2fB , enc tAct =4 +nc :)A:1c )6 ,7 encAc4=Ac +nc tAct a=+ :=mcAc\b )6 B, 1,f ,9 nca:c 2’h7 6h0 yc %a=E B 01 f 2B 2h6fd 69 6= d 2=0 fd 6ta\b hI g f O7 enca hf d9 nca:c 2Sh7 20 2016 =3, h enc(A)(.+4c( 59soc+9nt 2016 AMC – Intermediate Solutions

5555 2012016= 3,hen c(A) 06 .+4 59so4t =a \b=d5.6 \b4t 4B45. d6 w :5m .+4 .=.:1 59so4t =a \b=d5.6 p:d54m d6 r 7 (( 7 A u hln .+4t4 s96. +:B4 o445 l 4B45.6J 06 E4..g +:6 (( \b=d5.6 6+4 s96. +:B4 e Tt6. \b1:C46 :5m =54 64C=5m \b1:C4J k:.+g C=91m 5=. +:B4 D=5 :5 4B45.n =t +4t 6C=t4 D=91m o4 w =t s=t4n 6= 6+4 s96. +:B4 L96. =54 64C=5m \b1:C4J b= 0mtd4554 s96. +:B4 .D= 64C=5m \b1:C46n +45C4 2k)J 261 8+4 59so4t 2016 2 Dd11 o4 .+4 1:tp46. 6%9:t4 a:C.=t =a hf(wJ -:C.=td6d5p d5.= \btds46n hf(w u h 22 e 02vuh 2v2 2h 02 e) 0n 6= .+4 1:tp46. 6%9:t4 a:C.=t =a hf(w d6 (h 0:5m .+45 2u (ln +45C4 20)J 2=1 05g=54 .411d5p .+4 .t9.+ d6 o4.D445 .D= 1d:t6I O 8 O 05g=54 1gd5p d6 5=. o4.D445 .D= 1d:t6S.+4g :t4 4d.+4t o4.D445 .D= .t9.+’.4114t6 =t =54 1d:t :5m =54 .t9.+’.4114tI 8 O 8 =t 8 O O 8 O 8 O O b= .+4t4 C:5y. o4 .D= .t9.+’.4114t6 d5 : t=D 5=t .+t44 1d:t6 d5 : t=DJ k=564%945.1gn :. 14:6. =54 \b4t6=5 d6 .411d5p .+4 .t9.+n :5m d6 o4.D445 .D= 1d:t6J 8+4 t4s:d5d5p .D= \b4=\b14 C:5y. o=.+ o4 1d:t6 5=t o=.+ .t9.+’ .4114t6n 6d5C4 .+45 .+4t4 D=91m o4 a=9t 1d:t6 =t .D= .t9.+’.4114t6 d5 : t=DJ b= .+4g s96. o4 =54 1d:t :5m =54 .t9.+’.4114tJ q5 :11n .+4t4 :t4 .+t44 1d:t6 :5m .D= .t9.+’.4114t6n +45C4 2k)J 231 = 6 3 8+4 1d%9dm d5 .+4 p1:66 d6 4%9:1 .= : Cg1d5m4t =a +4dp+. A Cs :5m +:1a =a : Cg1d5m4t =a +4dp+. w CsJ 8+96 0u12 e 02A7 1 2212 e 02w u lA1 7 hv1u vh1Cs 1 bd5C4 ( Cs 1u ( sOn .+4 :s=95. =a D:.4t d6 vh 1sOn +45C4 2x)J 2,1 2016= b(A) b9\b\b=64 .+:. 6d6 .+4 :B4t:p4 59so4t =a C=tt4C. :56D4t6 og .+4 64B45 6.9m45.6 D+=64 s:t”6 D4t45y. 1d6.4mJ 8+45 D4 ”5=D .+:. 