There is such a vast historical and technical literature on the calendar, that we can only present a tiny part of it. Here we will discuss the history briefly, and describe some simple mathematics behind the Julian and the Gregorian calendars known also as old-style and new-style respectively.
In both styles, a normal year has 365 days and leap year 366 days (when 29 February is added). In the Julian calendar the rule is that the year is leap if it can be divided by 4. The error of this calendar is 1 day approximately in 128 years. Because of that, in the 16th century the astronomical equinox (when the lengths of the day and the night are equal) appeared 10 days earlier than the accepted date 21 March. Therefore Pope Gregory XIII made a calendar reform which skipped 10 dates and changed the rule for the leap years. In the new style not all years divisible by 4 are leap. The exception is that among the exact centuries (ending 00) only the years divisible by 400 are leap. For example, the years 1600, 2000, 2400 are leap, but 1700, 1800, 1900, 2100, 2200 are not leap. In the new style an error of 1 day accumulates approximately in 3280 years, which is reasonably close to the real astronomical year for the usual practical purposes.
In the different countries the new style was introduced in different centuries. The first introduction was in Italy in 1582, when the date 4 October (Thursday) was followed by 15 October (Friday) with 10 dates skipped. The Catholic part of Europe accepted the new style in the same century and in the beginning of the next one. But the Orthodox countries only changed the style at the beginning of our century. For example, in Bulgaria the date 31 March 1916 (Thursday) was followed by the date 14 April 1916 (Friday). Perhaps you have noticed that, by the 20th century, the number of skipped dates is already 13, because the years 1700, 1800, and 1900 are leap years in the old, but not in the new style, which gives three additional dates to be removed.
It is interesting that in general there is a repetition of the centuries in the new style every 400 years, while for the old style such repetition appears every 700 years. We hope you will be able to check these facts after reading the rest of the paper.
Now, look at the calendar for 1999, and pay attention to the months February,
March, and November. They all start on Monday and only the number of the days
is different (28, 31, and 30).
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Let us assign a number to each day of the week: Mon(1), Tue(2), Wed(3), Thu(4), Fri(5), Sat(6), Sun(0). For Sunday we deliberately take 0 because we want to deal with the remainder when something is divided by 7. With that correspondence we can determine the day of the week in the above mentioned months simply by taking the remainder modulo 7, that is, after dividing by 7; we use notation (mod 7) (m º n (mod p) means that m-n isdivisible by p, i.e. m and n have the same remainder when divided by p.). For example 26 November 1999 is Friday because 26 º 5 (mod 7). This is a short expression for 26 = 3×7+5 or (26/7) = 3+5/7.
Now we want to use the same method for other months in 1999. Let us start from March, in which there are 31 days, and note that 31 º 3 (mod 7) (31 = 4×7+3). So we have four full weeks and three extra days. These extra days move the beginning of April to Thursday, and in fact any date in April is moved by 3 days of the week with respect to the same date in March. If you take the date 1 and add 3, the result is 4, which corresponds to Thursday. If you take an arbitrary date in April, for example, 20 April 1999, it is Tuesday, because 20+3 = 23 º 2 (mod 7). Let us call the number 3 the correction for April 1999 and we can take the number 0 as the correction for February, March, and November 1999.
What is the correction for May 1999? Using April as a basis, there are 2 extra days in it after excluding the full weeks (30 º 2 (mod 7)), so we have to add 2 to the April correction 3 and therefore the correction for May 1999 is 5. Then you can check that 15 May 1999 is Saturday (15+5 º 6 (mod 7)).
In this way, you can continue calculating consecutively the corrections for all months until the end of 1999 - use the number of the days in each month: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 - or (easier) just use their remainders (mod 7). As an exercise we suggest that you do this, with the hint that, if the correction becomes bigger than 6, you have to subtract 7 in order to have a number between 0 and 6. You can check if your calculations are correct when you reach November, because the correction should be 0 as mentioned earlier.
You can make a table with the month corrections for each separate year and memorize it for use. In general only 14 different tables do exist - 7 for the normal years and 7 for the leap years. But it is not worth changing your table every year if there is something easier. Here we suggest that you memorize only two tables - one for normal years and one for leap years. In fact, we can even combine them in one table as follows:
| Month correction MC (easy to remember): |
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Now, let us use this fixed table for 1999. It is obvious that the corrections do not coincide with the real ones, but we can involve a new correction for the year, which must be 3 for 1999. For example, November has correction 4 in the fixed table. If we add 3 for 1999, then the sum 7 can be reduced to 0 by subtracting 7 (we need the remainder), which corresponds to the correction 0 for November 1999 discussed earlier.
Next step is to determine the corrections for different years when only the fixed table above is to be used. Here we can work as for finding the consecutive corrections for the months above. Let us start with the case when the next year is a normal year. Then we have to add only one extra day, because 365 º 1 (mod 7), i.e. we remove the 52 full weeks and equal dates (in the months from March to December) differ by one day of the week. If the next year is leap, we can add 2, because 366 º 2 (mod 7), and then the special smaller corrections for January and February from the fixed table must be used. Therefore, going to 2000 (a leap year) will add 2 to our correction 3 for 1999, so the correction for 2000 is 5.
