Some St. Patrick’s Day math

At today’s Gathering 4 Gardner social, Colm Mulcahy presented on two Irish figures in recreational mathematics with whom Martin Gardner corresponded: Victor Meally and Owen O’Shea. Owen O’Shea is the natural successor to Gardner’s “Professor I.J. Matrix” as a prolific generator of numerological coincidences — see for example The Magic Numbers of the Professor (2007). Victor Meally shows up occasionally in Gardner’s Mathematical Games columns, and also in “Problem 3.14” (appropriate for Pi Day!) in The Colossal Book of Short Puzzles and Problems (2006):

One of the satisfactions of recreational mathematics comes from finding better solutions for problems thought to have been already solved in the best possible way. Consider the following digital problem that appears as Number 81 in Henry Ernest Dudeney’s Amusements in Mathematics. (There is a Dover reprint of this 1917 book.) Nine digits (0 is excluded) are arranged in two groups. On the left a three-digit number is to be multiplied by a two-digit number. On the right both numbers have two digits each:

$158\times 23 = 79\times 46$

In each case the product is the same: 3,634. How, Dudeney asked, can the same nine digits be arranged in the same pattern to produce as large a product as possible, and a product that is identical in both cases? Dudeney’s answer, which he said “is not to be found without the exercise of some judgment and patience,” was 5,568:

$174\times 32 = 96\times 58$

Victor Meally of Dublin County in Ireland later greatly improved on Dudeney’s answer with 7,008:

$584\times 12 = 96\times 73$

This remained the record until a Japanese reader found an even better solution. It is believed, although it has not yet been proved, to give the highest possible product. Can you find it without the aid of a computer?

With the aid of a computer (code), it’s easy to confirm that that Japanese reader’s solution is indeed the best of the 11 basic solutions (and Meally’s the runner-up):

134 * 29 = 58 * 67   = 3386
158 * 23 = 46 * 79   = 3634
138 * 27 = 54 * 69   = 3726
174 * 23 = 58 * 69   = 4002
146 * 29 = 58 * 73   = 4234
259 * 18 = 63 * 74   = 4662
186 * 27 = 54 * 93   = 5022
158 * 32 = 64 * 79   = 5056
174 * 32 = 58 * 96   = 5568
584 * 12 = 73 * 96   = 7008
532 * 14 = 76 * 98   = 7448


But now consider the ruminations of another Irishman, James Joyce’s Leopold Bloom, who in Chapter 17 of Ulysses recounts how he became aware of

the existence of a number computed to a relative degree of accuracy to be of such magnitude and of so many places, e.g., the 9th power of the 9th power of 9, that […] 33 closely printed volumes of 1000 pages each of innumerable quires and reams of India paper would have to be requisitioned in order to contain the complete tale of its printed integers […] the nucleus of the nebula of every digit of every series containing succinctly the potentiality of being raised to the utmost kinetic elaboration of any power of any of its powers.

There’s some confusion as to what number Joyce was really talking about (if any); but the mathematical community has apparently settled on $$9^{9^9}$$, as seen in e.g. the “Ulysses sequence” OEIS A054382: $$\lceil\log_{10} 1^{1^1}\rceil, \lceil\log_{10} 2^{2^2}\rceil, \lceil\log_{10} 3^{3^3}\rceil,\ldots$$

The ninth element of this sequence — the number of decimal digits in $$9^{9^9}$$ — is $$369\,693\,100$$. If “closely printed” at the resolution of A Million Random Digits with 100,000 Normal Deviates, printed by the RAND Corporation (no relation) in 1955, the recording of the entire expansion of $$9^{9^9}$$ would take up 148 thousand-page volumes. (Or, if you postulate a thousand physical pages, each printed on both sides: 74 volumes.)

Suppose we allow solutions of Dudeney’s problem to contain powers: not merely $$abc\times de = gh\times ij$$, but for example $$ab^c\times d^e = g^h\times i^j$$. Then there are five more basic solutions possible, the largest of which is

$48^3 \times 9^1 = 2^7 \times 6^5$

The solutions are:

329 * 8^1 = 47 * 56     = 2632
574 * 9^1 = 63 * 82     = 5166
49^2 * 8^1 = 56 * 7^3   = 19208
1^67 * 8^5 = 2^9 * 4^3  = 32768  (also with 1^76)
48^3 * 9^1 = 2^7 * 6^5  = 995328


Clement Wood — compiler of The Best Irish Jokes (1926) — asserts, in the same Book of Mathematical Oddities (1927) which we previously mined in “Mathematical golf” (2023-03-23), that there are only two solutions to the double equality

29 * 6 = 58 * 3 = 174
39 * 4 = 78 * 2 = 156


Wood is correct, and the administration of powers produces no further solutions to that puzzle.

Posted 2024-03-17