The first day of the year whose beginning is shown at the point of contact of the two wheels in figure 21 is 2 Ik 0 Pop, that is, the day 2 Ik which occupies the first position in the month Pop. The next day in succession will be 3 Akbal 1 Pop, the next 4 Kan 2 Pop, the next 5 Chicchan 3 Pop, the next 6 Cimi 4 Pop, and so on. As the wheels revolve in the directions indicated, the days of the tonalamatl successively fall into their appropriate positions in the divisions of the year. Since the number of cogs in A is smaller than the number in B, it is clear that the former will have returned to its starting point, 2 Ik (that is, made one complete revolution), before the latter will have made one complete revolution; and, further, that when the latter (B) has returned to its starting point, 0 Pop, the corresponding cog in B will not be 2 Ik, but another day (3 Manik), since by that time the smaller wheel will have progressed 105 cogs, or days, farther, to the cog 3 Manik. The question now arises, how many revolutions will each wheel have to make before the day 2 Ik will return to the position 0 Pop. The solution of this problem depends on the application of one sequence to another, and the possibilities concerning the numbers or names which stand at the head of the resulting sequence, a subject already discussed on page 52. In the present case the numbers in question, 260 and 365, contain a common factor, therefore our problem falls under the third contingency there presented. Consequently, only certain of the 260 days can occupy the position 0 Pop, or, in other words, cog 2 Ik in A will return to the position 0 Pop in B in fewer than 260 revolutions of A. The actual solution of the problem {58} is a simple question of arithmetic. Since the day 2 Ik can not return to its original position in A until after 260 days shall have passed, and since the day 0 Pop can not return to its original position in B until after 365 days shall have passed, it is clear that the day 2 Ik 0 Pop can not recur until after a number of days shall have passed equal to the least common multiple of these numbers, which is (260/5)×(365/5)×5, or 52×73×5 = 18,980 days. But 18,980 days = 52×365 = 73×260; in other words the day 2 Ik 0 Pop can not recur until after 52 revolutions of B, or 52 years of 365 days each, and 73 revolutions of A, or 73 tonalamatls of 260 days each. The Maya name for this 52-year period is unknown; it has been called the Calendar Round by modern students because it was only after this interval of time had elapsed that any given day could return to the same position in the year. The Aztec name for this period was _xiuhmolpilli_ or _toxiuhmolpia_.[34]