Canadian Open Mathematics Challenge 2017

Transcription

1 Cndin Open Mthemtics Chllenge 017 Officil Solutions Presented by the Cndin Mthemticl Society nd supported by the Acturil Profession. The COMC hs three sections: A. Short nswer questions worth 4 mrks ech. A correct nswer receives full mrks. Prtil mrks my be wrded for work shown if correct nswer is not provided. B. Short nswer questions worth 6 mrks ech. A correct nswer receives full mrks. Prtil mrks my be wrded for work shown if correct nswer is not provided. C. Multi-prt full solution questions worth 10 mrks ech. Solutions must be complete nd clerly presented to receive full mrks. Some novel solutions were selected from students ppers in order to showcse other possible wys of resoning. Dvid Rowe of Holy Hert of Mry Regionl High School Hneul Shin of Bergen County Acdemies Victor Wng of Sir Winston Churchill Secondry School Freddie Zho of Indus Center for Acdemic Excellence COMC exms from other yers, with or without the solutions included, re free to downlod online. Plese visit c 017, 018 Cndin Mthemticl Society Jnury 11, 018

2 Officil Solutions COMC 017 Section A 4 mrks ech A1. The verge of the numbers, 5, x, 14, 15 is x. Determine the vlue of x. Correct nswer: 9. Solution 1: The verge of the numbers is 36 + x = 5x, we hve x = x = x + 36 ; this must be equl x. Solving 5 Solution : Since the x being there would not ffect the verge of the other 4 numbers, x = = = 9. A. An equilterl tringle hs sides of length 4cm. At ech vertex, circle with rdius cm is drwn, s shown in the figure below. The totl re of the shded regions of the three circles is π cm. Determine. Correct nswer:. Solution: The re of the intersection of ech circle nd the tringle is 4π/6 cm. The three circles do not overlp, thus the totl re is π cm. Pge c 017, 018 Cndin Mthemticl Society

3 COMC Officil Solutions A3. Two 1 1 squres re removed from 5 5 grid s shown. Determine the totl number of squres of vrious sizes on the grid. Correct nswer: 39 squres. Solution 1: There re squres, 1 squres, nd squres. This gives = 39 squres. Solution : Number of squres without missing prts is = 55. Missing 1 1 squres is. Missing squres is 4. Missing 3 3 squres is 5. Missing 4 4 squres is 4. Missing 5 5 squres is 1. Totl missing squres is = 16. Thus the number of squres in the digrm is = 39. c 017, 018 Cndin Mthemticl Society Pge 3

4 Officil Solutions COMC 017 A4. Three positive integers, b, c stisfy Determine the sum of + b + c. 4 5 b 6 c = Correct nswer: 36. Solution: The prime fctoriztion of the left-hnd side is nd the prime fctoriztion of the right-hnd-side is 5 b c 3 c = +c 3 c 5 b, = Since the prime fctoriztion of n integer is determined uniquely, we hve, + c = 34, c = 18 nd b = 10. We now find. Since + c = 34 nd c = 18, + 18 = 34, yielding = 8. Therefore, + b + c = = 36. Pge 4 c 017, 018 Cndin Mthemticl Society

