Australian Intermediate Mathematics Olympiad 2017

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1 Austalian Intemediate Mathematics Olympiad 207 Questions. The numbe x is when witten in base b, but it is 22 when witten in base b 2. What is x in base 0? [2 maks] 2. A tiangle ABC is divided into fou egions by thee lines paallel to BC. The lines divide AB into fou equal segments. If the second lagest egion has aea 225, what is the aea of ABC? [2 maks] 3. Twelve students in a class ae each given a squae cad. The side length of each cad is a whole numbe of centimetes fom to 2 and no two cads ae the same size. Each student cuts his/he cad into unit squaes (of side length cm). The teache challenges them to join all thei unit squaes edge to edge to fom a single lage squae without gaps. They find that this is impossible. Alice, one of the students, oiginally had a cad of side length a cm. She says, If I don t use any of my squaes, but eveyone else uses thei squaes, then it is possible! Bob, anothe student, oiginally had a cad of side length b cm. He says, Me too! If I don t use any of my squaes, but eveyone else uses theis, then it is possible! Assuming Alice and Bob ae coect, what is ab? [3 maks] 4. Aimosia is a county which has thee kinds of coins, each woth a diffeent whole numbe of dollas. Jack, Jill, and Jimmy each have at least one of each type of coin. Jack has 4 coins totalling $28, Jill has 5 coins woth $2, and Jimmy has exactly 3 coins. What is the total value of Jimmy s coins? [3 maks] 5. Tiangle ABC has AB = 90, BC = 50, and CA = 70. A cicle is dawn with cente P on AB such that CA and CB ae tangents to the cicle. Find 2AP. [3 maks] 6. In quadilateal P QRS, PS = 5, SR = 6, RQ = 4, and P = Q = 60. Given that 2PQ = a + b, whee a and b ae unique positive integes, find the value of a + b. [4 maks] PLEASE TURN OVER THE PAGE FOR QUESTIONS 7, 8, 9 AND AMT Publishing

2 7. Dan has a ja containing a numbe of ed and geen sweets. If he selects a sweet at andom, notes its colou, puts it back and then selects a second sweet, the pobability that both ae ed is 05% of the pobability that both ae ed if he eats the fist sweet befoe selecting the second. What is the lagest numbe of sweets that could be in the ja? [4 maks] 8. Thee cicles, each of diamete, ae dawn each tangential to the othes. A squae enclosing the thee cicles is dawn so that two adjacent sides of the squae ae tangents to one of the b + c cicles and the squae is as small as possible. The side length of this squae is a + 2 whee a, b, c ae integes that ae unique (except fo swapping b and c). Find a + b + c. [4 maks] 9. Ten points P,P 2,...,P 0 ae equally spaced aound a cicle. They ae connected in sepaate pais by 5 line segments. How many ways can such line segments be dawn so that only one pai of line segments intesect? [5 maks] 0. Ten-dig is a game fo two playes. They ty to make a 0-digit numbe with all its digits diffeent. The fist playe, A, wites any non-zeo digit. On the ight of this digit, the second playe, B, then wites a digit so that the 2-digit numbe fomed is divisible by 2. They take tuns to add a digit, always on the ight, but when the nth digit is added, the numbe fomed must be divisible by n. The game finishes when a 0-digit numbe is successfully made (in which case it is a daw) o the next playe cannot legally place a digit (in which case the othe playe wins). Show that thee is only one way to each a daw. [5 maks] Investigation Show that if A stats with any non-zeo even digit, then A can always win no matte how B esponds. [4 bonus maks] 207 AMT Publishing

3 Austalian Intemediate Mathematics Olympiad 207 Solutions. We have x = b 2 + b + and x = 2(b 2) 2 +(b 2) + 2 = 2(b 2 4b +4)+b =2b 2 7b + 8. Hence 0 = (2b 2 7b + 8) (b 2 + b + ) = b 2 8b +7=(b 7)(b ). Fom the given infomation, b 2 > 2. So b = 7 and x =49+7+= Method Let B C, B 2 C 2, B 3 C 3, be the lines paallel to BC as shown. Then tiangles ABC, AB C, AB 2 C 2, AB 3 C 3 ae equiangula, hence simila. Region B 3 C 3 C 2 B 2 has aea 225. A B C B 2 C B 3 C 3 B C Since the lines divide AB into fou equal segments, the sides and altitudes of the tiangles ae in the atio :2:3:4. So thei aeas ae in the atio :4:9:6. Let the aea of tiangle AB C be x. Then 225 = AB 3 C 3 AB 2 C 2 =9x 4x =5x and the aea of tiangle ABC is 6x = = 6 45 = AMT Publishing

