ELEMENTARY PROBLEMS AND SOLUTIONS

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1 ELEMENTARY PROBLEMS AND SOLUTIONS Edited by A. P. HILLMAN Uiversity of Mew Mexico, Albuquerque, IMew Mexico Sed all commuicatios regardig Elemetary P r o b l e m s ad Solutios to P r o f e s s o r A, P. Hillma, Departmet of Mathematics ad Statistics, Uiv e r s i t y of New Mexico, Albuquerque, New Mexico, Each problem o r solutio should be submitted i legible form, preferably typed i double s p a c - ig, o a separate sheet o r sheets i the format used below. Solutios should be received withi three moths of the publicatio date, Cotributors (i the Uited States) who d e s i r e ackowledgemet of receipt of their cotributios a r e asked to eclose self-addressed stamped p o s t c a r d s. B-172 Proposed by Gloria C. Padilla, Albuquerque High School, Albuquerque, New Mexico. that Let F 0 u = 0, ' i Fi = 1,» ad F l 0 = F + F ( 1 for = 0, 1, ' e. Show p3 _ -p3 + -p3 -j- 3F F F B-173 Proposed by Gloria C. Padilla, Albuquerque High School, Albuquerque, New Mexico. F Q = F* - F 3 i - 3F F L 1F B-174 Proposed by Mel Most, Ridge field Park, New Jersey. Let a be a o-egative iteger., i the sequece 2 F a + l > ***** 2 3 F a + 3 ' - " all differeces betwee successive t e r m s m u s t ed i the s a m e digit. 545

2 546 ELEMENTARY PROBLEMS AND SOLUTIONS [Dec. B-175 Composed from the Solutio by David Zeitli to B-155. qu. ^ Let r ad q be costats ad let UQ = 0, Ui = 1, U 9 = ru U i H a U + b - U + a + b U = 1 U a U b - B-176 Proposed by Phil Maa, Uiversity of New Mexico, Albuquerque, N M. Let deote the Fiboomial Coefficiet F F - 1 F - r + 1 / F i F F T Fi s - [" J 2 ] - 4 3'] - E B-177 Proposed by Phil Maa, Uiversity of New Mexico, Albuquerque, N M. Usig the otatio of B-176, show that - C*:"] - * [ " : * ] - -[ t t : x ] * B: for some iteger a ad fid a. SOLUTIONS A VERY MAGIC SQUARE B-154 Proposed by S. K L. Rug, Jacksoville Uiversity, Jacksoville, Fla. What is special about the followig "magic" s q u a r e? ^

3 1969] ELEMENTARY PROBLEMS AND SOLUTIONS 547 Solutio by the Proposer. (a) The sum of all the umbers cotaied i a row, o r i a colum, o r i a diagoal is a p r i m e. (b) The sum of the s q u a r e s of all umbers cotaied i a row, colum, o r diagoal i s also a p r i m e, Solvers Guy A. Guillottee (Quebec, Caada) ad Michael Yoder listed (a) above. observatio A P E L L NUMBERS IDENTITY B-155 Composite of proposals by M. N. S. Swamy, Nova Scotia Techical College, Halifax, Caada, ad Carol Ae Vespe, Uiversity of New Mexico, Albuquerque, New Mexico. 2P _ + P. +1 Let the Pell umber be defied by P 0 = 0, P A = 1, ad P 2 = P, P,TU - P,,wP = ( - D P Px.. +a +b +a+b a b Solutio by Wray G. Brady, Uiversity of Bridgeport, Bridgeport, Co. Oe fids that P = r s - 2N/2 9 where r = 1 + A/2, s = 1 - <s/2, ad r s = - 1. The oz-ri TI -o -r, \ 2+ab +b +a +a +b, 2+a+b 8(P, P - - P _^,,, P ) = r - r s - r s + s +a +b +a+b 0,,,,,,,, 0,,,, 2+a+b +a+b +a+b, 2+a+b - ( r - r s - r s + s v ) / ^ / a+b, a+b a b b a = (-1) (r + s - r s - r s x ) ad the desired result follows. = ( - l ) ( r a - s a ) ( r b - s b ) = 8 P a P b ( - l ), EDITORIAL NOTE: Let f (x) = 0, t t {x) = 1 ad f, 9 (x) = xf (x) + f (x); the f (2) = P. N Also solved by Herta T. Freitag, Guy A. Guillottee (Quebec, Caada), Serge Hameli (Quebec, Caada), Bruce W. Kig, C B. A. Peck, A. G. Shao (Boroko, T.P.N.G.), Gregory Wulczy, Michael Yoder, David Zeitli, ad the Proposers.

