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AMC PRACTICE QUESTIONS AND SOLUTIONS Senior Copyright � 2014, 2019 Australian Mathematics Trust AMTT Limited ACN 083 950 341 (Set1)                                \b     
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Due to the right-angled triangleP QS , Pythagoras’ theorem gives QS= 4\b Then  QRS is isosceles, so its altitude RTbisects QS\b R S x P 5 Q 3 T 2 2 30 ◦ 60 ◦ || || Now,S RT is standard 30 ◦,60 ◦,90 ◦triangle with RT:RS :ST =1:2: √ 3 so that x= RS = 2√3ST= 4√3= 4√ 3 3 , hence (D)\b Comment This problem can also be solved using trigonometry: x= 2 cos 30 ◦=4 √3. 4. 2014 S20 Given that f 1(x)= x x +1 and f n+1 (x)= f 1(f n(x)), then f 2014 (x) equals (A) x 2014 x+1 (B) 2014 x 2014x+1 (C) x x + 2014 (D) 2014 x x+1 (E) x 2014(x + 1) Alternative 1 f 2( x)= f x x+1  = xx+1xx+1 +1 = x x + x+1 = x 2 x +1 f 3( x)= x2x+1x2x+1 +1 = x x +2 x+1 = x 3 x +1 and in general, by induction f n( x)= x nx +1 = ⇒ f n+1 (x)= xnx +1xnx+1 +1 = x x + nx +1 = x ( n + 1)x +1, so f 2014 (x)= x 2014 x+1 , hence (A)\b Alternative 2 Consider 1 fn( x) \b 1 f1( x) = 1+ 1 x= ⇒ 1 fn +1 (x) = f 1(f n(x)) = 1 + 1 fn( x) = ⇒ 1 f2014 ( x) = 1+ 1 f2013 ( x) = 2+ 1 f2012 ( x) = ∙∙∙ ∙∙∙ = 2013 + 1 f1( x) = 2014 + 1 x= 2014 x+1 x Hence f 2014 (x)= x 2014x +1 , hence (A)\b AMC Practice Questions and Solutions — Senior

5.2014 S25 The sequence 2,2 2,2 22,2 222 ,... is defined by a 1= 2 anda n+1 =2 an for all n≥ 1. Wha\b is \bhe firs\b \berm in \bhe sequence grea\ber \bhan 1000 1000 ? (A) a 4=2 222 (B) a 5=2 2222 (C) a 6=2 22222 (D) a 7=2 222222 (E) a 8=2 2222222 We w a n \b a n>1000 1000 = 10 3000 . We know \bha\b a 1= 2,a 2=2 2= 4,a 3=2 4= 16 and a 4=2 16 = 65536, all less \bhan 10 3000 . Also 2 10 = 1024 >10 3, so \bha\b we can es\bima\be a 5, a 5=2 65536 = (2 10)6553 26> (10 3)6553 26= 64 ×10 19659 This is grea\ber \bhan 10 3000 , hence (B). 6. 2014 S26 Wha\b is \bhe larges\b \bhree-digi\b number wi\bh \bhe proper\by \bha\b \bhe number is equal \bo \bhe sum of i\bs hundreds digi\b, \bhe square of i\bs \bens digi\b and \bhe cube of i\bs uni\bs digi\b? Alternative 1 Le\b \bhe number be abc. Then 100a+ 10b +c = a+ b 2+c 3 99a+ 10b −b 2 = c(c 2− 1) 99a +b(10 −b)=( c− 1)c( c+ 1) Consider \bhe possibili\bies: 99 a b(10 −b) (c − 1)c( c+ 1) 99 ×1=99 1× 9=9 1× 2× 3=6 99 ×2=198 2× 8=16 2× 3× 4=24 99 ×3=297 3× 7=21 3× 4× 5=60 99 ×4=396 4× 6=24 4× 5× 6=120 99 ×5=495 5× 5=25 5× 6× 7=210 99 ×6=594 6× 4=24 6× 7× 8=336 99 ×7=693 7× 3=21 7× 8× 9=504 99 ×8=792 8× 2=16 8× 9× 10 = 720 99 ×9=891 9× 1=9 Looking a\b \bhe possibili\bies for 99 a+ b(10 −b)=( c− 1)c( c+ 1), we have \bwo: 99 + 21 = 120 = ⇒a=1 ,b=3or7 ,c=5= ⇒n= 135 or n= 175. 495 + 9 = 504 = ⇒a=5 ,b=1or9 ,c=8= ⇒n= 518 or n= 598. So, \bhere are four 3-digi\b numbers which sa\bisfy \bhe requiremen\bs and \bhe larges\b of \bhese four numbers is 598, hence (598). AMC Practice Questions and Solutions — Senior

Alternative 2 The numb er abcis equal to a+ b 2+c 3, and these are the possible values of b 2andc 3: Di\bit 
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  •• •t i  lt2t ernati2 ve e vve \biv i ve\b lt i\bt rt2  cfirst, then fillin\b in possible values for aand b. This trial-and-error search is presented here as a tree. a: a b 2: c3: + ab c A 729 a b9 • • • • b 9 8 2 729 8 b9 8+2+9 = 19 (no b 2) 7 72 9 7b 9 8 3 729 8 b9 7+3+9 = 19 (no b 2) a 512 ab8 6 51 2 6 b 8 6 00 5 12 608 6+0+2 =8 5 51 2 5b 8  5 81 512 598  5 01 512 518 5+1+2 =8 a 343 ab 7 abc < 598 Likewise for c =6 ,5 ,... The lar\best solution found is 598, and any solutions on branches c= 7, c= 6, ..., c = 1 must be less than this, hence (598). AMC Practice Questions and Solutions — Senior ATh ATh ATh ATh ATh ATh ATh ATh ATh ATh