Place this rather
It may be
multiplied by any number up to 46 inclusive to give the same order of
figures in the ring.
The number previously given can be multiplied by
number up to 16.
I made the limit 9 in order to put readers off the
The fact is these two numbers are simply the recurring decimals
that equal 1/17 and 1/47 respectively.
Multiply the one by seventeen
the other by forty-seven, and you will get all nines in each case.
In transforming a
vulgar fraction, say 1/17, to a
proceed as below, adding as many noughts to the dividend as we like
there is no remainder, or until we get a recurring series of figures,
until we have carried it as far as we require, since every additional
figure in a never-ending decimal carries us nearer and nearer to
Now, since all
powers of 10 can only contain
factors of the
powers of 2
and 5, it clearly follows that your decimal never will come to an end
any other factor than these occurs in the denominator of your vulgar
Thus, ½, 1/4, and 1/8 give us the exact decimals,
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
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
Now, in the
division sum above, the successive
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
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
be 235, etc.; if we multiply by 6, 352, etc.
We can therefore multiply
any number from 1 to 16 and get the desired result.
The kernel of the
puzzle is this: Any prime
number, with the
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,
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
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 under the other, you will find that they will add up all 9's; so
need only work out one half and then write down the
the ribbon above, if you add 05882352 to 94117647 the result is
and so with our long solution number.
Note also in the diagram above
not only are the opposite numbers on the outer ring complementary,
making 9 when added, but that opposite numbers in the inner ring, our
remainders, are also complementary, adding to 17 in every
perhaps to point out that in limiting our multipliers to the first nine
numbers it seems just possible that a short period circulator might
a solution in fewer figures, but there are reasons for thinking it