6d6 :5 d5.4p4t :5m (f 06 0hfJ 8+4 :B4t:p4 59so4t =a C=tt4C. :56D4t6 og :11 .45 6.9m45.6 d6 r7r7•7v6 (fu hA7v 6 (f b= hA 7 v6 d6 mdBd6do14 og (fn D+dC+ d6 \b=66do14 =51g da 6d6 :5 =mm s91.d\b14 =a AJ –=D4B4tn (f 06 0hfn 6= .+:. 6u (AJ 8+4t4a=t4n .+4 :B4t:p4 59so4t =a C=tt4C. :56D4t6 og :11 .45 6.9m45.6 d6 hA7v2(A (fu (e +45C4 2x)J 2016 =3, h enc(A)(.+4c( 59soc+9nt 22 2016 AMC – Intermediate Solutions

56 20120 166 =3,henc(h0A h) c.3 +345eA9 c.3 e30sc. h) c.3 o=Ac t.5(0 c.5c +5AA3A h,3= c.3 s35=A (A 166 2a6 \b a666 e(0dA9 504 Ah c.3 1Bwchhc. s35= =3,he,3A a666 01B \b :66 c(m3Ap r.30 :66 2a: \b 7u66 e(0dA h) c.3 A3th04 t.5(0 +5AA h,3= c.3 s35=A9 504 Ah c.3 +nm+ s35= =hc5c3A 7u66 0u6 \b 176 c(m3A9 .30t3 lJEp 261 2016=3,1he6 n rgh c50s30cA )=hm 5 +h(0c ch 5 t(=te3 .5,3 3Tn5e e30sc.A9 Ah 2C 0 \bk 2 C 1 \b7 0 C 1 \b 1 72Ck 2\b 16 : 2 C: 0C: 1\b :u 2 C 0C 1 \b 1: 2 2 00 1 1 r.30 2\b :9 0\b 7 504 1\b up Dn3 ch c.3 =(s.c 50se39 c.3 =54(nA h) c.3 t(=te3 (A :9 .30t3 lLEp 2016=3,1he6 c r.3 c=(50se3 .5A A(43A 79 k 504 16 504 5=35 2 02 72 k \b :up 2c t50 b3 tnc (0ch c.=33 c=(50se3A h) 5=35A u 69a 6504 B6 5A A.hg0p 6 6 6 k16 7 r.30 1: 6\b :u 504 6\b :9 .30t3 lLEp 221 83=3 (A c.3 A3Tn30t3% = = = 3222 =222 , 222 h 222 e 222 2 A(0se3 Ac3+ 22 4hnbe3 Ac3+ fh,(0s )=hm h03 e3cc3= ch c.3 03-c t50 h0ev b3 4h03 5A (0 c.3 4(5s=5mp Ihn0c(0s c.3 A(0se3 Ac3+A 5A C1 504 c.3 4hnw be3 Ac3+A 5A C:9 50v t(=tn(c mnAc 544 ch 5 mnec(+e3 h) Bp Oh ch s3c )=hm =ch =(0 c.=33 Ac3+A l)h= c.3 o=Ac u e3cc3=A (0 c.3 A3Tn30t3E =3Tn(=3A h03 A(0se3 Ac3+ 504 cgh 4hnbe3 Ac3+A ch 544 ch Bp r.3=3 5=3 c.=33 g5vA h) 4h(0s c.(A9 1 C : C :9 : C 1 C : 504 : C : C 19 g.(t. 5=3 c.3 c=(50se3A =,e9=h e 9=h 3 p 20 2016 =3, h enc(A)(.+4c( 59soc+9nt 2016 AMC – Intermediate Solutions

