2015 Fermat Contest (Grade 11)

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1 The CENTRE for EDUCATION i MATHEMATICS ad COMPUTING cemc.uwaterloo.ca 015 Fermat Cotet (Grade 11) Tueday, February 4, 015 (i North America ad South America) Wededay, February 5, 015 (outide of North America ad South America) Solutio 014 Uiverity of Waterloo

2 015 Fermat Cotet Solutio Page The average of the five give umber i = = 10. Alteratively, ice there are a odd umber of coecutive iteger, the average i the middle umber, which i 10. Awer: (E). Evaluatig, = = 10 5 =. Awer: (A) 3. Suppoe that the ditace betwee coecutive poit o the path i d. The, walkig from P to U, Emily walk a total ditace of 5d. Walkig back from U to P, he alo walk a ditace of 5d, for a total ditace walked of 10d. Sice 70% of 10 i 7, the he ha completed 70% of her walk after walkig a ditace of 7d (that i, after havig walked 7 egmet). She ha walked a ditace of 7d after walkig a ditace of d back from U, which get her to S. Awer: (D) 4. Evaluatig, (x 3) = ( 3 3) = ( 6) = 36. Awer: (B) 5. Baed o the diagram how, the order of the vertice of the y rectagle mut be P QRS. (I ay other cofiguratio, either QRP or QP R would be a agle i the rectagle, which i ot Q (3, 1) poible.) R (7, 1) Sice P ad Q have the ame x-coordiate, the ide P Q of the x rectagle i vertical. Thi mea that ide SR mut alo be vertical, ad o the P (3, ) x-coordiate of S i the ame a the x-coordiate of R, which i 7. Sice Q ad R have the ame y-coordiate, the ide QR of the rectagle i horizotal. Thi mea that ide P S mut alo be horizotal, ad o the y-coordiate of S i the ame a the y-coordiate of P, which i. Thu, the coordiate of S are (7, ). Awer: (B) 6. Sice MNP Q i a rectagle, the NMQ i a right agle ad thu o i NMZ. The um of the agle i NMZ i 180, o NZM = 180 ZNM ZMN = = N 68 X P Z M Q 55 Y Sice the um of the agle i ZXY i alo 180, the Y XZ = 180 XZY XY Z = = 103 Awer: (E)

3 015 Fermat Cotet Solutio Page 3 7. Violet tart with oe-half of the moey that he eeded to buy the ecklace. After her iter give her moey, he ha three-quarter of the amout that he eed. Thi mea that her iter gave her 3 1 = 1 of the total amout that he eed. 4 4 Sice he ow ha three-quarter of the amout that he eed, the he till eed oe-quarter of the total cot. I other word, her father will give her the ame amout that her iter gave her, or $30. Awer: (D) 8. Sice 15 = 5 ad 15 = 3 5, the 5 = 15 = (3 5) = 3 5. Therefore, x = ad y =, o x + y = 4. Awer: (B) 9. Solutio 1 The two team iclude a total of = 44 player. There are exactly 36 tudet who are at leat oe team. Thu, there are = 8 tudet who are couted twice. Therefore, there are 8 tudet who play both baeball ad hockey. Solutio Suppoe that there are x tudet who play both baeball ad hockey. Sice there are 5 tudet who play baeball, the 5 x of thee play baeball ad ot hockey. Sice there are 19 tudet who play hockey, the 19 x of thee play hockey ad ot baeball. Sice 36 tudet play either baeball or hockey or both, the (5 x) + (19 x) + x = 36 (The left ide i the um of the umber of thoe who play baeball ad ot hockey, thoe who play hockey ad ot baeball, ad thoe who play both.) Therefore, 44 x = 36 ad o x = = 8. Thu, 8 tudet play both baeball ad hockey. Awer: (B) 10. Sice Bruce drove 00 km at a peed of 50 km/h, thi took him 00 = 4 hour. 50 Aca drove the ame 00 km at a peed of 60 km/h with a top omewhere alog the way. Sice Aca drove 00 km at a peed of 60 km/h, the time that the drivig portio of her trip took wa 00 = 3 1 hour The legth of Aca top i the differece i drivig time, or = hour. 3 3 Sice hour equal 40 miute, the Aca top for 40 miute. 3 Awer: (A) 11. For each of the three digit of uch a poitive iteger, there are three choice of digit (7, 8 or 9). Therefore, there are = 7 poible iteger that ue o digit other tha 7, 8 or 9. (We ote that there are 9 uch iteger begiig with each of 7, 8 ad 9. The 9 uch iteger begiig with 7 are 777, 778, 779, 787, 788, 789, 797, 798, 799.) Awer: (E)