5 COMC Officil Solutions Section B 6 mrks ech B1. Andrew nd Betrice prctice their free throws in bsketbll. One dy, they ttempted totl of 105 free throws between them, with ech person tking t lest one free throw. If Andrew mde exctly 1/3 of his free throw ttempts nd Betrice mde exctly 3/5 of her free throw ttempts, wht is the highest number of successful free throws they could hve mde between them? Correct nswer: 59 free throws. Solution 1: From their success rtes we conclude tht ech of them must hve mde multiple of 15 throws. Specificlly, from Andrew s success rte, his number of throws must be multiple of 3. Since the totl number of throws (105) is lso multiple of 3, Betrice s number of throws must be multiple of 3 too. From Betrice s success rte, her number of throws must be multiple of 5, nd thus must in fct be multiple of 15. Similrly, since 105 is multiple of 5, Andrew s number of throws must be multiple of 5 nd thus multiple of 15 too. Since 1/3 < 3/5, to mximize the result we should ssume tht Andrew mde the lest possible number of throws, tht is 15. Then Betrice mde 90 throws. Then the number of successful free throws they could hve mde between them is = = 59. The mximum possible number of successful free throws they could hve mde between them is 59. Solution : Suppose Andrew mde free throws nd Betrice b free throws, then + b = 105, > 0, b > 0. Let M be the number of successful free throws. We hve M is mximl when 4 15 M = 3 + 3b 5 = 3(105 ) = = is miniml. Tht is, = 15 nd so M = 59. The mximum possible number of successful free throws they could hve mde between them is 59. c 017, 018 Cndin Mthemticl Society Pge 5

6 Officil Solutions COMC 017 B. There re twenty people in room, with men nd b women. Ech pir of men shkes hnds, nd ech pir of women shkes hnds, but there re no hndshkes between mn nd womn. The totl number of hndshkes is 106. Determine the vlue of b. Correct nswer: 84. Solution 1: Since there re twenty people in the room, + b = 0. Now we use the fct tht there re no hndshkes between mn nd womn. In group of m people, there re m(m 1)/ pirs of people. Hence, the number of hndshkes tht took plce is ( 1) + b(b 1) = 106, which simplifies to + b ( + b) = b = 1 + ( + b) = 3. Substituting b = 0 into this eqution yields + (0 ) = 3 + ( ) = = 0 ( ) = 0. This fctors s ( 14)( 6) = 0. Therefore, = 14 or = 6. Since + b = 0, (, b) = (14, 6) or (6, 14). Hence, b = 84. Solution : Since there re 0 people in the room, there re 0 19 = 190 pirs of people. Out of them exctly b pirs do not shke hnds. Thus, we hve 190 b = 106, nd so b = = 84. Pge 6 c 017, 018 Cndin Mthemticl Society

7 COMC Officil Solutions B3. Regulr decgon (10-sided polygon) ABCDEF GHIJ hs re 017 squre units. Determine the re (in squre units) of the rectngle CDHI. Correct nswer: squre units. Solution 1: If O is the centre of the decgon nd we cut the decgon into equl isosceles tringles, s shown, we cn see tht the re of the decgon is 10 times the re of one tringle. Since digonls CH nd DI re equl in length nd bisect ech other, qudrilterl CDHI is rectngle, with the sme bse s ech of the tringles, but twice the height. Thus the re of CDHI is 4 times the re of ech of the tringles nd hence 40% of the re of the decgon or = squre units. B A J C I D H E F G The re of the rectngle is squre units. A second solution is provided on the next pge. c 017, 018 Cndin Mthemticl Society Pge 7

8 Officil Solutions COMC 017 Solution : Let s = AB be the side length of the regulr decgon. Since it is regulr polygon with 10 sides, ll its ngles re equl, in prticulr ABC = JAB = 180 (10 )/10 = 144. Let M be the midpoint of CI. Drw AM, which will be perpendiculr to CI. Plce point P on AM nd point Q on CM such tht BP is perpendiculr to AM nd BQ is perpendiculr to CM. C s B Q s A M P J I D H E F G The re of ABCM could be found s sum of res of two right tringles BP A nd BQC nd the rectngle BP MQ. Note tht ngle BAP is 144/=7 nd so ngle ABP is 90 7 = 18. Now, ngle CBQ is = 36, nd so ngle BCQ is = 54. The re of the right tringle BP A is 1 s sin 18 cos 18 = 1 4 s sin 36. The re of the right tringle BQC is 1 s sin 54 cos 54 = 1 4 s sin 108 = 1 4 s cos 18. B 18 s 7 A P J s 36 C 54 Q M I The re of the rectngle BP MQ is s sin 54 cos 18 = 1 s (sin 36 + sin 7 ) = 1 s (sin 36 + cos 18 ). Therefore, the re of ABCM is 3 4 s (sin 36 + cos 18 ). Next, the re of CDHI is s (sin 36 + cos 18 ) = 8 3. The totl re of the decgon is = 0 3 = 017. Thus, the re of CDHI is = Pge 8 c 017, 018 Cndin Mthemticl Society

