We have thus gone far in advance of the first problem, but a second always presents itself in these series, it is that of using these periods for larger numbers, which refer to a not too remote past or to a future not too distant. The first numbers are, as a rule, in the neighborhood of 1,252,680, the close of the eleventh Ahau-Katun, and the latter in the neighborhood of 1,480,440, the close of the thirteenth Ahau-Katun. The Manuscript presents the following:-- 1,272,544 1,268,540 1,538,342. XIII Akbal XIII Akbal XIII Akbal 121 17 51,419 IV Ahau IV Ahau 8 Cumhu 8 Cumhu IV Ahau. In connection with this it should be noted first that I have restored the 8 in the statement of the months, and second that the two numbers on the right were found with the aid of page 63 only by an easy conjecture. For with the reading of the Manuscript 10, 13, 3, 13, 2, I do not agree, but read instead 10, 13, 13, 3, 2; the number below, however, is given in the Manuscript as 7, 2 and then a black 14 joined to a red 5; I read this 7, 2, 14, 19. The three numbers nearest the bottom have red circles around them, indicating subtraction, or, according to my present point of view, addition. Now let us see how the computer arrived at the large numbers. Day XIII Akbal, the New Year's day of the 1 Kan years, is given; also the differences of the series 91 and 104, therefore also in the proportion of 7 to 8. If we combine these last two numbers by addition and then by multiplication with 260, the result is (7 + 8) × 260 = 3900. If, however, 7, 8 and 3900 be combined by multiplication the product is 7 × 8 × 3900 = 218,400 = 2400 × 91 = 2100 × 104 = 840 × 260 = 600 × 364 = 1120 × (91 + 104). We have already met with the 218,400 on page 24, which was obtained by the addition of 33,280 + 185,120. My opinion is as follows:--First 11 Ahau-Katuns = 1,252,680, were taken as a point of departure, and to this sum was added 15,600 = 4 × 3900, and 243 as the interval between the normal date IV Ahau and XIII Akbal. The result was 1,268,523. The position of this day, however, is XIII Akbal 11 Xul (1 Ix). Then the 3900 mentioned above was added to this number and the result was 1,272,423 = XIII Akbal 16 Pop (12 Muluc). Then to the 1,268,523 was added the 218,400 and the sum was 1,486,923 = XIII Akbal 1 Kankin (1 Kan), the very place in that year where a Tonalamatl ends. The following numbers were thus obtained:-- 1,272,423 1,268,523 1,486,923. These numbers are suppressed in the Manuscript. But if the encircled numbers are added to them, viz:--121 (interval between XIII Akbal and IV Kan), 17 (interval between XIII Akbal and IV Ahau), and 51,419 (= 197 × 260 + 199; 199, however, is the interval between XIII Akbal and IV Ik), the result is the three large numbers set down in the Manuscript, which have the following properties:--