4 Method 2 Let B C, B 2 C 2, B 3 C 3, be the lines paallel to BC. Daw lines paallel to AB as shown. This poduces 4 small conguent tiangles and 6 small conguent paallelogams. A B C B 2 C 2 B 3 C 3 B C Dawing the diagonal fom top left to bottom ight in any paallelogam poduces two tiangles that ae conguent to the top tiangle. Thus tiangle ABC can be divided into 6 conguent tiangles. The egion B 3 C 3 C 2 B 2 has aea 225 and consists of 5 of these tiangles. Hence 225 = 5 6 ABC and ABC = = Method Fistly, we note that the combined aea of the 2 student cads is = 650. (Altenatively, use n 2 = n(n + )(2n + )/6.) Accoding to Alice and Bob, 650 x 2 = y 2 fo some integes x and y, whee x 2. So y = 506 and y = 649. Theefoe 23 y 25. If y = 23, then x =. If y = 24, then x is not an intege. If y = 25, then x = 5. Thus a = 5 and b = o vice vesa. So ab =5 = 55. Method 2 Fistly, we note that the combined aea of the 2 student cads is = 650. (Altenatively, use n 2 = n(n + )(2n + )/6.) Accoding to Alice, 650 a 2 = c 2 fo some intege c. Since 650 is even, a and c must both be even o odd. If a and c ae even, then a 2 and c 2 ae multiples of 4. But 650 is not a multiple of 4, so a and c ae odd. We ty odd values fo a fom to = 649, which is not a pefect squae = 64, which is not a pefect squae = 625, which is 25 2, giving one of the solutions = 60, which is not a pefect squae = 569, which is not a pefect squae = 529, which is 23 2, giving the second solution. Thus a = 5 and b = o vice vesa. So ab =5 = AMT Publishing 2

5 4. Method Let the value of the thee types of coin be a, b, c and let Jack s collection be 2a + b + c = 28. Then, swapping b with c if necessay, Jill s collection is one of: 3a + b + c, 2a +2b + c, a +2b +2c, a +3b + c. Since 3a + b + c and 2a +2b + c ae geate than 28, Jill s collection is eithe a +2b +2c o a +3b + c. If a +2b +2c = 2, then adding 2a + b + c = 28 gives 3(a + b + c) = 49, which is impossible since 3 is not a facto of 49. So a +3b + c = 2. Subtacting fom 2a + b + c = 28 gives a =2b + 7, which means a is odd and at least 9. If a = 9, then b = and c = 9. But a, b, c must be distinct, so a is at least. Since b + c 3, we have 2a 25 and a 2. Hence a =, b = 2, c = 4 and a + b + c = 7. Method 2 Let the value of the thee types of coin be a, b, c. Then Jill s collection is one of: And Jack s collection is one of: 2a +2b + c, 3a + b + c. 2a + b + c, a +2b + c, a + b +2c. Suppose Jill s collection is 2a+2b+c = 2. Since 2a+b+c and a+2b+c ae less than 2a+2b+c, Jack s collection must be a + b +2c = 28. Adding this to Jill s yields 3(a + b + c) = 49, which is impossible since 3 is not a facto of 49. So Jill s collection is 3a + b + c = 2. Since 2a + b + c is less than 3a + b + c, Jack s collection must be a +2b + c = 28 o a + b +2c = 28. Swapping b with c if necessay, we may assume that a +2b + c = 28. Subtacting 3a + b + c = 2 gives b =2a + 7 and c = 4 5a. So a 2. If a =, then b =9=c. Hence a = 2, b =, c = 4 and a + b + c = 7. Method 3 Let the value of the thee types of coin be a, b, c, whee a<b<c. Then Jack s collection is one of: 2a + b + c, a +2b + c, a + b +2c. And Jills collection is one of: 3a + b + c, 2a +2b + c, 2a + b +2c, a +3b + c, a +2b +2c, a + b +3c. All of Jill s possible collections exceed 2a + b + c, so Jack s collection is a +2b + c o a + b +2c. All of Jill s possible collections exceed a+2b+c, except possibly fo 3a+b+c. If 3a+b+c = 2, then subtacting fom a+2b+c = 28 gives b = 7+2a 9. But then a+2b+c +8+0 > 28. So Jack s collection is a+b+2c = 28. Then a+b is even, hence b 3, a+b 4, 2c = 28 a b 24, and c 2. Of Jill s possible collections, only 3a + b + c, 2a +2b + c, and a +3b + c could be less than a + b +2c. If a +3b + c = 2, then subtacting fom a + b +2c = 28 gives c =7+2b, which means c 3. If 2a +2b + c = 2, then subtacting fom 2a +2b +4c = 56 gives 3c = 35, which means c is a faction. So 3a + b + c = 2. Subtacting fom a + b +2c = 28 gives c = 7+2a, which means c is odd and at least 9. If c = 9, then a = and b =9=c. So c =, a = 2, b = 4 and a + b + c = AMT Publishing 3