4 015 Fermat Cotet Solutio Page 4 1. Sice co 60 = 1 ad co 45 = 1, the the give equatio co 60 = co 45 co θ become 1 = 1 co θ. Therefore, co θ = = 1. Sice 0 θ 90, the θ = 45. Awer: (D) 13. We make a table of the total amout of moey that each of Steve ad Waye have at the ed of each year. After the year 000, each etry i Steve colum i foud by doublig the previou etry ad each etry i Waye colum i foud by dividig the previou etry by. We top whe the etry i Steve colum i larger tha that i Waye colum: Year Steve Waye 000 $100 $ $00 $ $400 $ $800 $ $1600 $65 Therefore, 004 i the firt time at which Steve ha more moey tha Waye at the ed of the year. Awer: (C) 14. Solutio 1 Sice P QRS i a quare, the it diagoal SQ cut it ito two equal area. Therefore, the ratio of the area of P QS to the area of quare P QRS i 1 :. P QS ca be viewed a havig bae P S ad height P Q. MQS ca be viewed a havig bae MS ad height P Q. (Thi i becaue P Q i perpedicular to the lie cotaiig MS.) Sice MS = 1 P S, the the area of MQS i oe-half of the area of P QS. Sice the ratio of the area of P QS to the area of quare P QRS i 1 :, the the ratio of the area of QMS to the area of quare P QRS i 1 : 4. Solutio Suppoe that the ide legth of quare P QRS i a. The the area of quare P QRS i (a) = 4a. Sice M i the midpoit of ide P S, the P M = MS = a. The QMS ca be ee a havig bae MS ad height P Q. (Thi i becaue P Q i perpedicular to the lie cotaiig MS.) Sice M S = a ad P Q = a, the the area of QM S i 1 (MS)(P Q) = 1a(a) = a. Therefore, the ratio of the area of QMS to the area of quare P QRS i a : 4a which equal 1 : 4. P a M a S a Q a a R Awer: (B)

5 015 Fermat Cotet Solutio Page Solutio 1 Zolta awered 45 quetio. If all of hi awer had bee correct, hi core would have bee 45(4) = 180 poit. Sice hi core wa 135 poit, the he lot = 45 poit. For each icorrect awer, he lot 5 poit compared to a correct awer, ice a correct awer add 4 poit ad a icorrect awer ubtract 1 poit. Therefore, Zolta had 45 5 = 9 icorrect awer. (We ca check that 36 correct, 9 icorrect, ad 5 uawered give a core of 4(36) 1(9)+0(5) or 135 poit.) Solutio Suppoe that Zolta awered x quetio icorrectly. Sice he awered 45 quetio i total, the he awered 45 x quetio correctly. Sice the tet icluded 50 quetio ad he awered 45, the he did ot awer 5 quetio. Uig the markig cheme, hi core wa 4(45 x) 1(x) + 0(5). We are told that hi core wa 135 poit. Hece, 4(45 x) 1(x) + 0(5) = 135 ad o 180 4x x = 135. Thu, 5x = 45 or x = 9. Therefore, Zolta awered 9 quetio icorrectly. (We ca check that 36 correct, 9 icorrect, ad 5 uawered give a core of 4(36) 1(9)+0(5) or 135 poit.) Awer: (A) 16. Sice P ( 4, 0) ad Q(16, 0) are edpoit of a diameter of the emi-circle, the the legth of the diameter i 16 ( 4) = 0. Sice a diameter of the emi-circle ha legth 0, the the radiu of the emi-circle i 1 (0) = 10. Alo, the cetre C i the midpoit of diameter P Q ad o ha coordiate ( 1 ( ), 1 (0 + 0)) or (6, 0). y R (0, t) P ( 4, 0) C (6, 0) Q (16, 0) x Now the ditace betwee C(6, 0) ad R(0, t) i 10, ice CR i a radiu. Therefore, (6 0) + (0 t) = t = 100 t = 64 (Alteratively, we could have oted that if O i the origi, the ROC i right-agled with RO = t, RC = 10 ad OC = 6 ad the ued the Pythagorea Theorem to obtai t +6 = 10, which give t = 64.) Sice t > 0, the t = 8. Awer: (C)

