Puzzle Corner . Allan Gottlieb
How's Your enue, Stafford Staffs ST17 OEF, England) taken as 100, what is the distance DX? has started a newsletter called Puzzle (The answer is a definite number, not a Calculus? World devoted to mechanical puzzles. formula.) Judith Longyear has suggested an in formal poll of our readers to see which OCT 5 How's your calculus? Harvey kinds of problems (chess, bridge, Elentuck asks for the area of the loop of Allan Gottlieb studied geometry, cryptarithmetic, etc.) are most Y2 = (X + 4)(X2 - X + 2Y - 4). mathematics at M.I.T. (and least) appreciated. You are all wel A noncalculus solution to this would be (S.B. 1967) and Brandeis come to respond, and significant prefer very impressive, but calculus is permitted. (A.M. 1968, Ph.D. 1973); ences will effect future problem selections. he is now Assistant Pro Severalreaders noted an error in the Speed Department fessor of Mathematics and published solution to NS9. That problem OCT SD 1 Ruth Duffy asks usto name a Coordinator of Computer is now reopened. word in the English language with seven Mathematics at York College of the City letters, five of which are the vowels a, e, i, University of New York. Send problems, o, and u (but not necessarily in alpha solutions, and comments to him at the Problems betical order). Department of Mathematics, York Col OCT 1 William Butler wonders how lege, Jamaica, N.Y., 11451. normal is normal (in bridge, at least): OCT SD 2 The Editor of the Review dis Conventional point counting gives four covered the following problem being dis points for each ace, three per king, two tributed as part of a tongue-in-cheek per queen, andone per jack. The average As a new volume begins, let me review the bridge hand has ten such points. What is "quiz" prepared by M.I.T. students for exhibitors in the 1978 Massachusetts Sci ground rules of "Puzzle Corner" for new theprobability of receiving a hand with ence Fair: readers. exactly ten points? Each issue we publish five regular prob Translate into a limerick: lems and two "speed" problems. Three is OCT 2 Sebastian Batac would like to find (12 + 144 + 20 + 3V?)/7 +5X11 sues later the solutions to the regular two positive rational numbers the sum of 92 problems appear. This month, for exam whose cubes is 6. In other words, find ple, we are printing the solutions to prob positive integers a, b, c, and d satisfying Solutions lems published last May. Challenges to (a/b)3 + (c/d):l = 6. NS 10 (This was first published as 1974 published solutions and acknowledge MIA 2 and never solved; it was published ment of late responses appear in the "Bet OCT 3 The following cryptarithmetic again as NS 10 in February, 1978: ter Late Than Never" department. The problem from Avi Ornstein consists of Find a closed form for "speed" problems are not to be taken too two mathematical statements which are l' + 22+ ... + n". seriously. Often whimsical, their solutions correct in base 10 when digits are substi areusually given the same issue as the tuted for letters and is alsotrueas read for When this was first published Leo Epstein problem is posed, and they rarely appear modulo 9 mathematics: supplied some asymptotic formulas. He in the "Better Late Than Never" depart SIX + TWO + TWO = ONE has improved these, but we still have no ment. SIX + SIX = TWO + ONE exact closed form. Perhaps none exists. Here is some news from our readers: 1 remember that during my senior year OCT 4 J. Friedman sends me a number NS 12 A standard deck of 52 cards is at M.I.T., many of the graduating seniors of problems published by Calibrom Prod shuffled and placed face down upon the were considering their chances for ac ucts as advertising in Technology Review; table. The cards are then turned face up ceptance at various graduate schools. One this one appeared in 1938: one at a time by flipping over the top card of my friends, Mike Rolle, decided to en of the face-down stack. As this is done, the hance his chances by solving the famous player simultaneously calls out the se .four-color conjecture. Since generations of quence A, 2,3,4,5, 6,7, 8, 9,10, J, Q, K, mathematicians had failed in this attempt, A, 2, etc., one call being made for each we didn't feel that Mike had much hope of card flipped over. To win the game, one success; but he was serious. During that mustgo through the deck without match year he actually obtained some impressive ing a card flipped over with the card partial results, but the conjecture was still called. Suits don't matter, so, for example, unsettled. The end of this story occurred any 4-spot flipped over on the 4th, 17th, this year after Appel and Haken finally 30th, or 43 rd turn results in a loss. "Since solved the problem. 