of the starting point, 0 Zip, in Table XV, the position in the year of the day of the terminal date will be found to be 0 Pax. Since 260 days equal just 13 uinals, we have only to count forward from 0 Zip 13 uinals in order to reach the year position; that is, 0 Zotz is 1 uinal; to 0 Tzec 2 uinals, to 0 Xul 3 uinals, and so on in Table XV to 0 Pax, which will complete the last of the 13 uinals (rule 3, p. 141). Combining the above values, we find that in counting forward 322,920 (or 260) from the starting point 13 Ik 0 Zip, the terminal date reached is 13 Ik 0 Pax. In order to illustrate the method of procedure when the count is _backward_, let us assume an example of this kind. Suppose we count backward the number 9,663 from the starting point 3 Imix 4 Uayeb. Since this number is below 18,980, no Calendar Round can be deducted from it. Dividing the given number by 13, we have: 9,663 ÷ 13 =743-4/13. Counting the numerator of the fractional part of this quotient, 4, _backward_ from the day coefficient of the starting point, 3, we reach 12 as the day coefficient of the terminal date, that is, 2, 1, 13, 12 (rule 1, p. 139). Dividing the given number by 20, we have: 9,663 ÷ 20 = 483-3/20. Counting the numerator of the fractional part of this quotient, 3, _backward_ from the day sign of the starting point, Imix, in Table I, we reach Eznab as the day sign of the terminal date (Ahau, Cauac, Eznab); consequently the day reached in the count will be 12 Eznab. Dividing the given number by 365, we have {147} 9,663 ÷ 365 = 26-173/365. Counting _backward_ the numerator of the fractional part of this quotient, 173, from the year position of the starting point, 4 Uayeb, the year position of the terminal date will be found to be 11 Yax. Before position 4 Uayeb (see Table XV) there are 4 positions in that division of the year (3, 2, 1, 0). Counting these _backward_ to the end of the month Cumhu (see Table XV), we have left 169 positions (173 - 4 = 169); this equals 8 uinals and 9 days extra. Therefore, beginning with the end of Cumhu, we may count _backward_ 8 whole uinals, namely: Cumhu, Kayab, Pax, Muan, Kankin, Mac, Ceh, and Zac, which will bring us to the end of Yax (since we are counting backward). As we have left still 9 days out of our original 173, these must be counted backward from position 0 Zac, that is, beginning with position 19 Yax: 19, 18, 17, 16, 15, 14, 13, 12, 11; so 11 Yax is the position in the year of the day of the terminal date. Assembling the above values, we find that in counting the number 9,663 _backward_ from the starting point, 2 Imix 4 Uayeb, the terminal date is 12 Eznab 11 Yax. Whether the count be forward or backward, the method is the same, the only difference being in the direction of the counting.