The third rule, given on page 141, tells how to find the position which the day of the terminal date occupied in the 365-day year. Dividing the given number by 365, we have: 5,799 ÷ 365 = 15-324/365. Counting forward the numerator of the fractional part of the resulting quotient, 324, from the year position of the starting date, 7 Tzec, in the sequence of the 365 year positions given in Table XV, the position 6 Zip will be reached as the position in the year of the day of the terminal date. The count by means of which the position 6 Zip is determined is given in detail. After the year position of the starting point, 7 Tzec, it requires 12 more positions (Nos. 8-19, inclusive) before the close of that month (see Table XV) will be reached. And after the close of Tzec, 13 uinals and the xma kaba kin must pass before the end of the year; 13 × 20 + 5 = 265, and 265 + 12 = 277. This latter number subtracted from 324, the total number of positions to be counted forward, will give the number of positions which remain to be counted in the next year following: 324 - 277 = 47. Counting forward 47 in the new year, we find that it will use up the months Pop and Uo (20 + 20 = 40) and extend 7 positions into the month Zip, or to 6 Zip. Therefore, gathering together the values determined for the several parts of the terminal date, we may say that in counting forward 5,799 from the starting point 2 Kan 7 Tzec, the terminal date reached will be 3 Akbal 6 Zip. For the next example let us select a much higher number, say 322,920, which we will assume is to be counted forward from the starting point 13 Ik 0 Zip. Since this number is above 18,980, we may apply our preliminary rule (p. 143) and deduct all the Calendar {146} Rounds possible. By turning to Table XVI we see that 17 Calendar Rounds, or 322,660, may be deducted from our number: 322,920 - 322,660 = 260. In other words, we can use 260 exactly as though it were 322,920. Dividing by 13, we have 260 ÷ 13 = 20. Since there is no fraction in the quotient, the numerator of the fraction will be 0, and counting 0 forward from the day coefficient of the starting point, 13, we have 13 as the day coefficient of the terminal date (rule 1, p. 139). Dividing by 20 we have 260 ÷ 20 = 13. Since there is no fraction in the quotient, the numerator of the fraction will be 0, and counting forward 0 from the day sign of the starting point, Ik in Table I, the day sign Ik will remain the day sign of the terminal date (rule 2, p. 140). Combining the two values just determined, we see that the day of the terminal date will be 13 Ik, or a day of the same name as the day of the starting point. This follows also from the fact that there are only 260 differently named days (see pp. 41-44) and any given day will have to recur, therefore, after the lapse of 260 days.[101] Dividing by 365 we have: 260 ÷ 365 = 260/365. Counting forward the numerator of the fraction, 260, from the year position