"Nine worthies were they called."
DRYDEN: _The Flower and the Leaf._
I give these puzzles, dealing with the nine digits, a class to
themselves, because I have always thought that they deserve more
consideration than they usually receive. Beyond the mere trick of
"casting out nines," very little seems to be generally known of the laws
involved in these problems, and yet an acquaintance with the properties
of the digits often supplies, among other uses, a certain number of
arithmetical checks that are of real value in the saving of labour. Let
me give just one example--the first that occurs to me.
If the reader were required to determine whether or not
15,763,530,163,289 is a square number, how would he proceed? If the
number had ended with a 2, 3, 7, or 8 in the digits place, of course he
would know that it could not be a square, but there is nothing in its
apparent form to prevent its being one. I suspect that in such a case he
would set to work, with a sigh or a groan, at the laborious task of
extracting the square root. Yet if he had given a little attention to
the study of the digital properties of numbers, he would settle the
question in this simple way. The sum of the digits is 59, the sum of
which is 14, the sum of which is 5 (which I call the "digital root"),
and therefore I know that the number cannot be a square, and for this
reason. The digital root of successive square numbers from 1 upwards is
always 1, 4, 7, or 9, and can never be anything else. In fact, the
series, 1, 4, 9, 7, 7, 9, 4, 1, 9, is repeated into infinity. The
analogous series for triangular numbers is 1, 3, 6, 1, 6, 3, 1, 9, 9. So
here we have a similar negative check, for a number cannot be triangular
(that is, (n² + n)/2) if its digital root be 2, 4, 5, 7, or 8.