7. arithmetic & number theoretic recreations a. Fibonacci numbers



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To Reveal a Person's Thoughts: an excellent Card Trick, pp. 87-89. The UK version: To Tell Cards Thought Of, p. 4, is much briefer. MUTUS DEDIT NOMEN COCIS.

To Guess which of Twenty-Four Cards have been Noted, pp. 82-83. Extension to eight triples using LIVINI LANATA LEVETE NOVOTO.


Elliott. Within Doors. Op. cit. in 6.V. 1872. Chap. 3, no. 9: The pairs repaired, pp. 83-84. MUTUS DEDIT NOMEN COCIS.

Boy's Own Book. The ten pairs of cards distinguished. 1868: 638. NOMEN MUTUS DEDIT RORIS.

Hanky Panky. 1872. To tell two cards out of twenty, p. 254. MISAI TATLO NEMON VESUL.

Berkeley & Rowland. Card Tricks and Puzzles. 1892. Simple Tricks by Calculation No. XII: To name several different pairs of cards chosen by one or more persons, pp. 44-46. SHEER CHAFF USUAL COLOR; MUTUS DEDIT NOMEN COCIS; MISAI TATLO NEMON VESUL (where V = U).

Ball. MRE. 1st ed., 1892. Determination of a selected pair of cards out of ½n(n+1) given pairs, pp. 98-101. Gives general rule for pairs. MATAS DEDIT NOMEN COCIS. Mentions 8 triples: LANATA LEVETE LIVINI NOVOTO, as in Secret Out.

Ahrens. MUS I. 1910. Pp. 148-152. MUTUS DEDIT NOMEN COCUS. Discusses generalizations, including examples with triples, etc. Cites Unger, 1838, for the first idea of finding triples and quotes the mnemonic from Ball.

Indoor Tricks and Games. Success Publishing, London, nd [1930s??].

The wandering pairs, pp. 38-40. MUTUS DEDIT NOMEN COCIS.

The re-united triplets, pp. 40-41. LIVINI LANATA LEVETE NOVOTO.


Phyllis Fraser & Edith Young. Puzzles Quizzes and Games. Bantam Books, NY, 1947. No. 2: Card trick, pp. 97-99. THIGH ATLAS GOOSE BIBLE.

Stewart Judah. The divertisements of Stewart Judah. The New Phoenix, No. 319 (Nov 1954) 83. UNDUE GOANO TETRA RIGID.

Will Dexter. The Illustrated Book of Magic Tricks. Abbey Library, London, nd, but Introduction dated 1957. The Romans had a word for it!, pp. 73-74. DAVID LOVEL IN YON ABBEY; MUTUS NOMEN COCIS DEDIT ("Mutus gave a name to Cocis").

Martin Gardner. Mutus Nomen. Pallbearers Review 5:7 (May 1970) 338. Reprinted in his: Martin Gardner Presents; Richard Kaufman & Alan Greenberg, 1993, pp. 237-238. For 30 cards: LIVELY RHYTHM MUFFIN SUPPER SAVANT. For 20 cards: BIBLE ATLAS GOOSE THIGH.

Walter Gibson. Big Book of Magic for All Ages. Kaye & Ward, Kingswood, Surrey, 1982. Twenty Cards & New Deal Twenty Card Trick, pp. 162-165. Mentions MUTUS ... and says there are many English possibilities and gives RUFUS STEEL TIARA FOLIO. The New Deal version uses a more complicated laying-out process starting with ten piles of two using MAGIC STORE, then laying out according to AGORA METER SCOTS GIMIC.

Karl Fulves. Mutus Nomen. The author, 1998. A 90pp booklet on variations of this classic trick. ??NYR.

At a lunch in 2002, Rob Eastaway did the trick and said he used BIBLE GOOSE ATLAS THIGH and Jeremy Wyndham said he used DAVID LOVEL IN YON ABBEY. I had never met anyone who used anything other than MUTUS ... before!
7.BA. CYCLE OF NUMBERS WITH EACH CLOSER TO TEN THAN

THE PREVIOUS
New section, due to discovering Lewis Carroll's version. I have no idea if this is older than Carroll.
Lewis Carroll. A Tangled Tale. (1885) = Dover, 1965. The pigs. Knot VIII: De omnibus rebus, pp. 52-57 & 132 134. (In the answers, this part of the Knot is denoted §1. The pigs.) "... place twenty four pigs in those four sties, so that, as she goes round the court, she may always find the number in each sty nearer to ten than the number in the last." Answer: 8, 10, 0, 6 -- the key is that "nothing is nearer ten than 10".
7.BB. ITERATED FUNCTIONS OF INTEGERS
See 7.T and 7.AY for some special forms.

Let f(n) be a function of an integer n. The problems in this section concern the behaviour of the sequence of iterations of f: n, f(n), f2(n) = f(f(n)), f3(n), .... There are a great number of behaviours that can occur.

1. The sequence can diverge to infinity.

2. The sequence can converge to a fixed point.

3. The sequence can become cyclic.

4. The sequence can behave chaotically.

Further the behaviour may be different for different starting points.

If f(n) > n in all cases, then the iteration will diverge to infinity and is not very interesting. Hence we are normally only interested in functions which have f(n) < n for some non-zero proportion of the integers.

