ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P. HILLMAN University of New Mexico, Albuquerque, New Mexico

Transcription

1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P. HILLMAN Uiversity of New Mexico, Albuquerque, New Mexico Sed all commuicatios regardig Elemetary Problems ad Solutios to Professor A. P. Hillma, Departmet of Mathematics ad Statistics, Uiversity of New Mexico, Albuquerque, New Mexico, Each problem or solutio should be submitted i legible form, preferably typed i double spacig, o a separate sheet or sheets, i the format used below. Solutios should be received withi three moths of the publicatio date. Cotributors (i the Uited States) who desire ackowledgemet of receipt of their cotributio are asked to eclose self-addressed stamped postcards. B-160 Proposed by Robert H. Atigli, Da River Mills, Daville, Virgiia. + y 2 = z 2. Show that, if x = F F + 3, y = S F ^ F ^, ad z = F ^, the x* B-161 Proposed by Joh Ivie, Studet at Uiversity of Califoria, Berkeley, Califoria Give the Pell umbers defied by P 2 = 2P P.., P 0 = 0, P* = 1, show that for k > 0; «W [(k-l)/2] P k= ( 2 r\l 2 r - r=0 k P 2k=E(r) 2 r p r r = l B-162 Proposed by V. E. Hoggatt, Jr., Sa Jose State College, Sa Jose, Califoria Let r be a fixed positive iteger ad let the sequeces Uj,U2,* * satisfy u = u - + u., + + u for > r ad have iitial coditios Ui J m r * = 2J"" 1 for j = l,2j***,r. Show that every represetatio of U as a sum 218

2 Apr ELEMENTARY PROBLEMS AND SOLUTIONS 219 of distict u. must be of the form u Itself or cotai explicitly the terms u - l s U -2 J " " ' u r+1 a(^ s o m e represetatio of u. B-163 Proposed by Phil Maa, Uiversity of New Mexico, Albuquerque, New Mexico Let be a positive iteger. Clearly (1 x + N/5) = a + b ^ 9-1 with a ad b itegers. Show that 2 is a divisor of a ad of b. B-164 Proposed by J. A. H. Huter, Toroto, Caada. A Fiboacci-type sequece is defied by: with Gi = a ad G 2 = b* G. 0 = G, - + G ad b 9 subject to a beig odd, to satisfy: Fid the miimum positive values of itegers a G - G ^ - G 2 = - l l l l l ( - l ) for > 1, B-165 Proposed by Douglas Lid, Uiversity of Virgiia, Charlottesville, Va. Defie the sequece b{)j by b(l) = b(2) = 1, b(2k) = b(k), ad b(2k + 1) =b(k + 1) + b(k) for k > 1. For > l f show the followig: (a) b ( [ (-l) ]/3) = F + 1, (b) M p. ^ + H) 1 1 ]^) = L.

3 220 ELEMENTARY PROBLEMS AND SOLUTIONS [Apr. SOLUTIONS A MULTIPLICATIVE ANALOGUE B-142 Proposed by William D. Jackso, SUNY at Buffalo, Amherst, N.Y. Defie a sequece as follows: A A = 2, A 2 = 3, ad A = A -A 2 for > 2. Fid a expressio for A. Solutio by J. L. Brow, Jr., Pesylvaia State Uiversity, State College, Pa. Let B = IA for > 1. The B 4 = I 2, B 2 = I 3 ad B = B - * * B Q for > 2. Clearly & B = F. 0 I 2 + F - I for > 2, o r A = 2 ~ 2 3 " 1 for > 2. Also solved by Christie Aderso, Richard L. Breisch, Timothy Burs, Herta T. Freitag, J. A. H. Huter (Caada), Joh Ivie, Bruce W. Kig, Leslie M. Klei, Arthur Marshall, C. B. A. Peck, Joh Wesser, Gregory Wulczy, Michael Yoder, David Zeitli, ad the proposer. THE DETERMINANT VANISHES B-143 Proposed by Raphael Fikelstei, Tempe, Arizoa. u m b e r s : Show that the followig determiat vaishes whe a ad d a r e atural F a F a + d F a+2d F a+3d I F a+6d F a+4d F a+7d F a+5d F a + 8 d What is the value of the determiat oe obtais by replacig each Fiboacci u m b e r by the correspodig Lucas u m b e r?

4 1969] ELEMENTARY PROBLEMS AND SOLUTIONS 221 Solutio by Michael Yoder, Studet, Albuquerque Academy, Albuquerque, New Mexico Let r = F 6 d / F 3 d ad a = F g d r F ^. The r F 3 d + S F 0 = F 6 d ad r F 3 d s F l = F Qd+1. It follows by iductio that F + 6d - r F + 3d + s F for all ; i particular, it is true for = a, = a + d, ad = a + 2d. Hece the three rows of the matrix are liearly depedet ad the determiat is zero. If each Fiboacci umber is replaced by the correspodig Lucas umber, the determiat will also be zero by similar reasoig. Editorial Note: It ca be show that r = L^, ad s = (-1) Also solved by F. D. Parker, C. B. A. Peck, David Zeitli ad the proposer. LUCAS ALPHABETIC B-144 Proposed by J. AsH. Huter, Toroto, Caada. I this alphametic each distict letter stads for a particular but differet digit, all te digits beig represeted here. It must be the Lucas series, but what is the value of the SERIES? ONE T H R E E START L S E R I E S Solutio by Charle s W. Trigg, Sa Diego, Califoria Sice they are the iitial digits of itegers, oe of 0 9 T, S, or L ca be zero. Proceedig from the left, clearly S = 1, E = 0, ad T is 8 or 9. I either evet, H + T > 10, so T = 8. The from the uits 1 colum, L = 3.

