USA Mathematical Talent Search PROBLEMS / SOLUTIONS / COMMENTS Round 4 - Year 13 - Academic Year solutions edited by Erin Schram

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1 USA Mthemticl Tlet Serch PROBLEMS / SOLUTIONS / COMMENTS Roud - Yer 13 - Acdemic Yer solutios edited y Eri Schrm 1//13. I strge lguge there re oly two letters, d, d it is postulted tht the letter is word. Furthermore, ll dditiol words re formed ccordig to the followig rules: A. Give y word, ew word c e formed from it y ddig t the righthd ed. B. If i y word sequece ppers, ew word c e formed y replcig the y the letter. C. If i y word the sequece ppers, ew word c e formed y omittig. D. Give y word, ew word c e formed y writig dow the sequece tht costitutes the give word twice. For exmple, y (D), is word, d y (D) gi, is word. Hece y (B) is word, d y (A) is lso word. Agi, y (A), is word, d so y (D), is lso word. Filly, y (C) we fid tht is word. Prove tht i this lguge is ot word. Commet: Our Prolem Editor, Professor George Berzseyi proposed this prolem. It is similr to prolem metioed i Mthemticl Chlleges II, pulished y the Scottish Mthemticl Coucil i Solutio 1 for 1//13 y Je Le (12/MN): (B) d (D) re the oly rules tht ffect the umer of times the letter ppers i your word. Suppose you hve word i which the ltter ppers times. If you pply rule (B), you will get ew word i which the letter ppers 3 times. If you pply rule (D) isted, you will get ew word i which the letter ppers 2 times. Uless is multiple of three, either 3 or 2 c e multiple of 3, y Lemms 1 d 2 elow. Therefore, uless you strt out with word i which the umer of times the letter ppers is multiple of three, you will ever e le to form word i which the letter ppers multiple of three times. You strt out with the word, which cotis oe letter, d oe is ot multiple of three. Therefore, you will ever e le to form the word, sice cotis the letter six times, d six is multiple of three. QED Lemm 1. If 3 divides 3, the 3 divides. Suppose 3 ( 3 ). The there exists iteger x such tht 3 = 3x. If follows tht = 3x + 3 = 3( x + 1 ), so 3. Lemm 2. If 3 divides 2, the 3 divides. Suppose 3 ( 2). The the Fudmetl Theorem of Arithmetic implies tht, sice 3 does ot divide 2 d 3 is prime, 3.

2 Solutio 2 for 1//13 y Muro Brustei (12/FL) We hve here xiomtic system with xiom d rules of iferece A, B, C, d D: AXIOM: is word. RULES: A. If x is word, so is x. B. If xy is word, so is xy. C. If xy is word, so is xy. D. If x is word, so is xx. I the ove rules, x d y std for strigs of letters, which could e the empty strig. Notice tht is ot postulted s word. We will show you tht it is t word t ll, d eve its dditio to the set of words would still keep from eig word. We will focus o the umer of s for our prolem, so let x ( ) e the umer of times occurs i x. For exmple, ( ) = 6. Rule A dds to word, so the umer of s clerly remis uchged. Rule B replces three s y, so the umer of s decreses y three. Rule C simply removes three s, so it lso leves the umer of s uchged. Rule D duplictes the word, therefore clerly doulig the umer of s. If we let = ( x), ( A( x)) = B( ( x)) = 3 C( ( x)) = D( ( x)) = 2 ( ) = 1, d from there we wt to crete cse where ( x) = 6. Wht order of multiplyig y two d sutrctig three will tke 1 to 6? Noe. Cosider (mod 3). Sutrctig 3 i rule B will leve (mod 3) uchged. Look t doulig. We re give 1; (mod 3), d (mod 3), so doulig will yield repetitive cycle: 1, 2, 1, 2,. However, 6 0 (mod 3), d o mout of doulig will yield 0 from 1. Sice ( ) = 6, it is impossile to costruct it from strig m such tht ( m ) = 1, prticulrly m =. Likewise, ( ) = 0, so cot e costructed either. If were ccepted s xiom, o rule would llow icrese i its s. A d C keep the umer of s fixed t 0, D doules the umer of s from 0 to 0, d B cot e used ecuse it eeds t lest three s, impossile requiremet sice we strted with zero s. Therefore, the word would ot e possile eve uder this ew system.

