On Rektor Chr. Zeller's 1882 Paper
"Die Grundaufgaben der Kalenderrechnung auf neue und vereinfachte Weise gelöst".

Let J be the century-number, k the remaining part of the year-number, e the residue, which remains when J is divided by 4, m the number of the month, q the day of the month.

Fraction- or division- expressions like k/4 etc. shall here not mean fractional numbers, but that whole number, which one by the carrying out of the specified division as quotient obtains.

Problem 1.   From the Date to find its Weekday.

Rule: One forms the total

a.  for the new calendar :-

        (m + 1) 26           k
  q  +  ----------  +  k  +  -  -  2 e     (or  + 5 e)
            10               4

b.  for the old calendar :-

        (m + 1) 26           A
  q  +  ----------  +  A  +  -  -  J       (or  + 6 J)
            10               4
                                   where A = k + 4

to be divided by 7, the remainder gives the desired Weekday-number.

January and February are to be seen as the 13th and 14th month of the preceding year and to be so taken in the calculation.

Now it is only necessary to add four easily obtainable items to the date q and to divide by 7; this can be done without written reckoning, especially if one omits multiples of seven during the addition of the sums or if one only adds the seven- remainders.

The first term changes every day, the second every month, the third every year, the fourth every leap-year, the fifth every century.

Now, if one uses instead of the last three the seven-remainder of k + k/4 -2e, which we will call = + t, then q + (m+1)×2,6 + t is the weekday-formula for the entire calendar year k; one must only add the number t + (m+1)×2,6, without fraction, to the date q in order to find the number, whose seven-remainder gives the day of the week.

Example 1.   1882, 11th Sept.   J = 18 = 4.4 + 2   k = 82   m = 9

                       e = 2             k/4 = 20   q = 11
      (9 + 1) 26
 11 + ---------- + 82 + 20 - 2.2 = 135 = 7.19 + 2
          10

thus the 11th of September 1882 was on the second day of the week or Monday.

Example 2.   1492, 12th Oct. Discovery of the New World.   A = 92 + 4 = 96

      (10 + 1) 26
 12 + ----------- + 96 + 24 - 14 = 146 = 7.20 + 6
           10
         = 28

thus the 12th of October 1492 was on the sixth day of the week or Friday.

Example 3.   1712, 24th Jan. Birthday of Frederick II.   k = 11   m = 13   e = 1

      (13 + 1) 26
 24 + ----------- + 11 + 2 - 2.1 = 24 + 36,4 + 11 + 2.2 = 71 = 7.10 + 1
           10

thus Frederick II was born on the 1st day of the week or Sunday.

Problem 2.   To find the date of Easter

  a.   in the old calendar.
1)  k + 5 J to be divided by 19, remainder a;
2)  19 a + 15 to be divided by 30, remainder b;
      b is the Easter-full-moon-number and gives how many days after 21 March the Easter full moon is,
3)  to b add k + k/4 - J, to be divided by 7, remainder d;
      then Easter is b + 7 - d days after 21 March, or 7 - d days after the Easter full moon.

  b.   in the new calendar.
1)  k + 5 J to be divided by 19, remainder a;
2)  to 19 a + 15 add the number h = J - J/4 - (8J+13)/25, which number often remains the same for successive centuries and amounts to 7, 8, 9 for years between 1583 and 1700, 1700-1900, 1900-2200; to be divided by 30, remainder b, the Easter-full-moon-number;
3)  to b add k + k/4 + 2 - 2 e, to be divided by 7, remainder d;
      then Easter is b + 7 - d days after 21 March, or 7 - d days after the Easter-full-moon.

When in 3) the division by 7 is exact, then put d=o, except in the case, when 1) also b=29 or 2) b=28 and a is greater than 10, then put d=7. Or what is the same thing, when the calculation for b=29 and d=o gives the 26th of April as the date of Easter, then instead put the 19th, and the 18th of April when one finds for a greater than 10 b=28 and d=o as Easter-date the 25th of April. This case will occur for the first time in 1954. --

Example 1.   Easter 1355. Coronation of Charles IV in Rome.

  55 + 5.13 = 120 = 19.6 + 6   a = 6 | 19.6 + 15 = 129 = 4.30 + 9  b = 9
  9 + 55 + 55/4 - 13 = 64 = 7.9 + 1   d = 1

Easter is 9 + 7 - 1 = 15 days after the 21st of March = 36th of March = 5th of April.

Example 2.   Easter 1886.  J = 18 = 4.4 + 2   e = 2

  86 + 5.18 = 176 = 19.9 + 5   19.5 + 15 + 8 = 118 = 30.3 + 28   b = 28
  28 + 86 + 86/4 + 2 - 2.2 = 133 = 7.19 + 0   d = 0  (because a equals
  smaller than 10)

Easter is 28 + 7 - 0 = 35 days after the 21st of March = 56th of March = 25th of April.