© J R Stockton, ≥ 2008-11-13

"Problema duplex Calendarii fundamentale".

Links within this site :-

**Merlyn Home Page**- Site Index, E-Mail, Copying- My Main Zeller Page :-
**This Page**:-- The Original 1883 Publication - Images (40kB)
- Other Papers published by Zeller :-
- The Date of Easter Sunday
- For more date/time pages, and links, see Date and Time Index and Links
- JavaScript Include Files

"Problema duplex Calendarii fundamentale;" Par M. Ch. Zeller,
*Bulletin de la Société Mathématique de France*,
vol.11, pp.59-61, (Séance du 16 mars 1883) - in Latin.

Scanned images of (my copies of) the original printed pages are in zel-83px.htm.

*** from the Latin by Tony Brook of
London SW20, Autumn 2002; edited by JRS ***

The Basic Double Problem of the Calendar

by M. Ch. Zeller

(Meeting of 16 March 1883)

**PROBLEM I. - How to find
the date or the appropriate day of the week from a given day, month,
year and century.**

Let there be

*I* the number of the century;

*e* its remainder after dividing by 4;

*k* the year within the century;

*m* the number of the month;

*q* the number of the day.

In the divisions *k/4, 26/10, ...* let the fractions be omitted,
leaving only the whole numbers resulting from the divisions.

*Rule.* - Let the total be calculated :-

1° For the Gregorian Calendar . . .
*q + (m+1)26/10 + k + k/4 - 2e (or +5e)*,

2° For the Julian Calendar . . . . . .
*q + (m+1)26/10 + k + k/4 - (I+2)*;

let it be divided by *7*, the remainder will be equal to
the number of the desired day of the week.

The months January and February are to be ascribed to the preceding year and to be understood as the thirteenth and fourteenth month of this.

If, in the summation, multiples of the number *7* are omitted,
as is permissible, the calculation is easier to do and will be
completed without difficulty.

The first term *q* varies with each individual day, the second
with the month, the third with the year, the fourth with the leap year,
the fifth with whatever century.

*Example.* - 12th Oct. 1492 (the discovery of the New World)

I = 14, k = 92, k/4 = 23, m = 10, q = 12, 12 + (10 + 1) × 2,6 + 92 + 23 - (14 + 2) = 12 + 28 + 92 + 23 - 16 = 139 = 7.19 + 6

therefore that day falls on the sixth date, *i.e.* Friday.

**PROBLEM II. - How to
calculate Easter.**

(a) for the Julian Calendar :-
*Rule*. -

1° let *k + 5I* be divided by *19*,
let *a* be the remainder;

2° let *19a + 15* be divided by *30*,
remainder *b*.

The number *b* provides the early limit of Easter,
indicating on which day after 21 March the Paschal Full Moon falls.

3° By adding to *b* the number
*k + k/4 - I* and by dividing by *7*,
you would have the remainder *d*;
then Easter will be *b + 7 - d* days after 21 March;
*i.e.* on the *(7-d) ^{th}* day after the Full Moon.

(b) for the Gregorian Calendar :-
*Rule*. -

1° Divide *k + 5I* by *19*,
remainder *a*.

2° Add to *19a + 15* the number
*h = I - I/4 - (8I+13)/25* , a number which does not
change over several centuries and is worth *7, 8, 9* for years in
*1583-1700, 1700-1900, 1900-2200*; by dividing the total by *30*
you will have the remainder *b*, the number for the Full Moon.

3° To *b* add *k + k/4 + 2 - 2e* ;
divide by *7*, *d* remains; then Easter will fall on the
*(b + 7 - d) ^{th}* day after 21 March;

If in step 3° the sum to be divided is an exact multiple of
*7*, let *d=0* be set, with this exception:
when either *b=29*, or *b=28* and also *a>10*,
for which case *d=7* must be taken.
The latter case will occur in the year 1954.

*Example* - Easter in 1886 :-

I = 18 = 4.4+2, e = 2, k = 86, k/4 = 21, 86+5.18 = 176 = 19.9+5, a = 5, 19.5+15+8 = 118 = 30.3+28, b = 28, 28+86+21+2-2.2 = 133 = 7.19+0, d = 0.

(because at the same time *a<10*).

Easter will fall on the

See Likely Errors in Implementations and Common Notes.

Read and adapt the 'Day of the Week Calculation' notes for the 1886 paper.

The dividend of the division by 7 is sometimes negative.

To avoid "mod of negative" error for the years 300, 702,
1101, 1503, 2600, 3401, ..., the term *-2e* in 3° can be
changed to *+5e*.