3 2//13. Let f( x) = x x x x for ll positive rel umers x, where y deotes the gretest iteger less th or equl to y. (1) Determie x so tht f( x) = (2) Prove tht f( x) = 2002 hs o solutio. Commet: This prolem ws suggested y Professor Bel Bjok of Corell Uiversity. It is dpted from Chech-Slovki prolem give few yers go. Solutio 1 to 2//13 y Dvid Brmore (10/IL) First, we will determie x such tht f( x ) = For x 1, f( x) is strictly icresig. Sice f (6.9965) = < 2001 d f (6.9966) = > 2001, if f( x) = 2001 f( x) the < x < Also, sice x = y defiitio of f( x), x is rtiol for x x x 2001 ll itegrl vlues of f( x). For our prolem, x = For < x < , x x x x x x = 286, so x could oly e 2001/286, which checks. Now, we will prove tht o x exists such tht f( x) = First, if there were some umer x such tht f( x) = 2002, the x = Clerly, sice f ( ) = 2001 < 2002 d x x x f ( 7 ) = 201 > 2002 d f( x) is strictly icresig o this itervl, < x < 7. For x o this itervl, x = 6, x x = 6x = 1 sice > -----, x x x = 1x = 286 sice > So x could oly e 2002/286 = 7. However, f ( 7) Sice we hve rrived t cotrdictio, our suppositio tht there exists x such tht f( x) = 2002 must e flse. Therefore, there does ot exist x such tht f( x) = Solutio 2 to 2//13 y Agusty Meht (10/OH) Prt 1. It is esy to see tht f( x) icreses s x icreses, ecuse x x x either remis the sme or icreses whe x icreses. We see tht f ( 6 ) = 6 = 1296 d f ( 7 ) = 7 = 201, so if f( x) = x x x x = 2001, the 6 < x < 7. Let m = x x x. m is iteger y the defiitio of. So, f( x) = x m = 2001 or x = 2001 m. Sice x < 7, we hve 286 m (Iequlity 1). Also sice 6 < x < 7, x = 6 x x < 2 x x 1 (Iequlity 2) x x x < 287 x x x 286

4 So y oth the iequlities, 286 m 286 ; thus, m = 286. Thus x = Prt 2. As we hve show i prt 1, f ( 6 ) = 6 = 1296, f ( 7) = 7 = 201, d 1296 < 2002 < 201. So if f( x) = 2002 hs solutio x, it must stisfy 6 < x < 7. This is the sme coditio tht gve us Iequlity 2 i prt 1: x x x 286. So x x x x < Thus, f( x) = 2002 hs o solutio. Solutio 3 y Diel McLury (11/OK) x must e rtiol umer, sice we re proposig to multiply it y iteger d get iteger result, so let x = + - -, where,, d re itegers d 0 < <. The we kow tht = 6, sice if we plug i x-vlue less th 6, we ll get umer less th 6 = 1296, d if x is 7 or more, we get 7 = 201 or more. Strt from iside the ested floor rckets. Oviously, rcket, we get 36 + ( 6) (-- ). Sice 0 < -- < 1, we kow tht 0 < ( 6) (-- ) < 6, so tht it ecomes iteger etwee 0 d 5. 5 Suppose ( 6) (-- ) is, which mes -- < - -. We would get 0 for 36 + ( 6) (-- ) d the 6 multiplyig y x would give 20 +( 0) (-- ). The floor of tht hs mximum vlue of 273, sice 5 -- < --. Now 2001 is more th 7 times 273, so oviously this is too smll. Therefore, y cotrdic 6 tio, we kow tht ( 6) ( -- ) = 5. x = = So we hve tht x x = 1, which mes tht x x x = 26 + ( 1) (-- ). Now 2001 divided y 7 is just less th 286, so ythig smller th tht wo t work. Sice - - < 1, ( 1) (-- ) < 1, so ( 1) (-- ) 0. But we eed 286, so it must e t lest 0. So it is 0. So filly, x x x x is simply ( )(286) = (286) (-- ). We c set = 286, d reduce -- lter if ecessry. We get = 2001, givig = 285. So our fil swer 285 is Lettig x e ythig etwee d 7 would still give f( x) = 286x, which could ot 286 e iteger result. Ad f ( 7) = 7 is much lrger th 2002, so it is impossile to get I do t like this prolem for some reso. There just seems to e wy too much rute force 6. At the ext floor