9 COMC Officil Solutions B4. Numbers, b nd c form n rithmetic sequence if b = c b. Let, b, c be positive integers forming n rithmetic sequence with < b < c. Let f(x) = x + bx + c. Two distinct rel numbers r nd s stisfy f(r) = s nd f(s) = r. If rs = 017, determine the smllest possible vlue of. Correct nswer: 9. Solution 1: Note tht r + br + c = s (1) s + bs + c = r () Subtrcting the second eqution from the first yields (r s ) + b(r s) = (r s) (r + s)(r s) + (b + 1)(r s) = 0 ((r + s) + b + 1)(r s) = 0. Since r s, r + s = b+1. Substituting s = b+1 r into the first eqution yields s into the second eqution yields s + bs + c = b+1 s s + (b + 1)s + c + b+1 = 0. Therefore, r, s re the roots to the eqution x + (b + 1)x + c + b+1 = 0. The product of the roots of this eqution is c + b+1 = 017. We lso know tht the coefficients form n rithmetic sequence, so let b = + k, c = + k. Then we = 017 nd so r + br + c = b+1 r r + (b + 1)r + c + b+1 = 0. Substituting r = b+1 hve +k + +k+1 k = = = ( ) Thus, in order for k to be integer, + 1 must be fctor of 1007 = Thus, = 9, = 6 or = 503. The smllest positive integer for which k is n integer is = 9. For completeness, we find k = 8594 nd so b = 8603 nd c = Then, r = nd s = stisfy the bove reltions. The nswer is 9. A second solution is provided on the next pge. c 017, 018 Cndin Mthemticl Society Pge 9

10 Officil Solutions COMC 017 Solution : Note tht r + br + c = s (1) s + bs + c = r () Subtrcting the second eqution from the first yields since r s, r + s = b+1. Adding (1) nd () we get Using r + s = b+1 nd then or equivlently Substituting c = b, we get nd so (r s ) + b(r s) = (r s) (r + s)(r s) + (b + 1)(r s) = 0 ((r + s) + b + 1)(r s) = 0, ((r + s) rs) + b(r + s) + c = r + s. nd rs = 017, we obtin b = (b + 1) 017 = (b + 1) (b + 1)(b 1) c = 0, b c = 0. b b = 0, ( + 1)(1009 1) c = 0 = Now, b is integer, so must be integer. Since gcd(, + 1) = 1, + 1 must be fctor of 1007 = 19 53, nd so = 9, = 6 or = 503. The smllest vlue is = 9. Pge 10 c 017, 018 Cndin Mthemticl Society

11 COMC Officil Solutions Section C 10 mrks ech Note: Section C questions require prticipnts to show ll their work. C1. For positive integer n, we define function P (n) to be the sum of the digits of n plus the number of digits of n. For exmple, P (45) = = 11. (Note tht the first digit of n reding from left to right, cnnot be 0). () Determine P (017). Solution: = 14. The nswer is P(017)=14. (b) Determine ll numbers n such tht P (n) = 4. Solution: Consider 1-digit numbers, then n = 3. Among -digit numbers we need those with the sum of their digits equl to ; so we hve n = 11 nd n = 0. Among 3-digit numbers we need those with the sum of their digits equl to 1; so we hve n = 100. For numbers with 4 or greter thn 4 digits, P (n) > 4, so we hve listed ll possible numbers with the required property. The nswer is 3, 11, 0, 100. (c) Determine with n explntion whether there exists number n for which P (n) P (n+1) > 50. Solution: If n nd n + 1 differ only in one lst digit the equlity P (n) P (n + 1) > 50 is not possible. Consider the cse when n consisits of k 9 s. Then P (n) = 9k + k = 10k nd P (n + 1) = 1 + (k + 1) = k +. So we hve P (n) P (n + 1) = 9k > 50, so k 6. For k = 6 we obtin n = 999, 999 nd P (999, 999) P (1, 000, 000) = 60 8 > 50. The nswer is Yes, for exmple n = 999, 999. c 017, 018 Cndin Mthemticl Society Pge 11

