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 kaba kin_ of 5 days; and further, that when the last position in the last division of the year (4 Uayeb) 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 {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 ÷ 365 = 86-351/365. Multiplying 86 by 365, we have 31,390, which 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 resulting 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