5 285 rithmetic. Ay prolem with swer like just seems to e severely cotrived d lcks 286 y rel euty. Commet y Eri Schrm o Mr. McLury s lst oservtio: We select our prolems for solutios of pproprite difficulty tht require mthemticl uderstdig. We would love to hve euty i every prolem, ut give the limited pool of prolems tht we c eg, orrow, or ivet, sometimes we hve to mke do without euty. However, our prolems re ot cotrived i the sese of deliertely selectig ugly swers to dd more difficulty. We wt the mthemtics of the prolems to e s old d cler s possile. 3//13. Let f e fuctio defied o the set of ll itegers, d ssume tht it stisfies the followig properties: A. f ( 0) 0 ; B. f ( 1) = 3 ; d C. f( x ) f( y ) = f( x + y ) + f( x y ) for ll itegers x d y. Determie f ( 7). Commet: This prolem is sed o similr prolem used i 1987 for the selectio of Hugry s IMO tem, d ws proposed y Prof. Berzseyi. Solutio 1 for 3//13 y George Khchtry (11/TX) f ( 1 ) f ( 0 ) = f ( 1 ) + f ( 1 ) 3 f ( 0 ) = f ( 0) = 2 f ( 1 ) f ( 1 ) = f ( 2 ) + f ( 0) 3 3 = f ( 2 ) + 2 f ( 2) = 7 f ( 2 ) f ( 1 ) = f ( 3 ) + f ( 1) 7 3 = f ( 3 ) + 3 f ( 3) = 18 f ( 3 ) f ( 1 ) = f ( ) + f ( 2 ) 18 3 = f ( ) + 7 f ( ) = 7 f ( ) f ( 3 ) = f ( 7 ) + f ( 1 ) 7 18 = f ( 7 ) + 3 f ( 7) = 83 so f ( 7) = 83. However, wht is iterestig out f( ) is tht it hs closed form. The sequece we sw ctully is specil. Over the positive itegers, f( 1 ) f ( 1 ) = f( ) + f( 2 ), which implies f( 1 ) 3 = f( ) + f( 2 ) d f( ) = 3 f( 1 ) f( 2 ). This is the recursive defiitio for the eve isectio of the Lucs umers, L ( ). The Lucs umer L ( ) is defied s L( 0 ) = 2, L( 1 ) = 1, d L ( ) = L ( 1 ) + L( 2 ). The sequece is: 2, 1, 3,, 7, 11,18, 29, 7, 76, 123,. (c.f. the Fiocci umers). The eve isectio of the Lucs umers, which re our f( ), hs closed form: f( ) = Settig = 7 ideed yields