6 Method 4 Let the value of the thee types of coin be a, b, c, whee a<b<c. Then Jack s collection is 28 = a + b + c + d whee d equals one of a, b, c. Since a + b 3, c + d 25. So d 25 c 25 d. Then 2d 25, hence d 2, which implies a + b + c 6. Jills collection is 2 = a+b+c+e whee e is the sum of two of a, b, c with epetition pemitted. So e 2a 2. Hence a + b + c 9. Fom a + b + c + d = 28 and 6 a + b + c 9, we get 9 d 2. If d = a, then a + b + c + d>4d 36. If d = b, then a + b + c + d>+3d 28. So d = c. Fom 2 = a + b + c + e 6 + e 6 + 2a we get 2a 5, hence a 2. The following table lists all cases. Note that each of x and y equals one of a, b, c. a a + b + c d c b e comment e =x + y e =x + y e =x + y b = c a = b e =2a e =x + y e =x + y So a = 2, b = 4, c =, d =, e = 4, and a + b + c = Method Let CA touch the cicle at R and CB touch the cicle at S. Let Q be a point on AB so that CQ and AB ae pependicula. C R S A P Q B Let be the adius of the cicle. Fom simila tiangles AQC and ARP, CQ/ = 70/AP. Fom simila tiangles BQC and BSP, CQ/ = 50/BP = 50/(90 AP ). Hence 7(90 AP )=5AP, 630 = 2AP,2AP = AMT Publishing 4

7 Method 2 Let CA touch the cicle at R and CB touch the cicle at S. C R S A P B The adius of the cicle is the height of tiangle AP C on base AC and the height of tiangle BPC on base BC. So atio of the aea of AP C to the aea of BPC is AC : PC =7:5. Tiangles AP C and BPC also have the same height on bases AP and BP. So the atio of thei aeas is AP : (90 AP ). Hence 5AP = 7(90 AP ), 2AP = 630, and 2AP = 05. Method 3 Let CA touch the cicle at R and CB touch the cicle at S. C R S A P B Since PR = PS, ight-angled tiangles P RC and PSC ae conguent. Hence CP bisects ACB. Fom the angle bisecto theoem, AP/P B = AC/BC = 7/5. Hence 5AP = 7(90 AP ), 2AP = 630, and 2AP = AMT Publishing 5

8 6. Method Let SS and RR be pependicula to PQ with S and R on PQ. Let RT be pependicula to SS with T on SS. 5 S T 6 R 4 P Q S R Since P = 60, PS =5/2 and SS =5 3/2. Since Q = 60, QR = 2 and RR =2 3. Hence ST = SS TS = SS RR =5 3/2 2 3= 3/2. Applying Pythagoas theoem to RT S gives RT 2 = = 4/4. So PQ = PS + S R + R Q = PS + TR+ R Q =5/2+ 4/ Hence a + b =2PQ =9+ 4. An obvious solution is a = 9, b = 4. Given that a and b ae unique, we have a + b = 50. Method 2 Let U be the point on PS so that UR is paallel to PQ. Let T be the point on RU so that ST is pependicula to RU. Extend PS and QR to meet at V. V S U T 6 R 4 4 P Q Tiangle P QV is equilateal. Since UR PQ, URV is equilateal and PU = QR = 4. So US =, UT = 2, ST = 3 2. Applying Pythagoas theoem to RT S gives RT 2 = = 4/4. We also have RT = RU UT = RV 2 = QV 9 2 = PQ 9 2. So 2PQ =9+ 4 = a + b. Given that a and b ae unique, we have a + b = AMT Publishing 6

