es to be considered. Thus,
taking 86, we may say that if 60000, etc., when divided by 76 leaves
either 22 or 60 (because 3x6 and 8x6 both produce 8), we get a solution.
But I reject the former on inspection, and see that 60 divided by 76 is
0, leaving a remainder 60. Therefore 8 x 86 = 688, the other example. It
will be found in the case of 71 that 100000, etc., divided by 61 gives a
remainder 42, (7 x 61 = 427) after producing the long dividend at the
beginning of this article, with the 7 added.

The other multipliers fail to produce a solution, so 83, 86, and 71 are
the only three possible multipliers. Those who are familiar with the
principle of recurring decimals (as somewhat explained in my next note on
No. 83, "The Ribbon Problem") will understand the conditions under which
the remainders repeat themselves after certain periods, and will only
find it necessary in two or three cases to make any lengthy divisions. It
clearly follows that there is an unlimited number of multiplicands for
each multiplier.

83.--_The Ribbon Problem._

The solution is as follows: Place this rather lengthy number on the
ribbon, 0212765957446808510638297872340425531914393617. 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 any
number up to 16. I made the limit 9 in order to put readers off the
scent. The fact is these two numbers are simply the recurring decimals
that equal 1/17 and 1/47 respectively. Multiply the one by seventeen and
the other by forty-seven, and you will get all nines in each case.

In transforming a vulgar fraction, say 1/17, to a decimal fraction, we
proceed as below, adding as many noughts to the dividend as we like until
there is no remainder, or until we get a recurring series of figures, or
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
exactitude.

17) 100 (.058823
85
----
150
136
----
140
136
----
40
34
----
60
51
----
9

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 if
any other factor than these occurs in the denominator of your vulgar
fraction. Thus, 1/2, 1/4, and 1/8 give us th