1 was reading their winter will surely come again," Mr. Con- important papers in the Illinois Journal of nine would like to know what arethe Mathematics and noticed an acknowl chances of winning the game. How about edgement to one Michael Rolle for his a second solution for the same game with help. Congratulations, Mike; he who a 48-card pinochle deck? laughs last ... This problem is not trivial! Judith Congratulations, too, to Frank Rubin: Starting with any triangle, construct three Longyear gave a colloquium talk on her the Journal of Recreational Mathematics exterior triangles having base angles of results in 1974. The answer is not (12/ will have a special Frank Rubin issue next 30" and vertices at D, E, and X — as indi 13)52. Although the probability of success April. ... Dale Overy (27 Bodmin Av cated in the diagram. If the distance DE is for any one card is 12/13, the events are
76 Technology Review, October, 1978 Puzzle Comer
not independent. Bob Kimble and an un swer tothe problem is e~4 — i.e., about "oenanthaldehyde" are excluded, since named computer assert that of the 52! one in 53. (A copy of the notes on com they are simply variant spellings or variant possible decks exactly 1, 309, 302, 175, plete permutations may beobtained from pronunciations of the same word. 551, 177, 162, 931,045, 000,259,922, the editor on request.) Dennis Kluk submitted a list which will 525,308, 763, 433, 362, 019, 257, 020, behard to beat. (I must add that some of 678, 406, 144 are winners. MAY 1 White to play and win: his words are not in my vocabulary; They also solved the pinochle problem:of perhaps I should have specified a (small) the 48! decks 2, 173, 013,719,746, 911, dictionary in which all words were re 580, 113, 686, 677, 997, 894, 282, 336, quired to appear.) Mr. Kluk's list, which 936,761, 753,600,000,000 are winners. follows, comes from Word Ways, subti Since 52! = 80, 658,175,170, 943, 878, tled the journal of recreational linguistics: 571, 660, 636, 856, 403, 766,975, 289, Aquintocubitalism, Blithesomeness, 505, 440, 883,277, 824, 000,000,000, Chemotherapeutics, Demulsification(s), 000, they obtain a success probability of Emotionlessness, Frightfulness, Gas- about 1.62 per cent. trophotographics, Hedriophthalmous, Stephen Flaum and a Tl 58 used an Identification^), Japaconitine, Kineasthe- iterative technique: tic, Limitableness, Methylhydrocupreine, At each iteration the probability of losing Neopaleozoic, Oesophagostenosis, Pre- on that iteration is calculated. In addition, mosrepresentation(s), Quinta(s), Revo- the expected number of cards remaining lutionally, Selectiveness, Treasonableness, witheach face value after the iteration, Utopographer(s), Vindictiveness, Whence- conditional on the assumption that the forward, Xanthosiderite, Yourselves, and game is not lost on that iteration, is calcu Zoosporiferous. I lated. These expected valuesare used in Also solved by Harry Hazard, Jacob Bermann, Emmet Duffy,Paul Hertz, and subsequent calculations of the probability Several readers slipped up on this one. the proposer, Donald Forman. of failure. By playing 2K — B5 they allow a neat This method requires fractions to be draw: kept throughout. Flaum and Tl actually MAY 3 Given an n-by-n checkerboard divide out the fraction and use the ap 2. ... R—N8 and n2 checkers of n different colors, and proximating decimal. Perhaps this ex 3. 3P — B8 (Q) R — B8(:h given that there are n checkers of each plains their answer of 1.77 and .0225 per Joseph Seo, however, avoids this and finds color, is it possible to arrange all the n2 cent for pinochle. a solution with only one line: checkers on the board such that no two A. Walther claims the answer is: checkers of the same color lie in the same 1. P —B7 R —N3 ch row, column, or diagonal? (By "diagonal" (§M) Pl/P!)4 2. K—N5 R —N4 ch is meant all the diagonals, not just the two 3. K—N4 R —N5 ch main diagonals.) It turns out that for cer or approximately e~4 (i.e., over 1.8 per 4. K—N3 R —N6 ch tain values ofn it is possible to so arrange cent). His remarks follow: 5. K—B2 R —N7 ch thecheckers; in this case we say a solution 6. K—Q3 R —N6 ch Make on the table an array 13 blocks long exists — e.g., n = 1. But for certain other ch and four blocks wide. Label the four rows 7. K—Q4 R —N5 values of n such an arrangement is impos 8. K—Q5 R —N4 ch with the names of the four suits. Turn the sible — i.e., no solution exists. For which 9. K— Q6 ch cards over, one at a time, and place them R —N3 values of n does a solution exist? in the array, going from left to right and 10. N —K6 R —Nl For some unknown reason, 1 published placing eachcard in the row matching its 11. N—Q8 R —N3 ch this problem twice: once as FEB 3 and 12. K —K7 R —N2 