RECURRING DECIMALS 105

plied by 7, gives a number ending in the digit 5. This number
can only be 5. We now have:

7) 999,999 ( ---.57

9

el

39

35
49
49

The digit g in the number 35 is then taken (as shown) as the
previous remainder. g deducted from g leaves 6. The next
digit in the cycle, still working backwards, must therefore be
8, since 8 X 7 =56, and this is the only two-digit multiple of 7
which has 6 as its second digit.

Another method in which we may calculate the digits in
a cycle in a backwards order is as follows. The division from
which the cycle is to result is first shown in the form of a
fraction (thus, 1). Both the numerator and the divisor are
then multiplied by the smallest number which will make the
divisor 2 number whose last digitis 9. (Thus, 2 X 7 =%.) Then
the last digit of the cycle will be the same as the new numera-
tor (that is, in our example, 7). The other digits are then
found in a different way. First we increase the new divisor
by 1 so as to give a multiple of 10 (that is, 49+1=50). This
multiple of 10 is then divided by 10 or, in other words, the
zero sign in the end position is crossed out. The number
which finally emerges from these operations is then taken as
what we shall call the Relative of the cyclic recurrence. In
the case of 1, the Relative is 5.

   

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