5757 20 16= 3,0h2=0 2en c (=6A( )30, =.6 ,6h+enen1 46==6,(5 =.6 h94=eA46 03 s h9(= o6 ta\b +( =.6 =0=+4 e( += 46+(= c +nd += h0(= tBw 2.6n =.6,6 h9(= o6 B (=6A( =.+= +,6 :t +nd m (=6A( =.+= +,6 :Bw 2.6(6 p+n o6 4e(=6dr ttBBBB tBtBBB tBBtBB tBBBtB tBBBBt BttBBB BtBtBB BtBBtB BtBBBt BBttBB BBtBtB BBtBBt BBBttB BBBtBt BBBBtt 70 =.6,6 +,6 ts u+l( 03 p0hA46=en1 =.6 (6p0nd A+,= 03 =.6 (6J96np6w E4=6,n+=eg64l\b =.6 p0hoen+=0,ep 30,h94+ 2 2 00 T ts p+n o6 9(6dw Cn +44\b =.6 n9ho6, 03 A0((eo46 (6J96np6( e( k 2ts T ms\b .6np6 )D5w 201 7enp6 =.6l +,,eg6d .0h6 ta hen9=6( 6+,4e6, =.+n 9(9+4\b =.e( .+( =+L6n s hen9=6( 0b 6+p. de,6p=e0n =,+g6446d ol E4+nw 70 E4+n h9(= .+g6 h6= 8ln=.e+ += srBs Ah en(=6+d 03 srka Ahw E( 8ln=.e+ (=+,=6d u+4Len1 += sraa Ah\b (.6 .+( u+4L6d 30, Bs hen9=6(\b .6np6 )85w 261 )E4(0 %fB-\b vBI5 2.6,6 +,6 =.,66 p+(6( 30, .0u =.6 =,e+n146( +,6 p0409,6dr )e5 E44 =.,66 (ed6( +,6 =.6 (+h6 p0409,\b ue=. s A0((eoe4e=e6(w )ee5 2u0 (ed6( +,6 =.6 (+h6 +nd 0n6 (ed6 e( deb6,6n=\b ue=. s 2m T Ba A0((eoe4e=e6(w )eee5 E44 =.,66 (ed6( +,6 deb6,6n=\b ue=. 22 02 1 6 T ta A0((eoe4e=e6(w 70 =.6,6 +,6 s : Ba : ta T ks A0((eoe4e=e6( en +44\b .6np6 )E5w 2016= 2.6 p+4p94+=e0n en )eee5 ,64e6( 0n =.6 0o(6,g+=e0n =.+= e3 u6 +,6 p.00(en1 =.6 (ed6( en 0,d6,\b =.6,6 +,6 s A0((eoe4e=e6( 30, =.6 O,(= (ed6\b =.6n m 30, =.6 (6p0nd\b =.6n k 30, =.6 =.e,dw S0u6g6,\b en =.6(6 s 2m2 k T ca A0((eoe4e=e6(\b 6+p. (646p=e0n 016 ue44 +AA6+, c =eh6(r 016= 061= 106= 160= 601= 610 w 2.e( ed6+ +AA6+,( en =.6 16n6,+4 30,h94+ 30, 2 2 00 \b =.6 n9ho6, 03 u+l( 03 p.00(en1 30o ’6p=( 3,0h ,0o ’6p=(w 2=1 E (9A6,yqeo0n+ppe (6J96np6 e( d6=6,hen6d 6n=e,64l ol e=( O,(= =u0 =6,h(\b t +nd 0\b (+lw 2.6 (6J96np6 =.6n A,0p66d( +( 30440u(r t= 0= )t :05=B)t : 05=m)t : 05=x)t : 05= hhh 7enp6 +dden1 +44 A,6ge09( =6,h( +h09n=( =0 d09o4en1 =.6 4+(= =6,h\b +44 =6,h( 3,0h =.6 =.e,d 0nu+,d( +,6 03 =.6 30,h B 1)t : 05\b 30, (0h6 e0 aw C3 Batc T B 12ck u6,6 0n6 03 =.6(6 =6,h(\b =.6n eue44 o6 0n6 03 =.6 c g+496( eTa =hhh= s\b 1egen1 c A0((eo46 g+496( 30, 0w 2.6(6 +,6 0T cB =tBs =Bst =sak =taaI =Batsw 2.6,6 e( +4(0 =.6 A0((eoe4e=l =.+= 0T Batc\b (0 =.6,6 +,6 I (9p. (6J96np6( en =0=+4\b .6np6 )D5w 231 ”+p. 03 =.6 (6g6n (h+446, =,e+n146( .