ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A.P. HfLLIVSAgy Uiversity of New Mexico, Albuquerque, l\9ew Mexico Sed ail commuicatios regardig Elemetary Problems to Professor A.P. Hillma; 709 Solao Dr., S.E.; Albuquerque, New Mexico Each solutio or problem should be o a separate sheet (or sheets). Preferece will be give to those typed with double spacig i the format used below. Solutios should be received withi four moths of the publicatio date. DEFINITIONS The Fiboacci umbers F ad the Lucas umbers L satisfy F +2 = F +1 +F, F 0 «0, F x «/ ad L +2 = L +<i + L, L 0 = 2, L l = /. PROBLEMS PROPOSED IU THIS ISSUE B-316 Proposed by J. AH. Huter, Fu with Figures, Toroto, Ot, Caada. Solve the alphametic: TWO E I G H T Believe it or ot, there must be o 8 i this! B-317 Proposed by Herta T. Freitag, Roaoke, Virgiia. Prove that /-2A?-/ > S a e x a c t divisor of L.4 -j - 1 for = 1,2,. B-318 Proposed by Herta T. Freitag, Roaoke, Virgiia. Prove that F\ + 8F2 ( F 2 + F 6) «s a perfect square for/7 - /, 2, > >. B-319 Proposed by Wray G. Brady, Slippery Rock State College, Slippery Rock, Pesylvaia. Prove or disprove: J- + ± + JL+...= JJj ± + ±_...\ L.2 L 6 L x% s/5\f 2 F 6 F 10 J B-320 Proposed by George Berzseyi, Lamar Uiversity, Beaumot, Texas. Evaluate the sum: Yl F k F k+2m > B-321 Proposed by George Berzseyi, Lamar Uiversity, Beaumot, Texas. Evaluate the sum: 373

2 374 ELEMENTARY PROBLEMS AND SOLUTIONS [DEC. B-292 Proposed by Herta T. Freitag, Roaoke, Virgiia. J3 F kfk+2m+1- SOLUTIONS A COMBINATORIAL PROBLEM Obtai ad prove a formula for the umber S(,t) of terms i (x t + x 2 + ~' + x ) t, are itegers with > 0, t > 0. where ad t I. Solutio by Graham Lord, Secae, Pesylvaia. S(,t) is the umber of uordered selectios of size t ad a set of elemets, that is: S(,t)= { + \ - 1 ). This is a well kow result See for example H.H. Ryser, "Combiatorial Mathematics," Cams Moograph, America Math Associatio, //. Solutio by Frak Higgis, Naperville, Illiois. S(,t) = ( + t t - 1 ). For = 1, the formula clearly holds for all itegers t> 0. Suppose the formula holds for some iteger > 1 ad all itegers t > 0. Now, for ay iteger t>0, we have that (x x +x x + x H ) t = [(x % + x 2 +~+x )+x +1 ] t - ] ( ) fri t + x* + "'+x ) tmk x1. 1 ad hece, by the iductio hypothesis, that ^w-» + ;:*-').(Y') which completes the proof. Also solved by Paul S. Brue km a, Jeffrey Shall it, A C. Shao, Gregory Wulczy, ad the Proposer. THE FIRST SIX FIBONACCI TERMS B-293 Proposed by Harold Do Alle, Nova Scotia Teachers College, N.S., Caada. Idetify T, W, H, R, E, F, I, Vad? as distict digits I 1,2, -, 9 such that we have the followig sum (i which 1 ad 0 are the digits 1 ad 0): 1 1 TWO F I V E E I G H T

3 197B] ELEMENTARY FBOBLEIVSS AND SOLUTIONS 37i Solutio by George Berzseyi, Lamar Uiversity, Beaumot, Texas. It is easy to see that the digit carried from the thousads colum must be 1; cosequetly, T+ 1 = E. Applyig this fact to the oes colum yields the cogruece 2T + 4 = T (mod 10) whose oly solutio is T= 6. Therefore, E = 7 follows. O the basis of the thousads colum oe ca also easily deduce that / < 5. Furthermore, it is evidet that the values of Vad M^are iterchageable. The value of H determies the possible values for V ad W, resultig i the followig te cases: (1) H=1;V,W.<=: is,8\: (2) H=1;V,We {4,9\; (Z\ H = 2;V,W^ {13}; (4) H = 2;V,Wsz {5,9}; (5) H = 3;V,WG \l/\; (6) H = 4;V,W<= il5.\ : (7) H = 5;V,We {3,4}; (8) H = 5;V,W<a {8,9}; (9) H = 8;V,W<E {1,9}; (10) H = 9;V,We {3,8}. All but two of these lead to cotradictios. Case (4) yields oe solutio, from Case (9) two solutios are obtaied; they are give below As remarked earlier, upo iterchagig the values of fad W, three additioal solutios may be give, it may be of iterest to ote that the umber of essetially differet solutios, the possible values of F (commoly used to deote the Fiboacci umbers), as well as the possible values of H (ofte used to deote geeralized Fiboacci umbers) are all Fiboacci umbers. Also (partially) solved by Paul S. B rue km a, Warre Cheves, J. AH. Huter, Joh W. Miisom, Carl Moore, Jim Po A. C. Shao, ad the Proposer. A FORMULA SYMMETRIC IN &AND B-294 Proposed by Richard Blazej, Quees Village, New York. Show that Solutio by Frak Higgis, Naperville, Illiois. Usig the Biet formulas we have F L k + F k L = 2F +k. Also solved by George Berzseyi, Paul S. Bruckma, Warre Cheves, Herta T. Freitag, Mike Hoffma, Peter A Lidstrom, Graham Lord, Joh W. Miisom, Carl Moore, F.D. Parker, Jeffrey Shailit, A.C. Shao, Paul Smith, Gregory Wulczy, ad the Proposer. CONVOLUTION OR DOUBLE SUM B-295 Proposed by V.E. Hoggatt, Jr., Califoria State Uiversity, Sa Jose, Califoria. Fid a closed form for J^ (+1-k)F 2k = F 2 + (-1)F4 + -+F 2. =1