12 Officil Solutions COMC 017 C. A function f(x) is periodic with period T > 0 if f(x + T ) = f(x) for ll x. The smllest such number T is clled the lest period. For exmple, the functions sin(x) nd cos(x) re periodic with lest period π. () Let function g(x) be periodic with the lest period T = π. Determine the lest period of g(x/3). Solution: A period of g ( ) ( x 3 is 3π becuse g x+3π ) ( 3 = g x 3 + π) = g ( ) x 3. Note tht if g(x/3) hd period smller thn 3π, then g(x) would hve period smller thn π, contrdiction. The nswer is 3π. (b) Determine the lest period of H(x) = sin(8x) + cos(4x). Solution: The lest period of sin(8x) is π 4 becuse sin(8(x + π 4 )) = sin(8x + π) = sin(8x). The lest period of cos(4x) is π becuse cos(4(x + π )) = cos(4x + π) = cos(4x). We pick the lrger of the two periods for the sum of the functions. In generl, it is the lest common multiple of the two periods tht must be picked. Note tht if H(x) hd period smller thn π/, then cos(4x) = H(x) sin(8x) would lso hve period smller thn π/, nd so cos x would hve period smller thn π. Indeed, if cos(4x) = cos(4(x + T )) = cos(4x + 4T ) nd T < π/ then cos y = cos(y + 4T ) nd 4T < 4(π/) = π. This is contrdiction. The nswer is π/. (c) Determine the lest periods of ech of G(x) = sin(cos(x)) nd F (x) = cos(sin(x)). Solution in two prts: 1. G(x + π) = sin(cos(x + π)) = sin(cos(x)) = G(x). To prove tht T = π is the smllest positive number consider for exmple x = 0. Then G(0) = sin(1). For G(T ) = sin(cos(t )) = sin(1), we need either cos(t ) = 1 + πk or cos(t ) = 1 + (k + 1)π, k Z. In the first cse we see tht the inequlity πk 1 is true only for k = 0. This gives cos(t ) = 1, nd so the smllest positive T = π. In the second cse the inequlity (k + 1)π 1 is true for no integer k. Thus, the only solution is T = π.. F (x + π) = cos(sin(x + π)) = cos( sin(x)) = cos(sin(x)) = F (x). To prove tht T = π is the smllest positive number consider for exmple x = 0. Then F (0) = 1. Now, we need F (0 + T ) = cos(sin(t )) = 1. Thus, sin T = 0, nd so the smllest positive T = π. The nswers re π nd π. Pge 1 c 017, 018 Cndin Mthemticl Society

13 COMC Officil Solutions C3. Let XY Z be n cute-ngled tringle. Let s be the side-length of the squre which hs two djcent vertices on side Y Z, one vertex on side XY nd one vertex on side XZ. Let h be the distnce from X to the side Y Z nd let b be the distnce from Y to Z. () If the vertices hve coordintes X = (, 4), Y = (0, 0) nd Z = (4, 0), find b, h nd s. Solution: Subtrcting coordintes of corresponding points we hve b = z 1 y 1 = 4 0 = 4, h = x y = x z = 4 0 = 4. Note tht points (1, ) nd (3, ) lie on sides XY nd XZ respectively nd together with points (1, 0) nd (3, 0) they define squre tht stisfies the conditions of the problem. This squre hs side s =. (Alterntively, from similr tringles we hve s b = h s h tht is s 4 = 4 s 4, so s =.) y 4 3 X (1, ) (3, ) 1 Y Z x The nswer is b = 4, h = 4, s =. (b) Given the height h = 3 nd s =, find the bse b. Solution: Since P Q is prllel to Y Z, tringle XP Q is similr to XY Z. Becuse h = 3 nd corresponding height of XP Q is 3 = 1, we conclude tht the bse of XY Z is 3 = 6. The nswer is b = 6. c 017, 018 Cndin Mthemticl Society Pge 13