+( +,6+ nT =3,6 h T Bxx ph 0TB 2tB 0w 2.6 ,e1.= e(0(p646( =,e+n146 1c(A e( h+d6 9A 03 m 03 =.6(6\b (0 e=( +,6+ e( mn Tx2tB 0T 602mx 0 +nd e=( =u0 6J9+4 (ed6( +,6 cATc( T mx phw 2.6 +,6+ 03 1c). e( = 1 =.6 +,6+ 03 1c)A \b (0 =.+= c.T = 1cAT kB phw 2.6n en 1c.+ \b 46=0T c+ \b =.6n =.6 +,6+ e( Bxx T , =2kB0\b (0 =.+= 0T =ee ,6 T tx ph\b .6np6 )tx5w 2016 =3, h enc(A)(.+4c( 59soc+9nt 20 2016 AMC – Intermediate Solutions

58 201201 6=3,, 6he n1 c03(A0c 3) h, h .h=A1 ,+(h=1 45c0 )3(= 1+(h. ,9h.. ,+(h=1, =193s1o )=39 5c, 63=e1=,t a1c c01 .h=A1 ,+(h=1 0hs1 ,5o1 2heo c01 ,9h..1= ,+(h=1, ,5o1 0t 2 0 \b=1h d 2 22B0 2dw 2:m 0pw2 2m0 p d mr7u l1=591c1= d B2 J3 2:m 0heo 22 m0 h=1 )h6c3=, 3) mr7u c0hc o5E1= ng B0 t T1 4hec 2c3 n1 h, ,9h.. h, C3,,5n.1t 205, 45.. 366(= 401e 05, h, ,9h.. h, C3,,5n.1k ,3 c01 c43 )h6c3=, 3) mr7u h=1 h, 6.3,1 c3A1c01= h, C3,,5n.1t 201 6.3,1,c c43 h=1 Bm heo BDk n(c c05, o31, e3c A5s1 he 5ec1A1= sh.(1 )3= 0t L1bc n1,c 5, 8u heo %uk 40560 A5s1, 2:m 0d %u12 2m0 d 8uk ,3 c0hc m 2d fm heo c01 C1=591c1= 5, 7DBk 01e61 w7DBpt 261 -16h(,1 c01 91he 3) c01 7r ,63=1, 5e6=1h,1o ng rt% h)c1= c01 95,chv1 4h, 63==16c1ok c015= ,(9 5e6=1h,1o ng %t L3 3c01= ,63=1 60heA1ok ,3 Ih.63.9O, ,63=1 5e6=1h,1o ng %t 201 91o5he 3) c01 7r ,63=1, 43(.o e3c 0hs1 60heA1o 5) Ih.63.9O, =1s5,1o ,63=1 0ho n11e Df 3= .341= 3= 5) 05, 3=5A5eh. ,63=1 0ho n11e fm 3= 05A01=k ,3 05, 3=5A5eh. ,63=1 9(,c 0hs1 n11e n1c411e D% heo f7k 5e6.(,5s1t 201 C3,,5n5.5c51, 6he n1 ,(99h=5,1o h, )3..34,S 3=5A5eh. ,63=1 =1s5,1o ,63=1 3=5A5eh. 91o5he =1s5,1o 91o5he D% fr Dft% fr Duk D’k DDk Df f7k fmk f8k fB Dft% 93=1 c0he fr fr f% fr f7 f7 fu frt% f7 ye.g 5e c01 q=,c heo .h,c 6h,1, o31, c01 91o5he 5e6=1h,1 ng rt%t 201 ,(9 3) Ih.63.9O, c43 C3,,5n.1 63==16c ,63=1, 5, fr : fu d 7Duk 01e61 w7Dupt 2=1 w\b.,3 Jmup a1c c01 )3=9hc53e 0hs1 6=34, heo =63.