MOELBY] INTRODUCTION TO STUDY OF MAYA HIEROGLYPHS 



141 



counted from the starting point. This is true, because no matter 

 what position in the year the day of the starting point may occupy, 

 any multiple of 365 will bring the count back again to the same 

 position in the year. 



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

 highest multiple of 365 which it contains. This will be found by 

 dividing the number by 365 and multiplying the whole number part 

 of the resulting quotient by 365: 31,741 X 365 = 86||i. Multiplying 

 86 by 365, we have 31,390, wliich is the highest multiple that 31,741 

 contains. Hence it may be deducted from 31,741 without affecting 

 the position in the year of the resulting day; 31,741—31,390 = 351. 

 Therefore, in the present example, 351 is the number which, if 

 counted forward from the year position of the starting date in the 

 sequence of the 365 positions in the year, given in Table XV, will 

 reach the position in the year of the day of the resulting date. This 

 enables us to formulate the rule for determining the third and fourth 

 unknowns on page 138 (the position in the year of the day of the re- 

 sulting date) : 



Rule 3. To find the position in the year of the new day, divide 

 the given number by 365 and count forward the numerator of the 

 fractional part of the resulting quotient from the year position of 

 the starting point in the sequence of the 365 positions of the year 

 shown in Table XV, if the count is forward; and backward if the 

 count is backward, and the position reached will be the position in 

 the year which the day of the resulting date will occupy. 



Table XV. THE 365 POSITIONS IN THE MAYA YEAR 



Month.. 



Position. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 Do.. 

 •Do.. 



