CORRESPONDENCE

New Zealand Listener, Volume 3, Issue 53, 28 June 1940, Page 16

CORRESPONDENCE

‘H. G. Lambert (Taupo): sends, for S.G.E., a formula for giving a series of n consecutive prime numbers. He denies that he is scornful of simpler mathematics and agrees that calculus is neither difficult nor complicated; but a simple method for doing difficult work. He says that the briefest formula for the first member of a series of n consecutive non+prime numbers is (n-+-1) factorial plus 2; but the following rule is quicker to use and gives an infinite number of series: Multiply together the prime numbers up to, but not including, n-+2,. Then multiply the product by any whole number and add two, which gives you the first member of a series. Alternatively, if you multiply the product of the primes by any whole number greater than one, you can subtract two, which gives the last member of a similar series. §.G.E. had asked for a series of 1,000, says H.G.L., but, as the members of such a series would be 350-figure numbers, or bigger, H.G.L. gives only 25. This we have sent on to §.G.E. Before we take S.G.E.’s observations, we must acknowledge H.G.L.’s estimate of the conditions governing the suppositious sweepstake on the Rugby match. (See issue of May 24). He says there are only 39 possible scores fulfilling the conditions set, ‘and. thatthe chances are 38 to 1 against guessing the correct score, no matter how many participate. S.G.E. (Glenavy): writing about the nonprime numbers, gives his own views of a "fairly compact form of the rule’: Suppose it be required (he says) to find a series of 15 consecutive numbers none of which is prime. Let n be the number whose factors have to be determined. Then a series integers is n+2,n+3..... -n-+416. These will be divisible successively by 2,3 .... 16 if n contains those numbers as factors. ‘That is, ifn = 1X2X3.... X16 no member of the series can be prime. Now that H.G.L. and §.G.E. seem to be reaching agreement on how to find "any series of mn consecutive integers not one of_ which is prime" ,(issue of June 7) we, like the small boy at his first geometry class, want to know what they propose to do with it. The following correspondents are the latest to write showing by means of numerous diagtams and much argument that they know enough about geometry to see the fallacy of S.G.E.’s Saye about go ee geometry: T. Ashburton), William Howard (Hastings), F. D. Blackburn (Riccarton), J. Morice (Whakatane), D.D. (Hicks Bay), etc. No doubt all these correspondents are now basking «in. the ‘sunlight of self-satisfac-tion. They should thank S.G.E. for the opportunity. FIGURE MYSTERIES The Editor, "The Listener.’ Sir,-May I thank S.G.E. of ‘Gienavy for his correction of a statement in ef article, and offer this, explanation by way a slight excuse for my seeming rashness. The "article" was written early in the year in reply to Spohr S yommarseg of Mr. J. A. Reid, and. was intend erely as a letter to the Puzzle Page. However, it soon’ too unwieldy to appear there, and the result was the article as printed. As all the statements I made referred to facts I had been interested to find out myself, I ~~ careful, to put the reservation "as far as I can tell’ (or some other: cautious paren to anyI had not .actually verified. I assure S.G.E. that I do know that where mathematics are concerned, a little learning is a "very, very, very" dangerous thing, unless tempered with caution. I really have the greatest respect for both infinity and prime numIt was unfortunate that I stopped my investigations with 29. Had I persevered, the

case of 31 would have given me a clue to the puzzle of 13. But I do not see the significance of S.G.E.’s statement that "the figures in the period. of 1/7 are 12, 45, 78." % notice that these are all multiples of 3, and that 3, 6, and 9 are omitted, but why arrange the group of commencing figures, 1, 2, 4, 5, 7, 8, in the apparently arbitrary form of pairs? There is certainly a pattern in the commencing figures of seventeenths and twentythirds, and after a little investigation on my own account I find a similar sort of pattern in the case of nineteenths and twenty-ninths, But all I can gather from these exercises is this: The commencing figures of the decimal forms of the five fractions sevenths, seventeenths, nineteenths, twenty-thirds, and twentyninths are (naturally) in ascending order of the digits, and in each case there is a nonconformity with two or all of 0, 3, 6 and 9. With sevenths, none of these four appear at all; with seventeenths there are only one of each of these while there are two of all other digits; with nineteenths, there is only one 0 and one 9, but two of all others; with twentythirds there are three 3’s and three 6’s, but two of all others; and with twenty-ninths there are two 0’s, two 9’s, and three of all others. Which only goes to show that there is something even more peculiar about 3 and its multiples than I had thought. Yours, etc., R.W.C. This letter was submitted to S.G.E, before publication. His further comments are summarised as lending additional interest to a fascinating subject: "When I set down the figures in the decimal of 1/7th as 12,45,78, I meant, of course, that the commas should indicate blank spaces for 0,3,6,9. The symmetry in the commencing figures which I noticed when I read R.W.C.’s article really has a theoretical basis, as I realised soon after. The point is that if any decimal of 1/n repeats thus: O.abc...fg (a and g repeating) then the multiplication of this decimal by n does not give 1, but its equivalent 0.9 repeater. For example: 1/7th = 0.142857142857 etc. ke eS 7 Therefore 1 = 0.999999999999 etc. "Suppose then that we write 1 as 0.9 re peater and subtract 1/n from it. We get n-1/n, So that if n equals 1/17th, then 0.9-1/17th equals. 16/17. Or, in general, 1 = 0.9999999999..... 1/n = O.abc.....fg (a and g repeating) "Therefore 1-1/n = 0. .....-.where X is the commencing figure of 1-1/n. Clearly the top row is a row of 9’s and the abcg’s must be numbers either each equal to or each less than 9, Consequently there is never any carrying figure in the subtraction. xX is therefore the difference between 9 and a, under all circumstances. Thus, since 1/17 starts off 0.0 repeater, 16/17ths will start off 0.9 repeater, and if 2/17ths starts off 0.1 repeater 15/17ths will start off 0.8 repeater. This ee ew applies a the commencing figures of decimals repeat. I leave you to a for 1/13, 3/13, .12/13."