(9e,k ,3 c01 ,5x1 3) c01 nheo 5, 6=t 201 nheo 63ech5e, 8 6n3g, heo % =A5=.,k ,3 c01 ,5x1 3) c01 nheo 5, h.,3 86 :%=t 201e 6=286 2 %= d rk 3= 1+(5sh.1ec.gk w6 2%pw= 28p d 7%t 201 C3,5c5s1 5ec1A1= ,3.(c53e, )3= c01 3=o1=1o Ch5= w 62 %1= 28p h=1 w71 7%pk w81%pk w% 18pk heo w7%1 7pt 201=1 h=1 e1Ahc5s1 5ec1A1= )h6c3=5,hc53e, 3) 7%k n(c c01,1 0hs1 6 0 r 3= =0 rt 201 63==1,C3eo5eA ,3.(c53e, )3= w61 = p h=1 wu17Dpk wD1Dpk w7r 1upk heo wmr1 Bpk heo c01 63==1,C3eo5eA sh.(1, 3) 6=h=1 7rDk uBk urk heo Drt ”3= 1h60 3) c01,1 ,5x1,k c01=1 5, hc .1h,c 3e1 h==heA191ec 3) n3g, heo A5=.,S 20 2016 =3, h enc(A)(.+4c( 59soc+9nt 2016 AMC – Intermediate Solutions

5959 220 02 16 2 26 026 2 2 1= 020 2 2 3= 02, 2016 = 3, hen1nc31n (en A). 3c +66 43AA0=6n =+59 A0snA 0A ota \b dB \b dt \b at w :omp en5rn 7:omul 201 7J6A3 Emgu 2016=3,1he6 n T1+C (en dBk235 +59 +66 :t 90+235+6A 4+1+66n6 (3 + DLn9 A09np 90b09052 0( 05(3 :o (1+4ns0).Al 85 n+re (1+4ns0).p 91+C =3(e 90+235+6Al he0A 1n%)01nA dB \b :t \b m2 :o w ofd re319Al E3 (en .+L0.). 5).=n1 3c re319A 0A ofd 31 .31nl 85 c+r(p (en .+L0.). 0A ofdp +A 0A Ae3C5 =n63Cl -01A(6,p c31 24305(A 35 + r01r6np Cen1n 20A nbn5p (en A+.n +12).n5( (n66A )A (e+( 0c 0 10A (en .+L0.). 5).=n1 3c re319Ap (en5 0 102\b 2 21 m\bm2 7 2 21ou w 0 221 Bl c0,h(A -31 +66 nbn5 b+6)nA 3c 2p0 1w 0 221 Bl v6n+16, Cen5 2w Bp 0 1wdw 0 22B1 Bp A3 (en r6+0. 0A (1)nl -31 + 90+21+. C0(e + re319 1635 (en =3)59+1,p r35A09n1 (en c3663C052 (C3 43AA0=6n A(n4AI 70u On(Cnn5 1+59 6p +99 + 4305( =+59 (C3 re319A 1=+59 6=l 700u On(Cnn5 1+59 6p +99 (C3 4305(A =3 ,+59 Dbn re319A 1 =3 1 ,3 6=3 6,3 =, l 1 6 1 6 = 1 6 = , 70u 700u E(n4 700u 20bnA .31n re319A 4n1 4305(p A3 + r35A(1)r(035 (e+( A(+1(A C0(e (en Bk4305( 90+21+. +=3bn +59 =)069A )4 )A052 A(n4A 70u +59 700u C066 e+bn (en 21n+(nA( 5).=n1 3c re319A 0c 0( )AnA A(n4 700u +A .)re +A 43AA0=6nl -31 +5 nbn5 5).=n1 3c 4305(Ap 0( C066 356, )An A(n4 700ul E)re + 90+21+. C066 e+bn d \b f 2 1 272 1 Bu w 0 221 B re319Al hen 356, )5rn1(+05(, 0A Cen(en1 + 90+21+. C0(e (en .