6 015 Fermat Cotet Solutio Page Sice a + b = 3, the a + b = 3(a b) or a + b = 3a 3b. a b Thu, 4b = a ad o b = a or = a b. (Note that b 0, ice otherwie the origial equatio would become a a = 3, which i ot true.) Awer: (D) 18. The equatio x + kx + 7k 10 = 0 ha two equal real root preciely whe the dicrimiat of thi quadratic equatio equal 0. The dicrimiat,, equal = (k) 4(1)(7k 10) = 4k 8k + 40 For the dicrimiat to equal 0, we have 4k 8k+40 = 0 or k 7k+10 = 0 or (k )(k 5) = 0. Thu, k = or k = 5. We check that each of thee value give a equatio with the deired property. Whe k =, the equatio i x + 4x + 4 = 0 which i equivalet to (x + ) = 0 ad o oly ha oe olutio for x. Whe k = 5, the equatio i x + 10x + 5 = 0 which i equivalet to (x + 5) = 0 ad o oly ha oe olutio for x. The um of thee value of k i + 5 = 7. Awer: (E) 19. Suppoe the lope of the three parallel lie i m. The equatio of a lie with lope m ad y-itercept i y = mx +. To fid the x-itercept i term of m, we et y = 0 ad olve for x. Doig thi, we obtai mx + = 0 or x = m. Similarly, the lie with lope m ad y-itercept 3 ha x-itercept 3 m. Alo, the lie with lope m ad y-itercept 4 ha x-itercept 4 m. ( Sice the um of the x-itercept of thee lie i 36, the ) + m ( 3 ) ( + 4 ) = 36. m m Multiplyig both ide by m, we obtai 3 4 = 36m ad o 36m = 9 or m = 1 4. Awer: (E) 0. Firt, we factor a a 015 a a 014 (1 + a). If a = 5 or a = 10, the the factor a 014 i a multiple of 5, o the origial expreio i diviible by 5. If a = 4 or a = 9, the the factor (1 + a) i a multiple of 5, o the origial expreio i diviible by 5. If a = 1,, 3, 6, 7, 8, the either a 014 or (1 + a) i a multiple of 5. Sice either factor i a multiple of 5, which i a prime umber, the the product a 014 (1 + a) i ot diviible by 5. Therefore, there are four iteger a i the rage 1 a 10 for which a a 015 i diviible by 5. Awer: (C)

