Returning to the number 31,741, let us deduct from it the highest multiple of 20 which it contains, found by dividing the number by 20 and multiplying the whole number part of the resulting quotient by 20; 31,741 ÷ 20 = 1,587-1/20. Multiplying 1,587 by 20, we have 31,740, which is the highest multiple of 20 that 31,741 contains, and which may be deducted from 31,741 without affecting the resulting day sign; 31,741 - 31,740 = 1. Therefore in the present example 1 is the number which, if counted forward from the day sign of the starting point in the sequence of the 20 day signs given in Table I, will reach the day sign of the resulting date. In other words, after dividing by 20 the only part of the resulting quotient which is used in determining the new day sign is the numerator of the fractional part. Thus we may formulate the rule for determining the second unknown on page 138 (the day sign): _Rule 2._ To find the new day sign, divide the given number by 20, and count forward the numerator of the fractional part of the resulting quotient from the starting point in the sequence of the twenty day signs given in Table I, if the count is forward, and backward if the count is backward, and the sign reached will be the new day sign. Applying this rule to 31,741, we have seen above that its division by 20 gives us as the fractional part of the quotient, 1/20. Since the count was forward from the starting point, if 1 (the numerator of the fractional part of the quotient) be counted forward in the sequence of the 20 day signs in Table I from the day sign of the starting point, Ahau (4 Ahau 8 Cumhu), the day sign reached will be the day sign of the resulting date. Counting forward 1 from Ahau in Table I, the day sign Imix is reached, and Imix, therefore, will be the new day sign. Thus our second unknown is determined. By combining the above two values, the 12 for the first unknown and Imix for the second, we can now say that in counting forward 31,741 from the date 4 Ahau 8 Cumhu, the day reached will be 12 Imix. It remains to find what position this particular day occupied in the 365-day year, or haab, and thus to determine the third and fourth unknowns on page 138. Both of these may be found at one time by the same operation.