4 376 ELEMENTARY PROBLEMS AND SOLUTIONS [DEC. Solutio by Graham Lord, Secae, Pesylvaia. The sum of the first k odd idexed Fiboacci umbers is F 2^ aa " t a t of the first k eve idexed oes is F2k+1 ~ 1, where k> 1. Therefore, i X) (+1-k)F 2k * E F * = E < F 2i+l-1) k=l j=i i=7 M = F2(+1)-~ 1 - NOTE: Compare B-290. Also solved by George Berzseyi, Paul S. Bruckma, Herta T Freitag, Frak Higgis, Mike Hoffma, Pe Lidstrom, Carl Moore, Jeffrey Shallit, AC Shao, Paul Smith, ad the Proposer. A MOST CHALLENGING PROBLEM B-296 Proposed by Gary Ford, Vacouver, B.C., Caada. Fid costats a ad b ad a trascedetal fuctio G such that wheever y satisfies y +2 = ay +l + V- G(y +3) + G(y ) + G(y + 2 )G(y +1 ) I. Solutio by Carl F. Moore, Tacoma, Washigto. Two solutios are give by: (1) a = b = 1 ad G(u) = 2cosu, (2) a = b=1 ad G(u) = c u + c~ u (c*1). [Notice G(u) = 2 cosh u is a pleasig special case.] To show (1), GlY+3) + G(Y) = 2 cos (y +3) + 2 cos (y ) = 2 /"cos (y +3> + cos (y )) To show (2), * (2 cos (y + 2 ))-(2 cos (y +1 )) = G(y +2 )-G(y +1 ). G(y +3) + G(y ) = (c V+3 + C- V+3 ) + (c V +c- V y= cy+2+y+1 +c-y+2-v+1 + cv+2-y+1 + cv +1-V+2 = cy+2. c v "+1 + c' V+2, C' V+1 + C V " +2 -c' V+1 + C V+1 -c' V+2 = (c y " +2 + c- y +2 Hc V+1 +c- y+1 ) = G(y +2 ).G(y +1 ). II. Solutio by the Proposer. Let G(x) = c x + c~ x, with c ay (complex) costat ad let y I be a geeralized Fiboacci sequece (satisfy ig y +2 = Y+l + y^ad havig ay iitial coditios). There were o other solvers.

5 1975] ELEMENTARY PROBLEMS AND SOLUTIONS 377 PARTIAL FRACTIONS B-297 Proposed by Paul S. Bruckma, Uiversity of Illiois, Chicago Circle, Illiois. Obtai a recursio formula ad a closed form i terms of Fiboacci ad Lucas umbers for the sequece (G ) defied by the geeratig fuctio: (1-3x-x 2 +5x z +x A -x 5 )- 1 * G 0 + G t x + G 2 x G x + -. Solutio by David Zeitli, Mieapolis, Miesota. We ote that G +s - 3G +4 - G G G +f~ G = 0. Sice (1-3x-x 2 + 5x z + x 4 - x 5 ) = (I - 3x+x 2 )(1 -x - x 2 )(1+x), we obtai, usig partial fractios, if W +2 = aw +i + bw, the Thus, Thus, 10 s 18-7x _ 5(2+x) x-x 2 +5x 3 +x 4 -x 5 "" 1-3x+x 2 1-x-x 2 1+x ' WX = ^ - y. " I -ax-hx 2 =0 oo oo oo 2 + x 18-7x _ V * / * i-3x+x> - ^ i X ' T^T^ V /: «!/" 1 - V / i) v F»+3 X X - L. =0 =0 =Q G = jjf (L F (~1)V. Also solved by Frak Higgis, Carl F. Moore, AC. Shao, Gregory Wulczy, ad the Proposer. { - " X -