4 548 ELEMENTARY PROBLEMS AND SOLUTIONS [ Dec - PERIODIC REMAINDERS B-156 Proposed by Alla Scott, Phoeix, Arizoa. Let F be the Fiboacci umber, G = F 4, - 2, ad H be the remaider whe G is divided by 10. J (a) the sequece H 2, H 3, H 4,' is periodic ad fid the r e - peatig block. (b) The last two digits of G 9 ad G 14 give Fiboacci umbers 34 ad 89 9 respectively. Are there ay other cases? Solutio by Herta T. Freitag, Hollis, Virgiia. (a) Sice F t =, 1 (mod 10),, F 59 = 1 (mod 10), F 60 = 0 we have Fj = F - ^ 1 5, (mod 10), where k is ay positive Also G = F 4-2 = H (mod 10) ad 2 = 2, 4, 6, 8, 0 (mod 10) for = 1, 2, 3, 4, 0 (mod 5). Thus the repeatig block of 15 terms Hi, H 2,, H l5 is 1, 7, 8, 9, 5, 6, 7, 3, 4, 5, 1, 2, 3, 9, 0 ad H. = H. -,_, for itegers i ad k. (b) By studyig the correspodig patter modulo 100 we detect aother periodicity cycle such that all Fiboacci umbers smaller tha 100 must occur withi the last two digits of G provided 1 < < 150. More explicitly, we idicate the G correspodig to a give F i the followig table: Subscript o F1 l o r 2 Subscript o Gi 1 l o r 26 Subscript o F1 7 Subscript o G] j 103 or 128 I or or or or or 139 Also solved by Serge Hameli (Quebec, Caada), C. B. A. Peck, ad the Proposer. A TELESCOPING SUM B-157 Proposed by Klaus Guther Recke, Uiversity of Gottige, Germay. that Let F be the Fiboacci umber ad {g } ay sequece. Show

5 1969] ELEMENTARY PROBLEMS AND SOLUTIONS 549 E ^k+2 + Sk-fl - g k ) F k = g + 2 F + g + 1 F g ±. k=l Solutio by Joh E. Horer, Jr., Uio Carbide Corporatio, Chicago, Illiois. The sum is equivalet to Z g k ( F k-2 + F k - 1 " F k> + S + 2 F + W F + F - 1 ) + g 2 F l - g 2 F 2 - g l F l k=3 = + '+2 gr.4-9. g 4.lF 5 +l ' sice F k-2 + F k F k = yl/so solved by Wray G. Brady, Herta T. Freitag, Serge Hameli (Quebec, Caada), Bruce W. Kig, Peter A. Lidstrom, Joh W. Milsom, C. B. A. Peck, A. G. Shao (Boroko, T.P.N. GJ, Michael Yoder, David Zeitli, ad the Proposer. ANOTHER TELESCOPING SUM B-158 Proposed by Klaus Guther Recke, Uiversity of Gottige, Germay. Z (kf k ) 2 = [( )F^+2 - ( )F^+1 - ( )F^ ]/2 k=l Solutio by David Zeitli, Mieapolis, Miesota. Let H satisfy H = H - + H. Notig that H 2 = 2H 2 + 2H 2 - H 2, it is easy to show, usig mathematical iductio, that

6 550 ELEMENTARY PROBLEMS AND SOLUTIONS [Dec. 2 ] T k 2 H 2 = ( )H 2 - ( )H 2 - ( )H 2 + C, k=l where C = 6H? + 2H2-8HiH 2 We ote that C = 0 whe H, = F T o r H, = L,. k k k k Also solved by Herta T. Freitag, Serge Hameli (Quebec, Caada), Joh E. Homer, Jr., Bruce W. Kig, Peter A. Lidstrom, Joh W. Milsom, C B. A. Peck, A. G. Shao (Boroko, T.P.N.G.), Michael Yoder, ad the Proposer. THE EULER TOTIENT B-159 Proposed by Charles R. Wall, Uiversity of Teessee, Koxville, Te. Let T tl be the triagular umber ( + l ) / 2 ad let cp() be the Euler totaet. <P(T ) for = 1,2,*. Solutio by Michael Yoder, Studet, Albuquerque Academy, Albuquerque, N. M. We a s s u m e it is kow that <^(ab) = cp(a.)cp(h) if (a,b) = 1; cp{) is eve if > 2; ad cp(2 ) if k > 1. Let = 2 s 9 where 2 j[ s. The proof is i three c a s e s. Case 1. k = 0. The <p(t ) = cpl ^ ^ T / = (p()cp\^^j, sice / * + 1\,

7 1969] ELEMENTARY PROBLEMS AND SOLUTIONS 551 Case 2. k = 1. Note that so cp{) = (p(2)cp(s) = <p($) = <p I ^(T ) = < W ( + 1) = g()cp( + 1). Case 3. k > 1. Now cp() = cp{2 )cp(s) = 2 cp(s) 9 ad <p $S = 2 k " 2^(s). Also we obviously have + 1 > 2; so let cp( + 1) = 2m 9 where m is a iteger. The <p(t ) = <p / ] <p( + 1) = 2 k - 2 <p(s)2m = m 2 k ~ 1 cp(s) = mcp(). -4&o so/m/ by Herta T. Freitag, Guy A. Guillottee (Caada), Serge Hameli (Caada), Douglas Lid (Eglad), C B. A. Peck, Gregory Wulczy, ad the Proposer. [Cotiued from page 538. ] or * * 491 = x(x + 1) + y(y + 1) + z(z + 1) - This is impossible, sice x(x + 1), y(y + 1), ad z(z + 1) are all eve. *