9 Method 3 Extend PS and QR to meet at V. V S 6 R 5 4 P Q Tiangle P QV is equilateal. Let PQ = x. Then VS= x 5 and VR= x 4. Applying the cosine ule to RV S gives 36 = (x 4) 2 +(x 5) 2 2(x 4)(x 5) cos 60 =(x 2 8x + 6) + (x 2 0x + 25) (x 2 9x + 20) 0=x 2 9x 5 Hence 2x = = a + b. Given that a and b ae unique, we have a + b = 50. Comment We can pove that a and b ae unique as follows. We have (a 9) 2 = 4 + b 2 4b. So 2 4b is an intege, hence 4b is a pefect squae. Since 4 = 3 47 and 3 and 47 ae pime, b = 4m 2 fo some intege m. Hence a 9 = 4 m. If neithe side of this equation is 0, then we can ewite it as = 4s whee and s ae copime integes, giving 2 = 4s 2 =3 47 s 2. So 3 divides 2. Then 3 divides, 9 divides 2, 9 divides 3s 2,3 divides s 2, hence 3 divides s, a contadiction. So both sides of the equation ae 0. Theefoe a = 9 and b = AMT Publishing 7

10 7. Method Let thee be ed sweets and g geen sweets. We may assume 2. If Dan puts the fist sweet back, then the pobability that the two selected sweets ae ed is + g + g. If Dan eats the fist sweet, then the pobability that the two selected sweets ae ed is + g + g. The fist pobability is 05% of the second, so dividing and eaanging gives So <2. If = 20, then 20 = 20 +g, and + g = 400. If + g inceases, then and theefoe + g + g = = 2 20 ( ) ( ) + g 20 = 2 + g ( 20 ) ( = 2 ) + g 2 20 =+ + g > +g decease, so inceases. Since cannot exceed 20, + g cannot exceed 400. So the lagest numbe of sweets in the ja is 400. Method 2 Let thee be ed sweets and g geen sweets. We may assume 2. If Dan puts the fist sweet back, then the pobability that the two selected sweets ae ed is + g + g. If Dan eats the fist sweet, then the pobability that the two selected sweets ae ed is + g + g. The fist pobability is 05% of the second, so dividing and eaanging gives + g + g = = ( + g ) = 2( + g)( ) 20( + g) 20 = 2( + g) 2( + g) + 2g = ( + g) + g = + 2g/ If 2, then + g 2 + g and + 2g/ +g, a contadiction. So 20. If = 20, then 20 + g = + 2g/20, hence g = = 380 and + g = 400. We also have the equation (2 )g = ( ). If <20, then g<(2 )g = ( ) < 20 9 = 380, hence + g< = 400. So the lagest numbe of sweets in the ja is AMT Publishing 8

11 Method 3 Let thee be ed sweets and g geen sweets. We may assume 2. Let n = + g. Then the pobability of selecting two ed sweets if the fist sweet is put back is n n and the pobability if Dan eats the fist sweet befoe selecting the second is n n. The fist pobability is 05% of the second, so dividing and eaanging gives n n = = (n ) = 2n( ) 2n n 20 =0 (n + 20)(2 ) = 420 Since n + 20 is positive, 2 is positive. Hence n is lagest when 2 = and then n + 20 = 420. So the lagest numbe of sweets in the ja is 400. Comment Since 2 is a facto of 420 and 2 20, the following table gives all possible values of, n, g. 2 n + 20 n g AMT Publishing 9

12 8. Let WXY Z be a squae that encloses the thee cicles and is as small as possible. Let the centes of the thee given cicles be A, B, C. Then ABC is an equilateal tiangle of side length. We may assume that A, B, C ae aanged anticlockwise and that the cicle with cente A touches WX and WZ. We may also assume that WX is hoizontal. Note that if neithe YX no YZ touch a cicle, then the squae can be contacted by moving Y along the diagonal WY towads W. So at least one of YX and YZ must touch a cicle and it can t be the cicle with cente A. We may assume that XY touches the cicle with cente B. Z Y C A B W X If YZ does not touch a cicle, then the 3-cicle cluste can be otated anticlockwise about A allowing neithe YX no YZ to touch a cicle. So YZ touches the cicle with cente C. Method Let ADEF be the ectangle with sides though C and B paallel to WX and WZ espectively. Z Y F C E A B D W X Since AF = WZ =WX =AD, ADEF is a squae. Since AC ==AB, tiangles AF C and ADB ae conguent. So FC = DB and CE = BE. Let x = AD. Since AB = and tiangle ADB is ight-angled, DB = x AMT Publishing 0

