7. arithmetic & number theoretic recreations a. Fibonacci numbers

Prob. 169, pp. 60-61 & 120. 21 lambs in four pens, each with an odd number. Same solution idea as in prob. 40.

Gibson. Op. cit. in 4.A.1.a. 1963.

P. 68: Odd figuring. Use seven odd figures to add up to 20. Answer:  13 + 3 + 1 + 1 + 1 + 1.

P. 69 & 74: Cross-out groups. Make a row of 20 labelled marks and cross out marks in consecutive groups so that the last mark is odd each time. Easy -- just mark backward!

Doubleday - 1. 1969. Prob. 71: Ups and downs, pp. 87 & 170. = Doubleday - 5, pp. 97-98. Labourer has to carry 85 bricks up to a bricklayer in a hod that can hold 16, but he is instructed to always carry an odd number. How does he do it in six journeys? Solution is to carry up 15 on five trips and then bring one down and carry up the remaining 11. To me, the return stages should all be alike and one could do this by carrying up and carrying down one five times, then carrying up the remaining 15. Of course, the last stage cannot be expected to be the same as the others -- indeed he may be sent off on some other task.

[Henry] Joseph & Lenore Scott. Master Mind Brain Teasers. 1973. Op. cit. in 5.E. Three short problems -- no. 1, pp. 41-42. Have five odd numbers add to 14. 11 + 1 + 1 + 1  =  14.

7.AO. DIVINATION OF A PERMUTATION
There are simpler versions of this used to divine three numbers, e.g. to locate a ring on a person, finger and digit -- a common one uses: x *2 +5 *5 +10 +y *10 +z  =  350 + 100x + 10y + z -- cf Fibonacci, Tagliente (1515). The operations are performed from left to right, corresponding to the instructions given. I will make no attempt to trace this very common but very dull type of problem, but see 7.M.4.b, especially the Folkerts entry which cites six early sources for this type of divination.

If the a_i are a permutation of 1, 2, ..., n, the method of interest forms P  =  a₁m₁ + a₂m₂ + ... a_nm_n for multipliers m_i. By appropriate choice of the m_i, the value of P determines the permutation. Generally P is subtracted from some convenient constant. Sometimes the solution uses a division to yield a₁ and a₂. Some formulae work even if a_i are not a permutation, but are digits or dice values. If we have a permutation, one can ignore a_n since it is determined by the others, i.e. one can let m_n = 0.

NOTE. The standard form of this problem has n objects permuted among n people. The permutation (a_i) can be viewed in two ways.

1. The more natural view is that a_i = j means the i-th person has object j. Then the multipliers are associated with the people.

2. The inverse view is easier to implement and hence much more common in this problem. We view a_i = j as meaning that the i-th object is with person j. Then the multipliers are associated with the objects.

The most common form of the problem is with three people and 24 counters. You give the people 1, 2, 3 counters to start and leave the other 18 counters on the table. Let them secretly permute three items, say A, B, C among themselves. You tell the person with object A to take as many counters as he already has; the person with object B to take twice as many; the person with object C to take four times as many. In our inverse view, we are taking P = a₁ + 2a₂ + 4a₃ from 18.

NOTATION: M = (m₁, m₂, ..., m_n) denotes this process.

A sequence of multipliers, M, is suitable if the products P are all distinct. In the case that the (a_i) are a permutation, it is easy to see that the following processes preserve the suitability of M. (1) Permuting the m_i. (2) Shifting all the m_i by a constant. (3) Multiplying all the m_i by a non-zero constant. Thus we can arrange the m_i in ascending order and make m₁ = 0 and m₂ = 1. So for the case n = 3, any sequence can be brought to the form (0, 1, m₃). By subtracting from m₃ and scaling, we see that this sequence is equivalent to (0, 1, m₃/(m₃-1)). A little work shows that either m₃  2 or m₃/(m₃-1)  2, so we can assume m₃  2. m₃ = 2 gives an unsuitable M, but m₃ > 2 always gives a suitable M. So the simplest possible case is (0, 1, 3), which is equivalent to the most common case (1, 2, 4) and to (3, 4, 6). Cases equivalent to (0, 1, 4), (0, 1, 6) and (0, 1, 8) also occur.

For the case of permutations, subtracting P values from some constant S is equivalent to changing m_i to S/Σa_i - m_i.

As in the Josephus Problem, mnemonics were constructed. For the case n = 3, objects were labelled by the vowels a, e, i and mnemonics were constructed consisting of words (or phrases) with these three vowels in all six permutations (or having just the first two vowels of each permutation). See Bachet for an example. Gardner also gives mnemonics using consonants.

See Meyer for a slightly more complex multiplication process, which can be reduced to the usual form.

See Tropfke 648 651.

COMMON VOWEL MNEMONICS -- note that spelling and layout vary.
Angeli, Beati, Taliter, Messias, Israel, Pietas. Baker;

Angeli Beati Pariter Elias Israel Pietas. Hunt.

Anger, fear, pain, may be hid with a smile. Magician's Own Book; Boy's Own Conjuring Book;

Aperi, Premati, Magister, Nihil, Femina, Vispane, Vispena. Minguét.

Aperì Prelati Magister Camille Perina Quid habes Ribera. Alberti 76-77;

Avec éclat L'Aï brillant devint libre. Labosne, under Bachet; Lucas;

Brave dashing sea, like a giant revives itself. Magician's Own Book; Boy's Own Conjuring Book;

Graceful Emma, charming she reigns in all circles. Magician's Own Book; Boy's Own Conjuring Book;

Il a jadis brillé dans ce petit État. Lucas;

James Easy admires now reigning with a bride. Magician's Own Book; Boy's Own Conjuring Book;

Pallētis Evandri Sanguine Feritas Imane (the m should have an overbar) Vigebat. Schott.

Par fer César jadis devint si grand prince. Bachet; Ozanam 1725, prob. 46; Alberti 76-77; Les Amusemens; Hooper; Magician's Own Book; Boy's Own Conjuring Book; Boy's Own Book; Magician's Own Book (UK version); Lucas;

Pare ella ai segni; Vita, Piè. Alberti 76-77.

Salve certa anima semita vita quies. Van Etten; Schott (with one respelling); Ozanam 1725, prob. 46; Alberti 76-77; Manuel des Sorciers; Endless Amusement; Parlour Pastimes; Magician's Own Book; Boy's Own Conjuring Book; Magician's Own Book (UK version);

Take her certain anise seedlings Ida quince. Magician's Own Book (UK version);
Tabari. Miftāh al-mu‘āmalāt. c1075. P. 109, part IV, no. 17. ??NYS - described by Tropfke 648. As in Fibonacci, below, with a = 18 and M = (2, 17, 18).

Fibonacci. 1202.