7 015 Fermat Cotet Solutio Page 7 1. If Amia wi, he ca wi o her firt tur, o her ecod tur, or o her third tur. If he wi o her firt tur, the he wet firt ad toed tail. Thi occur with probability 1. If he wi o her ecod tur, the he toed head, the Bert toed head, the Amia toed tail. Thi give the equece HHT. The probability of thi equece of toe occurrig i = 1. (Note that there i oly oe poible equece of T ad H for which Amia 8 wi o her ecod tur, ad the probability of a pecific to o ay tur i 1.) Similarly, if Amia wi o her third tur, the the equece of toe that mut have occurred i HHHHT, which ha probability ( ) 1 5 = 1. 3 Therefore, the probability that Amia wi i = = 1 3. Awer: (A). Sice a, b ad c form a arithmetic equece i thi order, the a = b d ad c = b + d for ome real umber d. We ote that d 0, ice otherwie we would have a = b = c ad the abc = would tell u that b 3 = or b = , which i ot a iteger. Writig the term of the geometric equece i term of b ad d, we have 3a+b = 3(b d)+b = 4b 3d 3b+c = 3b+(b+d) = 4b+d 3c+a = 3(b+d)+(b d) = 4b+d Sice 3a + b, 3b + c ad 3c + a form a geometric equece i thi order, the 3b + c = 3c + a 3a + b 3b + c (3b + c) = (3a + b)(3c + a) (4b + d) = (4b 3d)(4b + d) 16b + 8bd + d = 16b 4bd 6d 1bd = 7d 1b = 7d (ice d 0) d = 1 7 b Therefore, a = b d = b ( 1b) = 19b ad c = b + d = b + ( 1b) = 5b Sice abc = , the ( 19b)(b)( 5 95 b) = or b3 = or b 3 = 961 ad o b = 1. Thu, a = 19b = 19( 1) = 57 ad c = 5b = 5 ( 1) = We ca check that a = 57, b = 1 ad c = 15 have a product of , that 57, 1, 15 i ideed a arithmetic equece (with commo differece 36), ad that 3a + b = 19, 3b + c = 48, ad 3c + a = 1 form a geometric equece (with commo ratio 1). 4 Therefore, a + b + c = ( 57) + ( 1) + 15 = 63. Awer: (A)

8 015 Fermat Cotet Solutio Page 8 3. Startig from the give equatio, we obtai the equivalet equatio 5x 4xy + x + y = 64 5x 4xy + x + y + 1 = 65 4x 4xy + y + x + x + 1 = 65 (x y) + (x + 1) = 65 Note that 65 = 5. Sice x ad y are both iteger, the the left ide of the give equatio i the um of two perfect quare. Sice ay perfect quare i o-egative, the each of thee perfect quare i at mot 65 = 5. The perfect quare from 0 to 5 are: 0, 1, 4, 9, 16, 5, 36, 49, 64, 81, 100, 11, 144, 169, 196, 5, 56, 89, 34, 361, 400, 441, 484, 59, 576, 65 The pair of perfect quare from thi lit that have a um of 65 are 65 = = = (We ca verify thi relatively quickly by checkig the differece betwee 65 ad each of the perfect quare i the lit from 34 to 65 to ee if thi differece i itelf a perfect quare. We do ot eed to check aythig le tha 34, ice 65 i odd ad o oe of the two quare addig to 65 mut be larger tha the other ad hece larger tha half of 65.) Therefore, (x y) ad (x + 1) equal 5 ad 0 i ome order, or 4 ad 7 i ome order, or 0 ad 15 i ome order. Thu, x y ad x + 1 equal ±5 ad 0 i ome order, or ±4 ad ±7 i ome order, or ±0 ad ±15 i ome order. Sice x 0, the x+1 1, o we eed to coider the poibilitie that x+1 = 5, 4, 7, 0, 15: If x + 1 = 5, the x = 4. If x y = 0 ad x = 4, the y = 48. If x + 1 = 4, the x = 3. If x y = 7 ad x = 3, the y = 39; if x y = 7 ad x = 3, the y = 53. If x + 1 = 7, the x = 6. If x y = 4 ad x = 6, the y = 1; if x y = 4 ad x = 6, the y = 36. If x + 1 = 0, the x = 19. If x y = 15 ad x = 19, the y = 3; if x y = 15 ad x = 19, the y = 53. If x + 1 = 15, the x = 14. If x y = 0 ad x = 14, the y = 8; if x y = 0 ad x = 14, the y = 48. From thi lit, the pair of o-egative iteger (x, y) that atify the coditio 0 x y are (x, y) = (4, 48), (3, 39), (3, 53), (6, 36), (19, 3), (19, 53), (14, 48). There are 7 uch pair. (We ca check by direct ubtitutio that each pair atifie the origial equatio.) Awer: (E)