14 Officil Solutions COMC 017 (c) If the re of the squre is 017, determine the minimum re of tringle XY Z. Solution 1: From the similrity of XP Q nd XY Z we hve s b = h s bh. Equivlently, s = h b + h. Thus, s = (bh) bh = K (b + h) (b + h). Here K = bh is the re of XY Z. By AM-GM inequlity (rithmetic men is greter or equl thn geometric men), we hve thus s K. We hve 017 K. Thus, 4034 K. Now, we show tht the minimum re is chieved for b = h = s = 017. Indeed, if b = h then s = bh b + h = b = h nd K = s = The minimum vlue for the re of XY Z is The nswer is bh (b + h) 1, Solution : From the similrity of XP Q nd XY Z we hve s b = h s sh. Thus b =. The re h h s [XY Z] = bh = h s (h s). Finding the minimum vlue of this expression is equivlent to finding the mximum of its reciprocl (h s) h mx. s Now, note tht the reciprocl is qudrtic function in the vrible 1, tht is h (h s) h s = ( ) 1 + h s ( ) 1, h so the mximum is chieved t 1 h = 1 1 s Then b = s nd the re [XY Z] = s = or equivlently, for h = s. Specil thnks to Hneul Shin of Bergen County Acdemies for providing the solution bove. Edited for clrity nd concision. Pge 14 c 017, 018 Cndin Mthemticl Society

15 COMC Officil Solutions C4. Let n be positive integer nd S n = {1,,..., n 1, n}. A perfect piring of S n is defined to be prtitioning of the n numbers into n pirs, such tht the sum of the two numbers in ech pir is perfect squre. For exmple, if n = 4, then perfect piring of S 4 is (1, 8), (, 7), (3, 6), (4, 5). It is not necessry for ech pir to sum to the sme perfect squre. () Show tht S 8 hs t lest one perfect piring. Solution: For n = 8 we hve pirs (1, 8), (, 7), (3, 6), (4, 5) ech of which sums to 9 nd (9, 16), (10, 15), (11, 14), (1, 13) ech of which sums to 5. The nswer is (1, 8), (, 7), (3, 6), (4, 5), (9, 16), (10, 15), (11, 14), (1, 13). (b) Show tht S 5 does not hve ny perfect pirings. Solution 1: Ech pir must hve sum 16, 9, 4. There re 5 pirs tht must sum to 55. There must be t lest two pirs tht sum to 16. If there re two such pirs, then the other three pirs must sum to 55-3 = 3, which cnnot be done with the numbers 9, 4. There cnnot be more thn two pirs tht sum to 16, since there re only two such pirs, nmely (6, 10), (7, 9). P.S. It is lso possible to construct n rgument by considering ll sums modulo 8: 55 7 mod 8, 16 0 mod 8, 9 1, mod 8, 4 4 mod 8. Thus, for five pirs we must hve 7 = But this does not give the correct sum: = 47 55, so perfect piring is not possible. Solution : Consider S 5 = {1,, 3, 4, 5, 6, 7, 8, 9, 10}. Note tht 10 needs 6 to mke perfect pir, which mens tht 3 cn only pir perfectly with 1. This leves 8 with no prtner with which to mke perfect pir. Specil thnks to Dvid Rowe of Holy Hert of Mry Reg. H.S. for providing the solution bove. Edited for clrity nd concision. c 017, 018 Cndin Mthemticl Society Pge 15