13 Since CBE is ight-angled isosceles with BC =, we have BE =/ 2. So x = DE = x 2 +/ 2. Squaing both sides of x / 2= x 2 gives x 2 =(x / 2) 2 = x 2 2x +/2 0=2x 2 2x /2 x =( 2 ± 2 + 4)/4 Since x>0, we have x =( 2+ 6)/4 =( )/2. Hence WX =+( )/2. We ae told that WX = a +( b + c)/2 whee a, b, c ae unique integes. This gives a + b + c = = 73. Method 2 Daw lines though A paallel to WX and WZ. Z Y C A z x B W X With angles x and z as shown, we have WX = 2 + AB cos x + 2 = + cos x WZ = 2 + AC cos z + 2 = + cos z Since WX = WZ, x = z. Since x z = 90, we have x = 5. So WX = + cos 5 = + cos(45 30 ) = + cos 45 cos 30 + sin 45 sin 30 = = =+ = We ae told that WX = a +( b + c)/2 whee a, b, c ae unique integes. This gives a + b + c = = AMT Publishing

14 9. Method Let the pai of intesecting lines be AC and BD whee A,B,C,D ae fou of the ten given points. These lines split the emaining six points into fou subsets S,S 2,S 3,S 4. Fo each i, each line segment beginning in S i also ends in S i, othewise AC and BD would not be the only intesecting pai of lines. Thus each S i contains an even numbe of points, fom 0 to 6. If S i contains 2 points, then thee it has only line segment. If S i contains 4 points, then thee ae pecisely 2 ways to connect its points in pais by non-cossing segments. If S i contains 6 points, let the points be Q,Q 2,Q 3,Q 4,Q 5,Q 6 in clockwise ode. To avoid cossing segments, Q must be connected to one of Q 2, Q 4, Q 6. So, as shown, thee ae pecisely 5 ways to connect the six points in pais by non-cossing segments. Q 3 Q 4 Q 2 Q 5 Q 3 Q 4 Q 2 Q 5 Q Q 6 Q Q 6 Q 3 Q 4 Q 2 Q 5 Q 3 Q 4 Q 2 Q 5 Q 3 Q 4 Q 2 Q 5 Q Q 6 Q Q 6 Q Q 6 In some ode, the sizes of S,S 2,S 3,S 4 ae {6, 0, 0, 0}, {4, 2, 0, 0}, o {2, 2, 2, 0}. We conside the thee cases sepaately. In the fist case, by otation about the cicle, thee ae 0 ways to place the S i that has 6 points. Then thee ae 5 ways to aange the line segments within that S i. So the numbe of ways to daw the line segments in this case is 0 5 = 50. In the second case, in clockwise ode, the sizes of the S i must be (4, 2, 0, 0), (4, 0, 2, 0) o (4, 0, 0, 2). In each case, by otation about the cicle, thee ae 0 ways to place the S i. Then thee ae 2 ways to aange line segments within the S i that has 4 points, and thee is way to aange the line segment within the S i that has 2 points. So the numbe of ways to daw the line segments in this case is = 60. In the thid case, in clockwise ode, the sizes of the S i must be (2, 2, 2, 0). By otation about the cicle, thee ae 0 ways to place the S i. Then thee is only way to aange the line segment within each S i that has 2 points. So thee ae 0 ways to aange the line segments in this case. In total, the numbe of ways to aange the line segments is = AMT Publishing 2

15 Method 2 The pai of intesecting lines patition the cicle into fou acs. In ode to allow the emaining points to be paied up without futhe cossings, we equie each such ac to contain an even numbe of points. So each line of a cossing pai patitions the cicle into two acs, each of which contain an odd numbe of points. Disegading otation of the cicle, a cossing line is one of only two types. So, disegading otations, thee ae only fou ways to have the pai of cossing lines. Undeneath each diagam we list the numbe of ways of joining up the emaining pais of points without intoducing moe cossings. (The numbe 5 is justified in Method.) So, counting otations, the numbe of paiings with a single cossing is 0 ( )= AMT Publishing 3

