Vagaries of Mathematics.

New Zealand Graphic, Volume XXXVI, Issue 23, 9 June 1906, Page 7

Vagaries of Mathematics.

"As dull as arithmetic” is a phrase that is familiar to almost every schoolboy and is a figure of comparison that is frequently evoked by those sages who hold down empty cracker-boxes in rural general stores. The fact is, however, that arithmetic is not always half so dull as it looks. Like some of those persons who earn a livelihood by teaching it to the young, it has a dry humour and a few vagaries of its own. One of these vagaries has to do with the figure 9, and it is thus described by Wiliam Walsh in his “Handy Book of Literary Curiosities”. It is a most romantic number, and a most persistent self-willed, and obstinate one. You cannot multiply it away or get rid of it anyhow. Whatever you do. it is sure to turn up again, as did the body of Eugene Aram’s victim. Mr. W. Green, who died in 1794, is said to have first called attention to the fact that all through the multiplication table the product of nine comes to nine. Multiply by any figure you like, and the sum of the resultant digits will invariably add up as nine. Thus, twice nine is 18; add the digits together, and 1 and 8 makes 9. Three times 9 is 27; and 2 and 7 is 9. So it goes on up to 11 times 9, wiiich gives 99. Very good. Add the digits together, and 9 ami 9 is 18. and 8 and 1 is 9.

Go on to any extent, ana you will find it impossible to get away from the figure 9. Take an example at random. Nine times 339 is 3,051; add the digits together, and they make 9. Or, again, 9 times 2,127 is 19,143; add the digits together, they make 18, and 8 and 1 is 9. Or still again, 9 times 5,071 is 45,639; the sum of these digits is 27; and 2 and 7 is 9.

This seems startling enough. Yet there are other queer examples of the same form of persistence. It was M. de Maivan who discovered that if you take any row of figures, and, reversing their order, make a subtraction sum of obverse and reverse, the final result of adding up the digits of the answer will always be 9. As, for example:

2941 Reverse, 1492 1449 Now, 1 plus 4 plus 4 plus 9 equals 18; and 1 plus 8 equals 9. The same result is obtained if you raise the numbers so changed to their squares or cubes. Start anew, for example with 62; reversing ii, you get 26. Now, 62 —26 equals 36, and 3 plus 6 equals 9. The squares of 26 and 62 are, respectively. 676 and 3844. Subtrac. on? from the other, and you get 3168 equals 18. and 1 plus 8 equals 9. So with th? cubes of 26 and 62. which are 17.576 and 238.328. Subtracting, the result is 220.752 equals 18, and 1 plus 8 equals 9. Again, you are confronted with the same puzzling peculiarity in another form. Write down any number, as. for example. 7.549,132, subtract therefrom the sum of its digits, and. no matter what gure you -tart with, the digits of the products will always come to 9. 7549132. sum of digits equals 31. 31 7549101 sum <4 digits equals 27, and 2 plus 7 equals 9. Again, et the figure 9 down in multiplica t on. thus: 1 multiplied by 9 equals 9 2 multiplied by 9 equals 18 3 multiplied by 9 equals 27 4 mult plied by 9 equals 36 5 multiplied by 9 equals 45 6 multiplied by 9 equals 54 7 multiplied by 9 equals 63 8 multiplied by 9 equals 72 9 multiplied by 9 equals 81 10 multiplied by 9 equals 90 Now you will see that the tens column reads down 1,2, 3, 4. 5,6, 7. 8, 9. and the units column up 1,2, 3. 4,5, 6. 7,8, Her? is a different property of the same number. If you arrange in a row the cardinal numbers from 1 to 9, with the single omiss.on of 8, and multiply ihe sum so represented by any one of the figures multiplied by 9, the’ result will present a succession of figures identical with that which was multiplied by 9. I. us, if you wish a series of ves "you take 5 multiplied by 9 equals 45 f or a multiplier, with this result: 12345679 45 (>1728395 49382716 555555555 A very curious number is 142.857 which, mult plied by 1. 2. 3. 4. 5. or 6. gives the same figure-, in the same order, beginning at a different point, but if multiplied by 7 gives all nines. Multipied by 1 it gives 142.857; multiplied by 2, equals 285.714: multiplied by 3. equals 428.571: multiplied bv 4. equals 571.428; multiplied bv 5, equals ' multiplied by 6. equals 857.142; multiplied by 7, equals 999 999 Multiply 142.857 by 8. and'you have 1.142.856. Then add the fit st figure to the last, and you have 142,857. the original number, the figures exactly the same as at the start. The number 37 has this strange peculiarity; multiplied by* 3. or by anv multiple of 3 up to 27. it gives three figures ;'ll alike. Thus, three times 37 will be 111. Twice three times (6 times) 37 will be 222; three times three times (9 times) 37 gives three threes; four times three times (12 times) 37. three foul s. and so on. The wonderfully procreative power of figures, or. rather, their accumulative growth, has been exemplified in that familiar story of the farmer, who. undertaking to pay his farrier one grain of

wheat for the first nail, two for the second, and so on. found that he had bargained to give the farrier more wheat than was grown in all England. My beloved young friends who love to frequent the roulette table, do you know that if you begin with a dime, ami were allowed to leave all your winnings on the table, five consecutive lucky guesses would give you £300,000. Yet that would be the result of winning 35 for one five times hand-running. Here is another example. Take the number 15. let us say. Mult ply that by itself, and you get 225. Now multiply 225 by itself, and so on until fifteen products have been niultipliol by themselves in turn. You don't think that is a difficult problem? Well, you may be a elever mathematician. but it would take you about a quarter of a century to work out this simple little sum. The final product called for contains 38.589 figures, the first of which are 1442. Allowing three figures to an inch. the answer would be over 1070 ft long. To perfoim the operation would require about 500.000.000 figures. If they can be made at the rate of one a minute, a person working ten hours a day for three hundred days in each year would be 28 years about it. If. in multiplying, he should make a row of ciphers, as he does in other figures, the number of figures would be mo;e than 523.939.228. This would be the precise number of figures used if the product of the left hand figure in each multiplicand by each figure of the multiplier was always a single figure, but. as it is most frequently, though not always. two figures, the method employed to obtain the foregoing results cannot be accurately applied. Assuming that the cipher is used on an average once in ten times. 475.000,000.000 approximates the actual number.