+L0.). 5).=n1 3c re319A r+5 =n =)06( c13. (en 2w B 90+21+. )A052 356, A(n4A 70u +59 700ul E)443An (en .+L0.). 5).=n1 3c re319A +1n 91+C5l J5, re319 05 (en 90+21+. 0A n0(en1 +5 n92n 3c (en 3)(n1 2k235p + r13AAn9 re319 31 +5 )5r13AAn9 re319l hen )5r13AAn9 re319A 90b09n (en 2k235 05(3 A.+66n1 436,235Al 8c (enAn Cn1n (10+526nA +59 %)+9106+(n1+6A (en5 (en 2k235 r3)69 =n =)06( )4 )A052 A(n4A 70u +59 700up +99052 35n 436,235 +( + (0.nl 2016 =3, h enc(A)(.+4c( 59soc+9nt 20 2016 AMC – Intermediate Solutions

60 2 0 1 6 2 2016=3,h 6enc ( A0).+01 24=5c ,(nc 6=3, (1 e1nh066,3 nc0h3 (13 4=5c 9 0h s0h, 6=3,6o t0s, 3=(+01() 21se65 a, ( nc0h3 =1 5c, 3=(+h(s\b 6=1n, 5c, 3=(+h(s c(6 5c, s(d=ses 1esa,h 0B nc0h36o wc,1 21se65 a, nh066,3 a. (105c,h nc0h3 60\b 0h ,)6, 5c, e1nh066,3 nc0h3 2140e)3 c(:, 6A)=5 2=150 6s()),h A0).+016o m1. nc0h3 A(66=1+ 5ch0e+c (1. ,3+, 0B pe(3h=)(5,h() 261 040e)3 nh066 210h60 \b 4c=nc 40e)3 5c,1 a, 30ea),rnh066,3\b 4c=nc =6 B0ha=33,1o t0 ,:,h. ,3+, 0B 261 0nh066,6 10 nc0h36\b 60 =5 se65 a, =1 5c, 3=(+h(s 73e, 50 s(d=s()=5.u 4c,h, =5 4=)) a, (1 e1nh066,3 nc0h3o l04\b 6=1n, 2c(6 9 0h s0h, 6=3,6\b 261 00J2 60 (5 ),(65 01, 6=3, 0B 261 0\b 6(. 26 \b =6 105 ( 6=3, 0B 2o wc,1 26=6 (1 e1nh066,3 nc0h3 5c(5 6A)=56 2=150 6s()),h A0).+016\b 4c=nc n(1105 c(AA,1o E,1n, 6enc ( A0).+01 2se65 c(:, g 0h B,4,h ,3+,6o T1 n01n)e6=01\b (1. 3=(+h(s 4=5c 5c, s(d=ses 1esa,h 0B nc0h36 n(1 a, ae=)5 Bh0s 5c, grA0=15 3=(+h(s e6=1+ 65,A6 7=u (13 7==uo Cc,1 ==6 ,:,1\b 5c, s065 nc0h36 (h, 0a5(=1,3 e6=1+ 01). 65,A 7==u\b 4c=nc +=:,6 3 2J 2 0=1 go 2016,pe,15). 3 20 J kDL1g J k9D\b c,1n, 7k9Duo 2016=3,1he6 n m6 =1 5c, bh65 60)e5=01\b k9D nc0h36 (h, A066=a),o w0 6,, 5c(5 5c=6 =6 5c, s065\b 4, bh65 105, ( 4,))r81041 h,6e)5\b 81041 (6 1=h,3c(0,1hA3 A) , .A0+cA3 % Cc,1 ( A0).