6 //13. A certi compy hs fulty telephoe system tht sometimes trsposes pir of djcet digits whe someoe dils three-digit extesio. Hece cll to x318 would rig t either x318, x138, or x381, while cll received t x0 would e iteded for either x0 or x0. Rther th replce the system, the compy is ddig computer to deduce which diled extesios re i error d revert those umers to their correct form. They hve to leve out severl possile extesios for this to work. Wht is the gretest umer of threedigit extesios the compy c ssig uder this pl? Commet: This prolem ws proposed y Dr. Peter Aspch of the Ntiol Security Agecy. Solutio 1 for //13 y Tmr Broderick (11/OH) How my three-digit extesios c e ssiged so tht the computer c deduce the correct extesio every time? For extesios of the form : All possile outputs of the fulty system: Therefore, every extesio of the form (10 totl) my cotiue to e used s it will ot e cofused with y other extesio. For extesios of the form : Possile extesios of form Possile outputs of the system,,,, Sice there re o two extesios tht hve o shred possile outputs, ll of these extesios could e cofused with the others, d oly oe c, thus, e used y the computer system. Therefore, 1/3 of ll extesios of this form c e ssiged. For extesios of the form c: Possile extesios of form c c c c c c Possile outputs of the system c, c, c c, c, c c, c, c c, c, c c, c, c c c, c, c Ech extesio c e cofused with two other extesios. Therefore, the mximum umer of extesios tht my e verified s uique y the ew computer out of 6 such extesios is 2. For exmple, c d c shre o possile outputs i commo. Thus, 2/6 or 1/3 of these extesios c e used y the computer.

7 Sice 1/3 of ll forms esides form re usle, 1/3 of their remiig 990 totl three-digit extesios (fter sutrctig ll 10 of form ) re usle. Therefore, the gretest umer of ssigle extesios is = 30. Solutio 2 for //13 y Dvid Glkowski (10/NY) 1. The possile extesio umers c e divided ito 3 groups: extesio umers with three differet digits, extesio umers with two differet digits with oe repeted, d extesio umers with oe digit used three times. 2. Extesio umers with three differet digits re the lrgest group. There re 10 choose 3 wys to pick three differet digits. Ech set of three digits hs two possile extesios tht will lwys produce differet umers whe grled. For exmple, if the digits 3, 2, d 1 were chose, we could chose x123 d x321. Whe x123 ws diled, the computer would receive 123, 213, or 132. Whe x321 ws diled, the computer would receive 321, 312, or 231. Sice either shre umers, yet ll possile comitios of 3, 2, d 1 re used, there re two possile extesios for ech extesio umer with three differet digits. the umer 10 of extesio umers with three differet digits is 2 = The umer of extesio umers with two differet digits, oe repeted oce, is There re 10 choices for the umer tht repets d 9 choices for differet umer. If two 3 s d 1 were chose for extesio, oly oe umer could e used, sice ll three possile extesios could shre possile outcome of 313. x133 could chge to 313 or 133. x313 could chge to 133, 313, or 331. x331 could chge to 313 or 331. The umer of extesio umers with two differet digits is 10 9 = 90.. The umer of extesio umers with 1 digit repeted twice more is 10 sice there re oly 10 digits d there is oly oe possile extesio for ech digit. 5. The gretest umer of three-digit extesios the compy c ssig is = 30 extesios. 5//13. Determie the smllest umer of squres ito which oe c dissect rectgle, d exhiit such dissectio. The squres eed ot e of differet sizes, their ses should e itegers, d they should ot overlp. Commet: We thk Prof. Berzseyi for proposig this iterestig prolem. For excellet itroductio to the topic of dissectig rectgles ito squres, the reder is referred to Chpter 17 of Scietific Americ s Secod Book of Mthemticl Puzzles d Diversios y Mrti Grder. Commet y Eri Schrm: To mthemtici, determie mes fid d verify. Provig tht six is the smllest umer of squres i dissectio is hlf the work i the prolem, d solutio would e icomplete without such verifictio.