16 0. Since the 2nd, 4th, 6th, 8th and 0th digits must be even, the othe digits must be odd. Since the last digit must be 0, the fifth digit must be 5. Let a be the 3d digit and b be the 4th digit. If b is 4 o 8, then 4 divides b but does not divide 0a since a is odd. Hence 4 does not divide 0a + b. So the 4th digit is 2 o 6. Now let a, b, c be the 6th, 7th, 8th digits espectively. If c is 8, then 8 divides 00a + c but does not divide 0b since b is odd. Hence 8 does not divide 00a + 0b + c. If c is 4, then 8 divides 00a but does not divide 0b + c = 2(5b + 2) since b is odd. Hence 8 does not divide 00a + 0b + c. So the 8th digit is 2 o 6. So each of the 2nd and 6th digits is 4 o 8. Since 3 divides the sum of the fist thee digits and the sum of the fist six digits, it also divides the sum of the 4th, 5th, and 6th digits. So the 4th, 5th, and 6th digits ae espectively o Thus we have two cases with a, b, c, d equal to, 3, 7, 9 in some ode. Case. a 4 b 258c 6 d 0 Since 3 divides a +4+b, one of a and b equals and the othe is 7. Since 8 divides 8 c 6, c is9. Sowehave d 0o d 0. But neithe no is a multiple of 7. Case 2. a 8 b 654c 2 d 0 Since 8 divides 4 c 2, c is 3 o 7. If c = 3, then, because 3 divides a +8+b, we have one of: d 0, d 0, d 0, d 0. Butnoneof , , , isamultipleof7. If c = 7, then, because 3 divides a +8+b, we have one of: d 0, d 0, d 0, d 0. Noneof , , isamultipleof7. Thisleaves astheonlydaw. Investigation Note that B must play an even digit on each tun. If A stats with 2, then B can only espond with 20, 24, 26, o 28. A may then leave one of 204, 240, 26, 285. B cannot espond to 26. The othe numbes foce espectively 20485, 24085, B cannot espond to any of these. bonus If A stats with 4, then B can only espond with 40, 42, 46, o 48. A may then leave one of 408, 420, 462, 480. B cannot espond to 408 and 480. Each of the othe numbes foce one of 42085, 46205, 46280, B cannot espond to any of these. bonus If A stats with 6, then B can only espond with 60, 62, 64, o 68. A may then leave one of 609, 62, 648, 684. B cannot espond to 62. The othe numbes foce espectively 60925, 64805, B can only espond with Then A may eply with , to which B has no esponse. bonus If A stats with 8, then B can only espond with 80, 82, 84, o 86. A may then leave one of 804, 825, 840, 864. B cannot espond to 804 and 840. The othe numbes foce espectively 82560, B cannot espond to eithe of these. bonus 207 AMT Publishing 4

17 Making Scheme. Establishing a elevant quadatic equation. Coect answe (57). 2. A valid appoach with a elevant diagam. Coect answe (720). 3. Coect total aea of the 2 cads. A useful equation in two vaiables and useful constaints on the vaiables. Coect answe (55). 4. A valid appoach with some pogess. Futhe pogess. Coect answe (7). 5. A useful diagam. Establishing useful atios. Coect answe (05). 6. A useful diagam. Some useful calculations. Moe calculations. Coect answe (50). 7. Establishing a elevant pobability expession. Establishing anothe elevant pobability expession. Establishing a simplified equation. Coect answe (400). 8. Establishing 4 squae sides touch a cicle. A useful constuction. Establishing a useful equation in one vaiable. Coect answe (73). 9. Useful appoach. Some pogess. Establishing elevant cases and coectly enumeating one case. Coectly enumeating anothe case. Completing the enumeation with the coect answe (20). 0. Establishing places fo odd digits and 0 and 5. Establishing elevant options fo the even digits. Establishing elevant cases. Coectly esolving one case. Completing the poof with the coect answe ( ). Investigation: Poving A wins with fist digit 2. bonus Poving A wins with fist digit 4. bonus Poving A wins with fist digit 6. bonus Poving A wins with fist digit 8. bonus 207 AMT Publishing 5

Online Mathematics Competition Wednesday, November 30, 2016

Math@Mac Online Mathematics Competition Wednesday, Novembe 0, 206 SOLUTIONS. Suppose that a bag contains the nine lettes of the wod OXOMOXO. If you take one lette out of the bag at a time and line them