9 015 Fermat Cotet Solutio Page 9 4. Let r be the radiu of the lower circle. We label the quare a ABCD, the cetre of the upper circle a U, ad the cetre of the lower circle a L. We call E the poit at which the upper circle touche the top lie, G the poit at which the lower circle touche the bottom lie, F the poit at which the two circle touch, ad H the poit at which the lower circle touche the quare. We joi EU, UF, F L, LG, LH, ad UA. We eed to ue two fact about circle: Whe the cetre of a circle i joied to a poit of tagecy, the reultig lie egmet i perpedicular to the taget lie. Thu, UE i perpedicular to the top lie, LG i perpedicular to the bottom lie, ad LH i perpedicular to AD. Whe two circle are taget, the lie egmet joiig their cetre pae through the poit of tagecy betwee the circle. Thu, UF L i a traight lie egmet. Exted EU dowward to meet LH at J. Sice EU ad AD are perpedicular to the parallel lie ad LH i perpedicular to AD, the EJ i perpedicular to LH. E Exted BA to meet EJ at K. Uig a imilar argumet, AK U A K B i perpedicular to EJ. We focu o UJL, which i right-agled at J. F Sice the radiu of the upper circle i 65, the we kow that L H EU = UA = UF = 65. J Sice the radiu of the lower circle i r, the we kow that F L = LH = LG = r. G D C Thu, UL = UF + F L = 65 + r = r Sice the top ad bottom lie are parallel, EJ ad LG are perpedicular to thee lie, ad LJ i parallel to thee lie, the EU + UJ + LG equal the ditace betwee the lie. Thu, 65 + UJ + r = 400 ad o UJ = 335 r. Next, we ote that LJ = LH JH = r JH = r AK, ice AKJH i a rectagle. Now UKA i right-agled ad ha UA = 65. Alo, UK = EK EU = EK 65. But EK equal the differece betwee the ditace betwee the lie ad the ide legth of the quare, or = 11. Thu, UK = = 56. Uig the Pythagorea Theorem i UKA, we obtai AK = UA UK = ad o AK = 1089 = 33. Sice AK > 0, the AK = 33. Thu, LJ = r 33. Fially, uig the Pythagorea Theorem i U JL, we obtai UJ + LJ = UL (335 r) + (r 33) = (r + 65) r 670r r 66r + 33 = r + 130r + 65 r 866r = 0 By the quadratic formula, r = 866 ± 866 4(1)(109089) 866 ± 560 = = 153 or 713. Sice r mut be le tha the ditace betwee the two lie, which i 400, the r = 153. Of the give awer, thi i cloet to 153. Awer: (C)

10 015 Fermat Cotet Solutio Page Thi olutio i writte i a mathematically complete a way a poible. Of coure, while doig a multiple choice cotet, thoe who attempt ad/or complete thi problem would be ulikely to work through all of thee detail. Step 1: Uig iformatio about decimal equivalet to rewrite fractio Coider a real umber x whoe decimal equivalet i of the form 0.g 1 g... g p r 1 r... r q for ome iteger p 0 ad q > 0 ad digit g 1, g,..., g p, r 1, r,..., r q. (Note that if p = 0, the 0.g 1 g... g p r 1 r... r q = 0.r 1 r... r q.) c The x = for ome poitive iteger c. 10 p (10 q 1) We demotrate thi i a pecific example ad leave the full algebraic derivatio to the ed of thi olutio. If x = , the x = x = x 1 = (10 x 1) = (10 x 1) 745 = (10 x 1) 745 = 10 x (10 x 1) (10 x 1) = 745 (10 3 1)(10 x 1) = x 1 = x = x = (10 3 1) x = (103 1) (10 3 1) Suppoe that a fractio m ha m ad poitive iteger with m <. The 0 < m < 1. Suppoe ow that a fractio m of poitive iteger with 0 < m < 1 ha the property that there i a equece of coecutive digit of legth 6 i it decimal equivalet that repeat coecutively ad idefiitely. That i, uppoe that m = 0.g 1g... g p r 1 r... r 6 for ome iteger p 0 ad digit g 1, g,..., g p, r 1, r, r 3, r 4, r 5, r 6. From above, m = c 10 p for ome poitive iteger c. (Note that = ) Step : Uig further coditio give to aalyze Cotiuig from above, c = 10 p m. Sice we are alo told that m i i lowet term, the m ad have o commo divior larger tha 1, ad o mut be a divior of 10 p Note that 10 p = p 5 p = p 5 p (3 3 37) (11 91) = p 5 p Sice i a divior of p 5 p , the caot cotai ay prime factor other tha, 3, 5, 7, 11, 13, 37. Sice i ot diviible by the quare of ay poitive iteger, the it caot be diviible by the