+01 4=5c =6=3,6 7=r+01u =6 ne5 =150 5h=(1+),6\b 4c,h, ,(nc 5h=(1+),f6 :,h5=n,6 (h, :,h5=n,6 0B 5c, 0h=+=1() =r+01\b 5c,h, (h,=1 - 5h=(1+),6o m)60\b 5c,h, (h, =1 v ne56\b ,(nc ()01+ ( 3=(+01() 0B 5c, =r+01o m bh65 n016,pe,1n, 0B 5h=(1+e)(5=01 =6 5c(5 5c, s(d=ses 1esa,h 0B 101r=15,h6,n5=1+ 3=(+01()6 5c(5 n(1 a, 3h(41 =16=3, (1 =r+01 =6=1 vo t,n013).\b B0h 5h=(1+),6 4c06, :,h5=n,6 (h, :,h5=n,6 0B 5c, 0h=+=1() =r+01\b 5c, s(d=ses 1esa,h 0B 101r0:,h)(AA=1+ 5h=(1+),6 5c(5 n(1 a, 3h(41 =16=3, 5c, =r+01 =6=1 -o wc=6 =6 a,n(e6, 4, n(1 (33 s0h, 5h=(1+),6 50 +,5 ( 5h=(1+e)(5=01o wc=h3).\b B0h pe(3h=)(5,h()6 4c06, :,h5=n,6 (h, :,h5=n,6 0B 5c, 0h=+=1() =r+01\b 5c, s(dr =ses 1esa,h 0B 101r0:,h)(AA=1+ pe(3h=)(5,h()6 5c(5 n(1 a, 3h(41 =16=3, 5c, =r+01 =6 2 1 0 0 J 2 0 1 ko wc=6 =6 a,n(e6, ,(nc pe(3h=)(5,h() n(1 a, 6A)=5 =150 540 101r 0:,h)(AA=1+ 5h=(1+),6o , h e n c ( A ) I,5eh1=1+ 50 5c, pe,65=01\b B0h ,:,h. A(=h 0B nh066r =1+ nc0h36 ,e(13 hn\b 6c(3, =1 5c, pe(3h=)(5,h() ,h e n o O0h 540 6c(3,3 pe(3h=)(5,h()6 ,h e n(13 c ( A) \b 1,=5c,h 3=(+01() ,e0hhn =15,h6,n56 cA 0h () \b 60 ,h e n (13c ( A) 30 105 0:,h)(Ao wc(5 =6\b 5c, 6c(3,3 pe(3h=)(5,h()6 (h, 101r 0:,h)(AA=1+\b (13 60 5c,h, (h, (5 s065 2 0 1 kJvk 0B 5c,so wce6 5c,h, (h, (5 s065 vk A(=h6 0B nh066=1+ nc0h36o O0h ,(nc A(=h 0B nh066=1+ nc0h36\b h,s0:, 01, nc0h3o wc,h, (h, (5 s065 vk h,s0:,3 nc0h36o wc, nc0h36 h,s(=1=1+ c(:, 10 nh066=1+6\b (13 (h, ,=5c,h 6=3,6 0B 5c, Dgr+01 7(5 s065 Dg 0B 5c,6,u 0h 3=(+01()6 0B 5c, Dgr+01 7(5 s065 Dk 0B 5c,6,uo 2016,pe,15). 5c, 1esa,h 0B nc0h36 0h=+=1()). 4(6 (5 s065 Dg S Dk S vk J k9D\b c,1n, 7k9Duo 20 2016 =3, h enc(A)(.+4c( 59soc+9nt 2016 AMC – Intermediate Solutions

6161 2016=3,1he6 n 20 16 =3, he0= 0nc(=1n6A ).+ 43ne50 496 s, 5e9o6t an 0,, =39= ).