"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 humor and a few vagaries of its own.
One of these vagaries has to do with the figure 9, and it is thus
described by William 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.
A mathematician named 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
9 is 18; add the digits together, and 1 and 8 make 9. Three
times 9 is 27; and 2 and 7 is 9. So it goes on up to 11
times 9, which gives 99. Very good. Add the digits together;
9 and 9 is 18, and 8 and 1 is 9.
Go on to any extent, and you will find it impossible to get
away from the figure 9. Take an example at random: 9 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 + 4 + 4 + 9 = 18; and 1 + 8 = 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 it, you get 26. Now, 62 - 26 = 36, and 3
+ 6 = 9. The squares of 26 and 62 are, respectively, 676 and
3844. Subtract one from the other, and you get 3168 = 18,
and 1 + 8 = 9.
So with the cubes of 26 and 62, which are 17,576 and
238,328. Subtracting, the result is 220,752 = 18, and 1 + 8
= 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 figures you start
with, the digits of the product will always come to 9.
7549132, sum of digits = 31.
31
-------
7549101, sum of digits = 27, and 2 + 7 = 9.
Again, set the figure 9 down in multiplication, thus:
1 x 9 = 9
2 x 9 = 18
3 x 9 = 27
4 x 9 = 36
5 x 9 = 45
6 x 9 = 54
7 x 9 = 63
8 x 9 = 72
9 x 9 = 81
10 x 9 = 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, 9.
Here is a different property of the same number. If you arrange in a row
the cardinal numbers from 1 to 9, with the single omission of 8, and
multiply the 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. Thus, if you wish a series of fives, you take 5 x 9 =
45 for a multiplier, with this result:
12345679
45
--------
61728395
49382716
--------
555555555
A very curious number is 142,857, which, multiplied by 1, 2, 3, 4, 5, or
6, gives the same figures in the same order, beginning at a different
point, but if multiplied by 7 gives all nines. Multiplied by 1, it equals
142,857; multiplied by 2, equals 285,714; multiplied by 3, equals 428,571;
multiplied by 4, equals 571,428; multiplied by 5, equals 714,285;
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 first 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 any
multiple of 3 up to 27, it gives three figures all 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 fours, 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, four for the third, 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, and were allowed to leave all your
winnings on the table, five consecutive lucky guesses would give you
almost a million and a half of dollars, or, to be exact, $1,450,625.52?
Yet that would be the result of winning thirty-five for one five times
hand-running.
Here is another example. Take the number 15, let us say. Multiply that by
itself, and you get 225. Now multiply 225 by itself, and so on until
fifteen products have been multiplied by themselves in turn.
You don't think that is a difficult problem? Well, you may be a clever
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 1,442. Allowing three figures to an inch, the answer would be more
than a thousand feet long. To perform 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 twenty-eight years about it.