If ¹/₅₉ and ¹/₆₁ be taken, the repeating parts will be found to contain 58 and 60 figures. Of these we write down only the first halves, as the reader may supply the rest by the complemental property just given.

01694915254237288135593220338, &c.
016393442622950819672131147540, &c.

Here, then, are two numbers, the first of which multiplied by any number under 59, and the second by any number under 61, can have the products formed by carrying certain of the figures from one end to the other.

But, b being still prime, it may happen that remainder 1 may occur before b - 1 figures are obtained; in which case, as shewn, the number of figures must be a measure of b - 1. For example, take ¹/₄₁. The repeating quotient, written as above, has only 5 figures, and 5 measures 41 - 1.

0₁₀2₁₈4₁₆3₃₇9₁

Now, this period, it will be found, has its figures merely transposed, if we multiply by 10, 18, 16, or 37. But if we multiply by any other number under 41, we convert this period into the period of another fraction whose denominator is 41. The following are 8 periods which may be found.

0₁₀2₁₈4₁₆3₃₇9₁ 1₉2₈1₃₉9₂₁5₅
0₂₀4₃₆8₃₂7₃₃8₂1₁₉4₂₆6₁₄3₁₇4₆
0₃₀7₁₃3₇1₂₉7₃2₂₈6₃₄8₁₂2₃₈9₁₁
0₄₀9₈₁7₂₃5₂₅6₄3₂₇6₂₄5₃₅8₂₂5₁₅

To find m/41, look out for m among the remainders, and take the period in which it is, beginning after the remainder. Thus, ³⁴/₄₁ is ·8292682926, &c., and ¹⁵/₄₁ is ·3658536585, &c. These periods are complemental, four and four, as 02439 and 97560, 07317 and 92682, &c. And if the first number, 02439, be multiplied by any number under 41, look for that number among the remainders, and the product is found in the period of that remainder by beginning after the remainder. Thus, 02439 multiplied by 23 gives 56097, and by 6 gives 14634.

The reader may try to decipher for himself how it is that, with no more figures than the following, we can extend the result of our division. The fraction of which the period is to be found is ¹/₈₇.