8 Solutio 1 for 5//13 y Pul H. Ryu (10/CA) This figures shows dissectio with six squres. Of the six squres, two re of size, d the other four squres re of sizes 1 1, 5 5, 6 6, d 7 7. Sice the dissectio is vlid, we oly hve to verify tht there re o possile dissectios with fewer th 6 squres. To do tht, we hve to first figure out the set of whole umers whose squres sum up to the re of the rectgle, which is 13. With the id of computer progrm, oe c fid sets of whole umers whose squres sum up to the re of the rectgle. There re o sets of two or three whole umers whose squres sum to 13. There re five sets of four whole umers whose squres sum to 13. They re {1, 5, 6, 9}, {2, 3, 3, 11}, {2, 3, 7, 9}, {3, 3, 5, 10}, d {3, 6, 7, 7}. However, oe of them ctully work, ecuse the squres do ot fit together. Tht is ecuse the comied legth of the two lrgest squres exceeds 13; thus, two of them cot e lid o the rectgle without exceedig the oudry. The two squres hve to e lid dow somehow, ut if two of them hve sides summig greter th 13, their sides opposite ech other will e more th 13 prt. Now we fid sets of five whole umers whose squres sum to 13. There re eight of them: {1, 1, 2,, 11}, {1, 1,, 5, 10}, {1, 2, 5, 7, 8}, {1,3,, 6, 9}, {2,, 5, 7, 7}, {2, 5, 5, 5, 8}, {3, 3, 3,, 10}, d {3, 3, 5, 6, 8}. Noe c ctully e rrged ito rectgle. The squres simply do ot fit. As efore, the sum of the lrgest two squres exceeds 13, except with the {2, 5, 5, 5, 8} cse, i which the 5 5 squres cot fill the gps left y the 8 8 squre. Thus, we see tht our dissectio, which uses the set {1,,, 5, 6, 7}, is optiml, d the smllest umer of squres ito which oe c dissect rectgle is six squres. Solutio 2 for 5//13 y Xu Wu (12/CA) Testig for smllest umer of squres: 1? No. Biggest squre tht will fit i is 11 11, which does ot cover the remiig ? No. Oly cofigurtio of two squres tht mke rectgle is the figure to the right. The legth hs to e 2 times loger th the width, L = 2W. However, 13 is ot 2 11, so rectgle cot e formed y 2 squres. 3? No. Two cofigurtios: L = 3W 2L = 3W 2 2

9 ? No. Five cofigurtios: L = W 3 3L = W L = W 2 3L = 5W L = 5W Cosidered to e the sme s the cse to the left; ytime the sme squres c e rerrged to form the sme dimesios, it will e cosidered the sme cofigurtio. 5? No. Eleve cofigurtios: L = 5W L = 2W L = 6W 2L = 7W 3L = 7W 3L = 8W

10 L = 5W 3 L = 7W L = 7W 3 2 5L = 7W L = 8W ? Yes. Oe cofigurtio, show elow, works. The smllest umer of squres ito which oe c dissect rectgle is L = 13W 5 5

11 Solutio 3 for 5//13 y Roert Cordwell (9/NM) This is the solutio with six squres It is reltively esy to prove tht t lest six squres re eeded. We kow tht there must e t lest oe squre t ech corer for miimum solutio, sice we kow tht solutio with 6 squres exists d tht solutio with oe squre of size 11 coverig two corers tkes 8 squres. We prove tht there must e 6 squres y showig tht solutio with five or fewer squres is ot possile Notice tht ll the sides of the rectgles must e covered completely y the sides of the squres, d ech corer squre of legth x covers x uits o oe of the rectgle s sides of legth 13 d x uits o oe of the rectgle s sides of legth 11. It should e firly cler tht with five or fewer squres t lest three sides of the rectgle must e covered solely y the corer squres, icludig t lest oe 13-log side. Let those two squres o the log side hve sides of legths d. I order to hve t most five squres, the other two corer squres must hve legths of 11 d 11, which sum up to 9, levig gp of o the other side. Thus, there must e squre of side legth (it could e doe with more squres, ut with fives squres we c use oly oe o-corer squre) o 13-log side. We must ow try to plce the four corer squres so tht they cover the rest of the rectgle. Notice tht squre opposite the squre of legth must hve side legth 7 to void plcig sixth squre etwee them. 11 = 7 11

12 This i tur mes tht we hve squres o the two djcet corers of side legths d Filly, this mes tht there must e squre with side legth 5 i the lst corer However, whe oe dds up the res of ll five lrge squres: , oe fids tht it is 12, oe short of the re of the rectgle, = 13. Thus, there is o solutio with five squres. By ddig the oe missig squre, we hve isted foud our solutio with six squres.