We divide by 4 because the intercalary period is four years; and as every fourth year contains the divisor 4 once more than any of the three preceding years, so there is one more added to the fourth year than there is to any of the three preceding years; and as every year consists of 52 weeks and one day, this additional year gives an additional day to the remainder after dividing by 7. For example, the year
| 1 of the era consists of | 52 w. 1 d. |
| 2 years consist of | 104 w. 2 d. |
| 3 years consist of | 156 w. 3 d. |
| (4 ÷ 4) + 4 = 5 years consist of | 260 w. 5 d. |
Hence the numbers thus formed will be 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, and so on.
We divide by 7, because there are seven days in the week, and the remainders show how many days more than an even number of weeks there are in the given year. Take, for example, the first twelve years of the era after being increased by one-fourth, and we have
| 1 ÷ 7 = 0 | remainder | 1 | Then | 3 - 1 = 2 = B |
| 2 ÷ 7 = 0 | " | 2 | " | 3 - 2 = 1 = A |
| 3 ÷ 7 = 0 | " | 3 | " | 10 - 3 = 7 = G |
| 5 ÷ 7 = 0 | " | 5 | " | 10 - 5 = 5 = F E |
| 6 ÷ 7 = 0 | " | 6 | " | 10 - 6 = 4 = D |
| 7 ÷ 7 = 1 | " | 0 | " | 3 - 0 = 3 = C |
| 8 ÷ 7 = 1 | " | 1 | " | 3 - 1 = 2 = B |
| 10 ÷ 7 = 1 | " | 3 | " | 10 - 3 = 7 = A G |
| 11 ÷ 7 = 1 | " | 4 | " | 10 - 4 = 6 = F |
| 12 ÷ 7 = 1 | " | 5 | " | 10 - 5 = 5 = E |
| 13 ÷ 7 = 1 | " | 6 | " | 10 - 6 = 4 = D |
| 15 ÷ 7 = 2 | " | 1 | " | 3 - 1 = 2 = C B |
From this table it may be seen that it is these remainders representing the number of days more than an even number of weeks in the given year, that we have to deal with in finding the dominical letter.
Did the year consist of 364 days, or 52 weeks, invariably, there would be no change in the dominical letter from year to year, but the letter that represents Sunday in any given year would represent Sunday in every year. Did the year consist of only 363 days, thus wanting one day of an even number of weeks, then these remainders, instead of being taken from a given remainder, would be added to that number, thus removing the dominical letter forward one place, and the beginning of the year, instead of being one day later, would be one day earlier in the week than in the preceding year.
Thus, if the year 1 of the era be taken from 3, we would have 3 - 1 = 2; therefore, B being the second letter, is dominical letter for the year 1. But if the year consist of only 363 days, then the 1 instead of being taken from 3 would be added to 3; then we would have 3 + 1 = 4; therefore, D being the fourth letter would be dominical letter for the year 1. The former going back from C to B, the latter forward from C to D; or which amounts to the same thing, make the year to consist of 51 weeks and 6 days; then 10 - 6 = 4, making D the dominical letter as before.
As seven is the number of days in the week, and the object of these subtractions is to remove the dominical letter back one place every common year, and two in leap-year, why not take these remainders from 7? We answer, all depends upon the day of the week on which the era commenced. Had G, the seventh letter been dominical letter for the year preceding the era, then these remainders would be taken from 7; and 7 would be used until change of style in 1582. But we know from computation that C, the third letter, is dominical letter for the year preceding the era; so we commence with three, and take the smaller remainders, 1 and 2 from 3; that brings us to A. We take the larger remainders, from 3 to 6, from 3 + 7 = 10. We add the 7 because there are seven days in the week. We use the number 10 until we get back to C, the third letter, the place from whence we started. For example, we have
| 3 - 1 = 2 = B |
| 3 - 2 = 1 = A |
| 10 - 3 = 7 = G |
| 10 - 4 = 6 = F |
| 10 - 5 = 5 = E |
| 10 - 6 = 4 = D |
| 3 - 0 = 3 = C |