Let us write the corrections for consecutive years, starting with a leap year numbered 0 with correction 0. There is a repetition of the correction for the leap years (shown in bold) every 28 years. At the same time every correction for normal years appears exactly 3 times in the 3 possible positions with respect to the leap years.
| Year correction YC in the 28-years cycle |
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Where is the year 1999 in this cycle? It has correction 3 and is a normal year before a leap year (the position is important!). You can check that it corresponds to the year with number 3 in our sequence, so the initial year 0 must be 1996. With this initial data, let us find when the new century will start. (Many people do not know that it starts officially on 1 January 2001, NOT 2000 !!!) So, the year 2001 is with number 5 in our cycle and therefore the correction for it is 6. Next, add the correction 1 for January (normal year) from the fixed table and the date 1. The result is 6+1+1 = 8 º 1 (mod 7), so 1 January 2001 is on Monday! We invite you to check if you understand the procedure by calculating the day of the week for 29 February 2008 (leap year). Your answer should be ``Friday". (Try not to look at the footnote at the bottom of the page.)
At the moment our table can serve for the period 1996-2024, but the 28 years cycle can be continued to the past and to the future. Let us note that only the years between 1901 and 2099 can be considered, because 1900 and 2100 are not leap in the new style. In such a case the initial years (with number 0) in the cycle for 1901-2099 are 1884 (for 1901-1911 only), 1912, 1940, 1968, 1996, 2024, 2052, and 2080 (for 2080-2099 only).
Let us find, for example, what day of the week it was when a person was born on 17 August 1987? The year is in the period 1968-1995, so, its number in the cycle is 1987-1968 = 19. The correction for the 19-th year is 2. Therefore we add YC, MC, and the date to get 2+3+17 = 22 º 1 (mod 7), which gives Monday. The same calculation can be made for 17 August in the years 1931, 1959, 2015, etc. (1931-1912 = 1959-1940 = 2015-1996 = ... = 19). Check the method using your date of birth or that of your brothers or sisters, and ask your parents if your answer is correct.
Up to now we have used ready-made tables with corrections for the month and for
the year, but it is possible to give formulas for the corrections, to remember
them or include them in computer programs. It is easier to start with a formula
for the year correction YC in the table above. If all years were normal, then
their number would give the correction (taking the remainder (mod 7)), but in
fact we have to increase additionally the correction by 1 every 4-th year.
Therefore
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(A short expression for the formula is N º Y-1884 (mod 28).)
Let us go to the formula for the months corrections. First note that the additional day for the leap years is between February and March and therefore in this formula we take the beginning of the years to be March (numbered by 3) and
| January | = 13-th month of the PREVIOUS year! |
| February | = 14-th month of the PREVIOUS year! |
This time we are trying to find an expression, which must change the correction
by 2 or 3 depending on the length of the month (February is now the last 14-th
month of our year and the number of its days is not important). Therefore our
formula must be of the form 2·M+Z, where Z depends on M and must
increase by 0 or 1 when the previous month has 30 or 31 days respectively. We
can check the expressions of the form [(A·M+B)/C] by substituting M = 3,4,...,14. It appears that the smallest
possible value of C is 5, and we have the proper increment if A = 3 and
B = 3. In the formulas below we include additionally the centuries with some
explanation after the formulas.
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OLD STYLE:
(Last date October 4, 1582) (In Bulgaria: last date March 31, 1916)
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NEW STYLE:
(First date October 15, 1582) (In Bulgaria: first date April 14, 1916)
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Note: If the calculations are made by hand, it is easier to take for every term in the sum its remainder (mod 7) (cf. the Example above).
If you look at the terms Y2+[(Y2)/4] in the above formulas, you will recognize the formula for YC, but now with respect to the year 00 in the century. In the formula for the old style the term 6·Y1 expresses the fact that each century has 6 extra days if the whole weeks are removed (check this by calculating the number of days in a century, or easier by taking 75 normal and 25 leap years which change the correction by 1 and 2 respectively). In the formula for the new style the terms 5·Y1+[(Y1)/4] (they look familiar, don't they?) express the fact that the centuries are usually shorter by 1 day compared to the old style (only 5 extra days) and every 400 years 1 extra day is added.
So, the above formulas contain the corrections for all components of a date - (1) the date in the month; (2) the month; (3) the year in the century (the last two digits); (4) the hundreds of the year (= the century - 1). In the beginning there are constants (6 and 1 resp.) to adjust the corrections to the real day of the week.
I will finish with the hope that this mathematical description of the calendar will be useful for you and give you fun, and I would encourage you to memorize some of the easy formulas in this article. The ideas presented here are the basis for many different calendar tables and constructions. The author invites you to contact him with any questions related to the calendar.
P.S. If you use a computer with Unix system, type ``cal'' to see the calendar for the current month or ``man cal'' to check all options.
Dr Valentin Z. Hristov