11 015 Fermat Cotet Solutio Page 11 quare of ay prime umber. Thu, mut be a divior of = Step 3: Coolidatig curret progre We ow kow that ay fractio m m ad are poitive iteger with m <, atifyig the propertie m i i lowet term, i ot diviible by the quare of ay iteger larger tha 1, ad the decimal equivalet of m iclude a equece of coecutive digit of legth 6 that repeat coecutively ad idefiitely, ca be writte i the form m = for ome poitive iteger with (We ote that we have ot yet determied whether the hortet equece of coecutive digit that repeat coecutively ad idefiitely ha legth 6.) Step 4: Every fractio with ca be writte a a fractio atifyig thee four bullet Each i betwee 0 ad 1, ca be writte i lowet term ad ha deomiator ot diviible by the quare of ay poitive iteger larger tha 1, o ay equivalet fractio i lower (or lowet term) hare thi property a factor of the deomiator will oly be removed i reducig, ot added. Furthermore, = = 1 ( z ) y for ome o-egative iteger y ad z with 0 y < 10 ad 0 z < (y ad z are the quotiet ad remaider, repectively, whe 9 i divided by ) Writig z = r 1 r r 3 r 4 r 5 r 6 for ome digit r 1, r, r 3, r 4, r 5, r 6 (ome or all poibly 0), the = 1 ( y + r 1r r 3 r 4 r 5 r ) 6 = (y.r 1r r 3 r 4 r 5 r 6 ) = 0.yr 1 r r 3 r 4 r 5 r 6 o every ca be writte a a decimal with a repeatig equece of legth Alo, each i differet ad o will produce a differet m. Therefore, the umber of uch fractio (which i ) will equal the umber of fractio m that atify the four bullet above. Notice that we have ot yet checked to ee if the equece of digit of legth 6 i the hortet uch equece. Step 5: Coiderig horter poible legth Sice 6 i to be the legth of the hortet equece of repeatig digit after the decimal poit, the there ca be o equece of repeatig digit of legth 1,, 3, 4, or 5. Uig a imilar approach to the firt derivatio above, we ee that m thu caot be writte

12 015 Fermat Cotet Solutio Page 1 c i ay of the form 10 p 9 or c 10 p 99 or c 10 p 999 or c 10 p 9999 or c 10 p Uig a imilar approach to the aalyi of prime factor above, we ee that m caot be t writte i ay of the form 5 3 = t 30 or t = t 330 or t = t 1110 or t = t or t = t It i poible that a m with propertie a above icludig a repeatig equece of legth 6 i it decimal equivalet ca alo be writte with a repeatig equece of legth 1, or 3. Thi i becaue, for example, 0.r 1 = 0.r 1 r 1 r 1 r 1 r 1 r 1 0.r 1 r = 0.r 1 r r 1 r r 1 r 0.r 1 r r 3 = 0.r 1 r r 3 r 1 r r 3 which all have equece of legth 6 that repeat. It i ot poible for a m with a repeatig equece of legth 6 i it decimal equivalet to alo be writte with a repeatig equece of legth 4 or 5, without a repeatig equece of legth 1 or. Thi i becaue if, for example, there i a repeatig equece of legth 4, the = = t for ome poitive iteger t ad o = t ad o 101 = t which tell u that 101 i divior of t ad o = t = t 330 for ome poitive iteger t. Therefore, the decimal equivalet to = t ha a equece of legth that repeat, o 330 ay fractio with a decimal equivalet that ha a repeatig equece of legth 4 will be dealt with amog thoe with equece of legth 1, or 3. I a imilar way, we ca rule out decimal equivalet with equece of legth 5 that repeat. Therefore, we eed to (carefully) remove fractio from our cout that have decimal equivalet with equece of legth 1, or 3 that repeat. Step 6: Coiderig overlap We have fractio m to coider, each of which atifie the four bullet above. Sice the decimal equivalet of m caot have equece of legth 1, or 3 that repeat, the m u caot be writte i ay of the form 1110 or v 330 or w for poitive iteger u, v ad w 30 with u < 1110, v < 330, ad w < 30. Let U be the et of the 1009 fractio of the form u, V be the et of the 39 fractio of 1110 the form v 330, ad W be the et of the 9 fractio of the form w 30. Ay fractio of the form w i alo of the form u (ice w 30 = 37w ) ad i alo of the form 1110 v 330 (ice w 30 = 11w 330 ). Therefore, every fractio i W i alo i U ad i V. I et otatio W U ad W V. Furthermore, ay fractio that i i both U ad V i alo i W :