16 Officil Solutions COMC 017 (c) Prove or disprove: there exists positive integer n for which S n hs t lest 017 different perfect pirings. (Two pirings tht re comprised of the sme pirs written in different order re considered the sme piring.) Solution 1: The esiest wy to get perfect piring from {, + 1,..., m} is for ll pirs to hve the sme sum, i.e. + m to be squre. The next esiest wy is if there re two possible sums, which hppens when the sets {,..., n} nd{n + 1,..., m} hve both + n nd n m being squres. We will first prove the following lemm: Lemm: Let 1 (mod 4) be positive integer. Then there exists n even positive integer m such tht m > nd {, + 1,..., m} hs t lest two perfect pirings into (m + 1)/ pirs. Proof of lemm: Let x, y be odd positive integers stisfying Then tking m = y x + 1, we hve x >, y > (x + 1). {,..., m} = {, + 1,..., x } {x + 1, x +,..., y x + 1}, where we pir up elements in the first set to ll hve sum x, nd in the second set to hve sum y. We would lso like to pir up elements so every pir hs the sme sum, i.e. we would like m + to be squre. This trnsltes to solving y x + 1 = z, subject to x, y odd positive integers, z positive integer, x >, nd y > (x + 1). The condition of m being even will be utomticlly stisfied since y x (mod 4) using x, y odd (nd so y x = (y x)(y + x) 0 (mod 4)) nd 1 (mod 4). We rerrnge our eqution into (y z)(y + z) = y z = x + 1. Picking x = r + 1 to be ny odd positive integer bigger thn, nd noting tht x = 4r(r + 1) (mod 8) nd (mod 8), we see tht x (mod 8). Thus we set y z =, y + z = x +1, nd we get y = x + 5 4, z = x 3. 4 Hence y, z re positive integers, y is odd, nd y, z stisfy y x + 1 = z. We ssumed tht x >, so we will be done s long s y > (x + 1). But y is qurtic in x, so this inequlity is stisfied for ll x sufficiently lrge, nd thus we cn pick x to be sufficiently lrge for this to hold. This completes the proof of the lemm. Pge 16 c 017, 018 Cndin Mthemticl Society

17 COMC Officil Solutions We now show tht for ny N there exists positive integer n for which S n hs t lest N different perfect pirings. We do it by induction on N, where we lso impose tht n must be even. The cse N = is given by the bove lemm with = 1. Assume we hve it up to N 1, nd m is n even number such tht {1,..., m} hs t lest N 1 perfect pirings. Then tke = m (mod 4) in the bove lemm, to get n n > m even with {m + 1,..., n} hving t lest two perfect pirings. Combining these with prtitions of {1,..., m}, we see tht {1,..., n} hs t lest (N 1) N perfect pirings. Therefore the result is true for N = 017. Solution : Let us first prove the following sttement. Lemm: There exist rbitrrily lrge n 0 mod (4) for which the set S n hs perfect piring. Proof of lemm: This proof is by induction. We know tht perfect piring exists for S 4. Suppose we hve perfect piring of the set S n for some n divisible by 4. We will construct perfect piring of the set S m with some m > n lso divisible by 4. The construction is strightforwrd generliztion of the method used in the numericl exmple of prt (). We will find m = n+r with r > 0 divisible by 4 such tht n+m+1 = k for some integer k. This yields perfect piring for S m consisting of ll pirs in perfect piring for S n nd new pirs (n + 1, m), (n +, m 1),,..., (n + r, n + r + 1). So we hve the eqution (n + m) + 1 = 4n + r + 1 = k, i.e. Put k = q + 1, where q n. Then r = k 4n 1. r = (4q + 4q + 1 4n 1)/ = (q(q + 1) n). ( ) Since both q(q + 1) nd n re even, r is divisible by 4 nd the construction is complete. This completes the proof of the lemm. Note tht in this construction ech perfect piring for S n gives rise to perfect piring for S m nd they ll re different. Suppose now, by wy of contrdiction, tht there is some number H (for the purpose of the problem, H < 017) such tht for ny i 16 divisible by 4 the set S i hs t most H perfect pirings. Let n 16 be divisible by 4 nd such tht S n hs exctly H perfect piring. Crrying out the bove construction, we will find H perfect pirings for S m. We demonstrte below how to obtin S m tht hs perfect piring different from those provided by our construction. Such n S m will hve t lest H + 1 perfect pirings, contrdiction. Since n is divisible by 4, Put q = n/4 in ( ), which yields r = n /8 + n/ n = n(n 1)/8. Here the role of the condition n 16 becomes cler: it ensures tht r > 0. The following simple clcultion shows c 017, 018 Cndin Mthemticl Society Pge 17