+ 10 =3, \b9d1\b(\bA 0(BBn0, =3, \b9d1\b(\b 6(\bs,e nw 43ne50 9e, 5e9o6: 6n \bne, 496 s, 955,5t m6 B9e=14(c9eA 96p Bn001sc, 43ne5 =39= 16=,e0,4=0 6n n=3,e 5e9o6 43ne5 \b(0= s, 5e9o6t a310 164c(5,0 9cc ,5r,0 nw =3, e,r(c9e +7urn6t m= 9c0n \b,960 =39= =3, n6cp Bncprn60 =39= 39l, 9cc l,e=14,0 n6 =3, 41e4c, 965 6n 43ne50 16015, 9e, =e196rc,0A 0164, n=3,eo10, 9 519rn69c 4n(c5 s, 5e9o6t 2 0 1 6 = J3,6 =on nw =3, 43ne50 21965 06 16=,e0,4= 9= 96 16=,e1ne Bn16= = A =3,e, 9e, 6n n=3,e 43ne50 16=,e0,4=16r 21965 06A 0n 6n n=3,e 43ne5 o1cc B900 16015, =3, E(95e1c9=,e9c 20 1 6t gn60,E(,6=cpA =3, 43ne50 20A01 A16 965625n 6n= 16=,e0,4= 96p n=3,e 43ne50A 0n =3,p \b(0= s, 164c(5,5t a3,6 20 1 69BB,9e0 16 =3, 519re9\b 90 9 c=(AA6) .+,)=h0,16=,0 T 9cc 015,0 965 sn=3 519rn69c0 9e, 5e9o6t Cn =39= o, 496 (0, k(c,eD0 wne\b(c9 3L ,b hL 8A o, 4n6015,e =3, hr(e, 90 9 Bc969e re9B3 o3,e, =3, l,e=14,0 164c(5, =3, +7 ne1r169c Bn16=0 965 =3, 16=,e0,4=1n6 Bn16=0A 965 =3, ,5r,0 164c(5, =3, 43ne50 =39= 9e,6D= 4(= sp 96n=3,e 43ne5 965 =3, =on B9e=0 nw =3, 43ne50 =39= 9e, 4(= sp 96n=3,e 43ne5t a3,e, 9e, =3e,, =pB,0 nw w94,0T =3, ,d=,e1ne nw =3, +7urn6A =e196rc,0 =39= 9e, B9e= nw 9 4en00,5 E(95e1c9=,e9cA 965 =e196rc,0 =39= 39l, 9cc l,e=14,0 n6 =3, 41e4c,t C(BBn0, =3,e, 9e, e4en00,5 E(95e1c9=,e9c0 965 n=e196rc,0t a3,6 2 a3, 6(\bs,e nw l,e=14,0 10 ,b +7 L eA =3, +7 161=19c l,e=14,0 Bc(0 n6, wne ,943 4en00,5 E(95e1c9=,e9ct 2 a3, 6(\bs,e nw w94,0 10 3b )L nL7 eA =3, n(=015, nw =3, +7urn6A =3, n=e196rc,0A 965 7 =e196rc,0 wne ,943 4en00,5 E(95e1c9=,e9ct 2 a3, 6(\bs,e nw l,e=14,0 10 ho3,e, 8h b +7L%f nL7 e-t a310 =n=9c 10 wen\b 95516r =3, 6(\bs,e nw ,5r,0 n6 ,943 w94,A o3143 4n(6=0 ,943 ,5r, =o14,t 2 a3, 6(\bs,e nw 43ne50 10 cb h0 8e b %8 L 2 0nL7 eA 0164, wne ,943 4en00,5 E(95e1c9=,e9c =3, 6(\bs,e nw ,5r,0 10 8 \bne, =396 =3, 6(\bs,e nw 43ne50t a3,6 16 k(c,eD0 wne\b(c9 3L ,b hL 8T vbf 3L ,- 0 fh L 8- b f+. L nL. e- 0 f%7 L 2 0nL+ e- b %) 0 1 0n0 e c b %8 L 2 0nL 7f%) 0 1 0n- b ).+ 0 1 0n I,64, =3, \b9d1\b(\b 6(\bs,e nw 43ne50 5e9o6 10 ).+A 9==916,5 o3,6 eb %) 965 nb vA 3,64, f).+-t 2016 =3, h enc(A)(.+4c( 59soc+9nt 20 2016 AMC – Intermediate Solutions