13 015 Fermat Cotet Solutio Page 13 u Suppoe that a fractio ca be writte i both form 1110 ad v 330. u The 1110 = v 330 or u 37 = v ad o 11u = 37v. 11 Sice 37v i a thu a multiple of 11 ad 37 i ot diviible by the prime umber 11, the v i a multiple of 11. v Thi mea that 330 = 11f 330 = f for ome poitive iteger f, ad o i i W. 30 I et otatio, U V W. Sice W U ad W V ad U V W, the U V = W ; i other word, the et of fractio i each of U ad V i preciely the et of fractio W. Step 7: Fial coutig We tart with fractio, a above, ad wat to remove all of the fractio i U, V ad W. Sice each fractio i W i i U ad V, it i eough to remove thoe U ad V oly. The total umber of fractio i U ad V (that i, i U V ) equal the umber of fractio i U plu the umber of fractio i V miu the umber of fractio i their overlap (that i, i U V = W ). Thi i becaue ay fractio i the overlap i couted twice whe iclude all fractio i U ad all fractio i V. Therefore, we eed to remove fractio from the et of Therefore, F, the umber of fractio havig the deired propertie, i F = ( ) = Sice F ha 7 digit, the G = F + 7 = The um of the quare of the digit of G i = = 181. Step 8: Geeral algebraic derivatio from Step 1 Coider a real umber x whoe decimal equivalet i of the form 0.g 1 g... g p r 1 r... r q for ome iteger p 0 ad q > 0 ad digit g 1, g,..., g p, r 1, r,..., r q. The x = 0.g 1 g... g p r 1 r... r q 10 p x = g 1 g... g p.r 1 r... r q 10 p x g 1 g... g p = 0.r 1 r... r q 10 q (10 p x g 1 g... g p ) = r 1 r... r q.r 1 r... r q 10 q (10 p x g 1 g... g p ) r 1 r... r q = 0.r 1 r... r q 10 q (10 p x g 1 g... g p ) r 1 r... r q = 10 p x g 1 g... g p 10 q (10 p x g 1 g... g p ) (10 p x g 1 g... g p ) = r 1 r... r q (10 q 1)(10 p x g 1 g... g p ) = r 1 r... r q 10 p x g 1 g... g p = r 1r... r q 10 q 1 10 p x = g 1 g... g p + r 1r... r q 10 q 1 x = g 1g... g p + r 1r... r q 10 p 10 p (10 q 1) x = (10q 1)g 1 g... g p + r 1 r... r q 10 p (10 q 1) Awer: (E)

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST. First Round For all Colorado Students Grades 7-12 November 3, 2007

UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Roud For all Colorado Studets Grades 7- November, 7 The positive itegers are,,, 4, 5, 6, 7, 8, 9,,,,. The Pythagorea Theorem says that a + b =