140 BUREAU OF AMERICAN ETHNOLOGY [bull. 57 



Returning to the number 31,741, let us deduct from it the liighest 

 multijile of 20 which it contains, found by dividing the number by 

 20 and multiplying the whole number ])art of the resulting quotient 

 by 20; 31,741^20 = 1,587-^. Multiplying 1,587 by 20, we have 

 31,740, which is the highest multiple of 20 that 31,741 contains, and 

 which maybe deducted from 31,741 wdthout affecting the resulting 

 day sign; 31,741—31,740 = 1. Therefore in the present exam])le 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 fmd the new day sign, divide the given number by 20, 

 and count forw^ard the numerator of the fractional part of the result- 

 ing 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, ^. 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 ])oint, 

 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. 



It was explained on pages 44-51 that the Maya year, at least in 

 so far as the calendar was concerned, contained only 365 days, divided 

 into 18 uinals of 20 days each, and the xma Icaha Icin of 5 days; and 

 further, that when the last position in the last division of the year 

 (4 TJayeb) was reached, it was followed without interruption by the 

 first position of the first division of the next year (0 Pop) ; and, 

 finally, that this sequence was continued indefinitely. Consequently 

 it Is clear that the highest multiple of 365 which the given number 

 contains may be subtracted from it without affecting in any way the 

 position in the year of the day which the number will reach when 