18 Officil Solutions COMC 017 tht m + 1 is n integer squre: m + 1 = (n + r) + 1 = n + n(n 1) = n 4n ( ) n =. Hence, in ddition to the perfect pirings for S m tht correspond to those for S n, we hve the perfect piring {(1, m), (, m 1),..., (m, m + 1)}. Solution 3: This solution is bsed on the fct tht there exists n rithmetic progression of length 3 consisting of integer squres; specificlly, {1, 5, 49}. Observe tht if n = 5N 1 for some odd integer N 1 then there exists t lest one perfect piring of the set S n, nmely (1, 5N 1), (, 5N ),..., ( 5N 1, 5N +1 ). Now, we cn choose lrge enough integer N for which there re t lest 016 pirs of integers (, b) such tht 1 < b n nd b = 4N. (Any odd N with N > 017 will do.) Then one cn tke ny pirs (, 5N ) nd (b, 5N b) from the perfect piring bove nd swp them with (, 5N b) nd (b, 5N ). Whenever b = 4N, this swp will produce new perfect piring becuse 5N + b = 5N + 4N = 49N nd 5N b + = 5N 4N = N. Thus we cn construct 016 new perfect pirings, which together with the initil perfect piring gives 017 different perfect pirings. Specil thnks to Victor Wng of Sir Winston Churchill S.S. for providing the solution bove. Edited for clrity nd concision. Solution 4: Let us split the set S n into two groups, 1... x nd x n. The prtition (1, x), (, x 1),..., (x + 1, n), (x +, n 1),... ) of S n is perfect piring provided x is even nd 1 + x = m, m + n = k for some odd integers k > m > 1. Conversely, ny pir of integers k, m such tht k m = n yields perfect piring of S n nd the perfect pirings corresponding to different pirs (m, k) re different. We will exhibit n n such tht the eqution n = k m hs t lest 017 solutions in positive integers k, m. Let us tke 017 distinct Pythgoren triples (r i, s i, t i ), 1 i 017 (so tht ri + s i = t i ), nd let = Π 017 i=1 s i. We my ssume tht t lest one of the s i s is even, so tht is even. Put n = /. Now, m i = r i /s i nd k i = t i /s i re both integers nd m i + = (ri +s i ) = k s i for ny 1 i 017. This i gives 017 solutions of the eqution k m = n nd hence t lest 017 perfect pirings of S n. Specil thnks to Freddie Zho of Indus Center for Acdemic Excellence for providing the solution bove. Edited for clrity nd concision. Pge 18 c 017, 018 Cndin Mthemticl Society

19 COMC Officil Solutions Premier Sponsors in ssocition with Sponsors: Aqueduct Bnff Interntionl Reserch Sttion Cndin Avition Electronics Centre de recherche mthémtiques The Fields Institute Mplesoft The McLen Foundtion Nelson The Pcific Institute for Mthemticl Sciences Populr Book Compny RBC Foundtion S.M. Blir Foundtion The Smuel Betty Fund University Prtners: University of British Columbi University of Clgry Dlhousie University University of Mnitob Memoril University University of New Brunswick University of Prince Edwrd Islnd Dept. of Mthemtics & Sttistics, (University of Ssktchewn) University of Toronto York University Government Prtners: Albert Eduction Mnitob New Brunswick Northwest Territories Nov Scoti Nunvut Ontrio Prince Edwrd Islnd Quebec c 017, 018 Cndin Mthemticl Society Pge 19