"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 labor. 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 it's 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, $(n^2+n)/2$) if its digital root is $2,$ $4,$ $5,$ $7,$ or $8.$
Solutions: 1
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