If f(n)  n in all cases, then the sequence must converge to a fixed point, though different starting points may lead to different fixed points. Then the problem is to identify the fixed points and the sets they attract.

So the most interesting cases have f(n) sometimes greater and sometimes smaller than n. Two classic examples are the following.

The proper sum of divisors, Σ(n), is σ(n) - n, where σ(n) is the classic number theoretic function of the sum of all divisors. The iterates of this have been studied since the 19C as the fixed points are the perfect numbers and 2-cycles are amicable numbers -- see 7.AB. Some larger cycles have been found -- I recall a 5 and a 17 cycle. No provably infinite sequence has been found.

The Syracuse function is the following. If n is even, halve it; if n is odd, form 3n + 1. Some of the sequences have been computed for hundreds of thousands of terms without termination.

Problems of this sort are genuine mathematical recreations but are a bit too elaborate for me to include here. What I will include are functions whose sequences are always finite. These generally are of two types. Functions such that f(n) < n for all n above some limit -- see, e.g., 7.AY, and functions such that f(n) has the same number of digits as n. If f(n) is the sum of some function g(ai) of the digits ai of n, set M = max {g(0), g(1), ..., g(9)}. Then for 10d-1  n < 10d, we have f(n)  (d+1)M and this will be less than n from some point on.
Joseph S. Madachy. Mathematics on Vacation. Op. cit. in 5.O, (1966), 1979. Narcissistic numbers: Digital invariants, pp. 163 165. Based on Rumney and Nelson. Carries on to consider other types of 'narcissistic numbers' which are numbers which are some function of their digits -- various forms are discussed on pp. 165-177.

George D. Poole. Integers and the sum of the factorials of their digits. MM 44:5 (Nov 1971) 278-279. The only integers equal to the sum of the factorials of their digits are: 1,  2,  145, 40585. He has to check up to 2,000,000 before a general method can be used.

In Dec 2001, Al Posamentier mentioned the following cycle of this function:

169  363601  1454  169. I essentially repeated Poole's work, but found five numbers which are equal to the sum of the factorials of their digits, excluding leading zeroes: 0, 1, 2, 145, 40585. Zero is exceptional -- I view it as having no digits, so the sum involved is empty! The largest number for which this function is larger than the given number is 1,999,999 and there are 208,907 such numbers. Using this, one has a limited search to find all the cycles - there are two other cycles, both 2-cycles: 871  45361 and 872  45362. Cf Schwartz and Kiss, below.

Benjamin L. Schwarz. Self-generating integers. MM 46:3 (May 1973) 158 160. Gives a simple condition for functions of the digits so that the number of PPDIs (= SGIs in his notation) must be finite. This applies to the sum of the factorials of the digits.

Peter Kiss. A generalization of a number-theoretic problem [in Hungarian, with English summary]. Mat. Lapok 25:1-2 (1974 (1977)) 145-149. ??NYS -- described in Reviews in Number Theory A62-201F (= MR 55, #12612). An apparent English translation is: A generalization of a problem in number theory; Math. Sem. Notes Kobe Univ. 5:3 (1977) 313-317; ??NYS - described ibid. A62-201G (= MR 57, #12362). He demonstrates that if f(n) is the sum of any function of the digits of n, then the iteration of f must lead to a cycle -- as sketched above. He finds all the cycles for three cases, including for the factorial function. The review mentions Steinhaus (cf section 7.AY) and Poole.

Roger Cook. Reversing digits. MS 31:2 (1998/9) 35-37. The initial problem was to take a four digit number, arrange the digits in ascending order and in descending order and subtract the first from the second. In all cases except aaaa, the iterations converge to 6174 (7641 - 1467 = 6174). He discusses other number of digits and other bases.
7.BC. UNUSUAL DIFFICULTY IN GIVING CHANGE
The typical problem here is that a customer offers a note to pay a bill and the shopkeeper says he can't give change, but when the customer offers a larger note, then the shopkeeper can give change.

Some of the problems in 7.P.1 could be considered as falling into this topic.

New section. I have seen a 1935 version somewhere.
Carroll-Collingwood. c1890 Prob. 2, pp. 317-318. = John Fisher; The Magic of Lewis Carroll; op. cit. in 1; p. 79. = Carroll-Wakeling II, prob. 31: Coins, pp. 49 & 72-73. A man has a half-sovereign (10s), a florin (2s) and a 6d and wants to buy goods worth 7s 3d. However the shopkeeper has a crown (5s), a shilling (1s) and a penny (1d), so cannot make change. A friend comes into the shop and he has a double florin (4s), a half-crown (2s 6d), a fourpenny piece (4d) and a threepenny bit (3d). Can they sort out the payment? I consulted a dictionary and found that the double-florin was first minted in 1887, which puts a lower bound on the date of this problem, so I have given the date as c1890 rather than c1890?

Clark. Mental Nuts. 1916, no. 4. The conductor's money. Passenger tenders a $1 note to pay 5¢ (= $.05) fare, but conductor cannot give change, but he can give change for a $5 note. The answer is that he can give a $2.50 gold piece, a $2 note and 45¢ in coins.



David Singmaster. No change! Written up as a problem for my Telegraph column in summer 1999. Shopkeeper cannot give change for a £5 note, but can for two £5 notes. How can this happen? How about in the US?

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