We have seen on pages 51-60 that in Maya chronology there were possible only 18,980 different dates--that is, combinations of the 260 days and the 365 positions of the year--and further, that any given day of the 260 could return to any given position of the 365 only after the lapse of 18,980 days, or 52 years. {143} Since the foregoing is true, it follows, that this number 18,980 or any multiple thereof, may be deducted from the number which is to be counted without affecting in any way the terminal date which the number will reach when counted from the starting point. It is obvious that this modification applies only to numbers which are above 18,980, all others being divided by 13, 20, and 365 directly, as indicated in rules 1, 2, and 3, respectively. This enables us to formulate another rule, which should be applied to the number to be counted before proceeding with rules 1, 2, and 3 above, if that number is above 18,980. _Rule_. If the number to be counted is above 18,980, first deduct from it the highest multiple of 18,980 which it contains. This rule should be applied whenever possible, since it reduces the size of the number to be handled, and consequently involves fewer calculations. In Table XVI are given 80 Calendar Rounds, that is, 80 multiples of 18,980, in terms of both the Maya notation and our own. These will be found sufficient to cover most numbers. Applying the above rule to the number 31,741, which was selected for our first example, it is seen by Table XVI that 1 Calendar Round, or 18,980 days, may be deducted from it; 31,741 - 18,980 = 12,761. In other words, we can count the number 12,761 forward (or backward had the count been backward in our example) from the starting point 4 Ahau 8 Cumhu, and reach exactly the same terminal date as though we had counted forward 31,741, as in the first case. Mathematical proof of this point follows: 12,761 ÷ 13 = 981-8/13 12,761 ÷ 20 = 638-1/20 12,761 ÷ 365 = 34-351/365 The numerators of the fractions in these three quotients are 8, 1, and 351; these are identical with the numerators of the fractions in the quotients obtained by dividing 31,741 by the same divisors, those indicated in rules 1, 2, and 3, respectively. Consequently, if these three numerators be counted forward from the corresponding parts of the starting point, 4 Ahau 8 Cumhu, the resulting terms together will form the corresponding parts of the same terminal date, 12 Imix 14 Kayab.