5 222 ELEMENTARY PROBLEMS ANB SOLUTIONS [Apr. The three iteger colums the establish the equalities: N + R + 1 = :R + A + 1 = I + 10 H l = R ' Whereupo, N + H = 10 ad (N,H) = (4,6)' or (6,4). But H = R + 1, so R = 5, H = 6, N = 4. The 0 + A = I + 4, ad 0 = 9, A = 2, 1 = 7. (0 ad A may be iterchaged.) Cosequetly, S E R I E S = Also solved by Richard R. Breisch, Timothy Burs, A. Gommel, Edgar Karst, Joh Milso^ C. B. A. Peck, Joh Wesser, Michael Yoder, ad the proposer. BINARY N-TUPLES B-145 Proposed by Douglas Lid, Uiversity of Virgiia, Charlottesville, Va. Give a ulimited supply of each of two distict types of objects, let f() be the umber of permutatios of of these objects such that o three cosecutive objectives are alike. Show that f () = 2F, i, where F is the th Fiboacci umber. Solutio by Bruce W. Kig, Adriodack Commuity College, Gle Falls, N.Y. Call a permutatio of the required type a Admissible permutatio," ad let A ad B be two of the distict types of objects. A list of admissible + 1 permutatios ca be costructed i the followig way: (a) For each admissible permutatio edig i A, adjoi B o the right; for each distict admissible permutatio edig i B, adjoi o the right. (b) For each distict admissible - 1 permutatio edig i A, adjoi BB o the right; for each distict* admissible - 1 permutatio edig i B, adjoi AA o the right. Certaily the resultig list cotais oly admissible + 1 permutatios. Furthermore, there is o possibility of duplicatio because the permutatios described i (b) ed with two idetical letters, but those described i (a) ed with two differet letters* Lastly, o + 1 permutatio is uobtaiable i

6 1969] ELEMENTARY PROBLEMS AND SOLUTIONS' 223 this way. For, if there were such a permutatio, either the - 1 permutatio excludig its last two letters 5 or the permutatio excludig its last letter would have to be admissible. Cosequetly, we see that f( + l) = f(^ 1) +f(). The rest is a easy proof by iductio. By direct eumeratio, f(3) = 6 = 2F 4. If f() = 2F - for itegers 1 N 9 the f(n + 1) = fou - 1) + f(n) = 2 F N + 2 F N + 1 = 2(F N + F N + i ) = 2 F N + 2 ad the proof is complete. Also solved by J. L. Brow, Jr., C. B. A. Peck, Michael Yoder, ad the proposer. ANGLES OF A TRIANGLE B-146 Proposed by Walter W. Homer, Pittsburgh, Pesylvaia Show that tr = Arcta ( l / F 2 ) + Arcta F Arcta F Solutio by C. B. A. Peck, Ordace Research Lab., State College, Pa. F r o m the solutio to H-82 (FQ ) 9 we get Arcta ( 1 / F 2 ) = Arcta ( l / F ) + Arcta d / F K The result ow follows from Arcta x + Arcta (1/x) = 7r/2. Also solved by Herta T. Freitag, Joh Ivie, Bruce W. Kig, Joh Wesser, Gregory Wulczy, Michael Yoder, ad the proposer, TWIN PRIMES B-147 Proposed by Edgar Karst, Uiversity of Arizoa, Tusco, Arizoa, i hoor of the 66th birthday of Hasraj Gupta o Oct. 9, 1968, Let

7 224 ELEMENTARY PROBLEMS AND SOLUTIONS Apr S = (1/3 + 1/5) + (1/5 + 1/7) + + (1/ /32719) be the sum of the sum of the reciprocals of all twi primes below Idicate which of the followig iequalities is true; (a) S K TT 2 /6 (b) 7r 2 /6..< S < Ve (c) *Je < S Solutios by Paul Sads, Studet, Uiversity of New Mexico, Albuquerque, New Mexico, ad proposer. (Both used electroic computers.) True iequality Number of pairs of primes ivolved S, to six decimal places * Proposer ( b ) 55 lo PaulJ3ads (b) (Cotiued from p. 210.) 6. T = -(-i) 11 T _,, = 5T - 6T T = 2 + S 11 " <s/29 2 s V <^29 with terms 1, 5, 26, 135, r + s with terms 5, 27, 140, 3+WIT s = 3 - in/11 r / W l l \ / i>v/ll\ = \ 5 ) * + { 55 ) S 10. T _,_- = 5 T + 2 T -; T< = 3, T 2 = * * * *