e exact decimals, .5, .25,
and .125; 1/5 and 1/25 give us .2 and .04; 1/10 and 1/20 give us .1 and
.05: because the denominators are all composed of 2 and 5 factors. But if
you wish to convert 1/3, 1/6, or 1/7, your division sum will never end,
but you will get these decimals, .33333, etc., .166666, etc., and
.142857142857142857, etc., where, in the first case, the 3 keeps on
repeating for ever and ever; in the second case the 6 is the repeater,
and in the last case we get the recurring period of 142857. In the case
of 1/17 (in "The Ribbon Problem") we find the circulating period to be
.0588235294117647.

Now, in the division sum above, the successive remainders are 1, 10, 15,
14, 4, 6, 9, etc., and these numbers I have inserted around the inner
ring of the diagram. It will be seen that every number from 1 to 16
occurs once, and that if we multiply our ribbon number by any one of the
numbers in the inner ring its position indicates exactly the point at
which the product will begin. Thus, if we multiply by 4, the product will
be 235, etc.; if we multiply by 6, 352, etc. We can therefore multiply by
any number from 1 to 16 and get the desired result.

[Illustration]

The kernel of the puzzle is this: Any prime number, with the exception of
2 and 5, which are the factors of 10, will exactly divide without
remainder a number consisting of as many nines as the number itself, less
one. Thus 999999 (six 9's) is divisible by 7, sixteen 9's are divisible
by 17, eighteen 9's by 19, and so on. This is always the case, though
frequently fewer 9's will suffice; for one 9 is divisible by 3, two by
11, six by 13, when our ribbon rule for consecutive multipliers breaks
down and another law comes in. Therefore, since the 0 and 7 at the ends
of the ribbon may not be removed, we must seek a fraction with a prime
denominator ending in 7 that gives a full period circulator. We try 37,
and find that it gives a short period decimal, .027, because 37 exactly
divides 999; it, therefore, will not do. We next examine 47, and find
that it gives us the full period circulator, in 46 figures, at the
beginning of this article.

If you cut any of these full period circulators in half and place one
half under the other, you will find that they will add up all 9's; so you
need only work out one half and then write down the complements. Thus, in
the ribbon above, if you add 05882352 to 94117647 the result is 99999999,
and so with our long solution num