ch suit. After we have gone through the en now again as MAY 3. More surprising tire deck, we have on the table four rows 13. N—B7ch K — any than this is the fact that no reader noticed of 13 cards, one for each suit in the deck. 14. P —B8 (Q) Resigns my error; as soon as I saw the May issue I To win the game no card must be in its Responses also received from Bob made ready for the slings and arrows. The proper place — i.e., each row of13 cards Kimble, William Butler, Rufus Franklin, responses to MAY 3 are consistent with must be a "complete permutation." A Cary Silvcrston, Robert Bart, Smith the published solutionto FEB 3 [seejunel complete permutation is defined as a per Turner, Jerome Taylor, Winthrop Leeds, July, page 27). In short, an algorithm mutation in which none of the elements is Roger and Paul Milkman, Jacob Ber- existsfor N = 6K ± 1 (i.e., N not divisible in its proper place. The theory of complete mann, T. Mahon, Darryl Hartman, by 2 or 3), and several readers assert (without proof) that no solution exists for permutations is developed on some sheets Richard Kandziolka, Richard Hess, saved for me by E. L. O'Neill which you Walther Fischer, and Harry Nelson. the remaining cases. I repeat my comment may want to share with interested readers. of June/July: this looks like an NS prob The ratio of the number of complete per MAY 2 "Dentification" and "identifica lem for the 1980s. mutations to the total number of permu tion" are both English words. For each Responses received from Judith Long- tations for m elements is English letter a, what is the longest string year, William Butler, T. Mahon, Ari Orn- /3 such that both /3 and a-jS (the concate stein, Richard Hess, Bob Leisy, and the nation of a followed by /8) are English proposer, Sheldon Razin. words? Pairs such as "allelujah" and "hal MAY 4 Assign numerical values to each This is very nearly e"1; therefore, the an lelujah"or "enanthaldehyde" and letter:
78 Technology Review, October, 1978 M= 12.9442
HEN AARON)PHARAOH AARON = 37.76° OOYPO BBNYZ AEYDNH ACEPHH ERPZC This is a base-12 cryptarithmetic problem, 8 = 22.239° and those solving it werereminded that MAY 5 A dog is lost in a square maze of duodecimal notation has two extra digits As a check, let the dog start at intersection corridors. At each intersection, he chooses following 9 before reading the radix. For 3. Then a direction at random and proceeds tothe uniformity, these were specified to be P3 = PJ4 + PV4 + 1/4. (4) "dek" (symbol X, numerical value equals next intersection or exits at one of the decimal 10) and "el" (symbol e, numerical sides. His walk is over when he reaches The relations (3) are consistent with (4). one of the sides. What is theprobability P k value equals decimal 11). Then the radix Also solved by Bob Kimble, John that the dog, starting at intersection k,will is "dozen," or "do" for short. Pierce, Jeff McGuire, Steve Rosenberg, exit at the south side? Cryptarithmetic problems tend to be William Butler, Smith Turner, Paul Hertz, popular, and this one, with its base-12 Peter McMenamin, Robert Bart, Jon twist, was no exception. Severalreaders Thaler, Winslow Hartford, Harry asked for more such problems; the best Zaremba, James Tanenbaum, Gerald way to achieve this is to send more in. The Blum, Douglas Szper, Frank Carbin, T. following solution is from Shirley Wilson: Mahon, Richard Hess, Marshall Fritz, 12e Rodney Weatherford, Judith Longyear, 1 2 33J9e)4135391 and the proposer, John Pressing. 3359e "59749 W 1977 DEC 5 Irving Hopkins is not happy 66e7x with Raymond Kinsey's solution. Mr. 327571 Hopkins' comments follow the drawing 302411 3 4 (see above) which shows his solution: 254x0 From my solution, given (0 + ft>) = 60°, 1. H= 1 K = M/a = cos 60° + (cos1 60° + 1)» 2. C =0 = 1.618034 3. A - N = P or 10 (do) + A- N = P, Given a = 8, M = 12.9442 but P = A + 1, so 10 + A - N = P h = a(sin 60°) = 6.928 and hence N = e 0 = arctan (6.928/16.9442) = 22.2390° 4. Z = x The following solution is from Vic 0 = 60° - 0 = 33.76° 5. O= 9 Elias, who was a graduate student in 8 = 120° 6. Since A-R = 9orlO + A-R = 9, physics at Santa Cruz the year I taught y = 180° - 120° = 60° and since A * x and A * e, 10 + A mathematics there. His solution seems to a = arctan (h/(M - 4)) = 37.76° -R = 9 and hence 10 + H - 1 -A me to look like a physicist's; he finds an J8 = 180° - 60° - y = 82.24°. = 9. Therefore, A = 3 and R = 5. extra symmetry most people miss. Of From this, a= 9 + <£ = 60° 7. P = 4, E = A - 1 = 2, D = 8, Y = 7, course, they solve the problemanyway, 5 = a = 0 = 120°, and B = 6. but that's beside the point. He writes: not ( 7) = (a + P) and ( 8 ) = (0 +
Schumacher. P:! = P., = 3/8. (3) more.
Technology Review, October, 1978 79