Is January 1, 2001, a Monday?

(Calendar Maths)

(Adapted for 2021)

by Dr Valentin Hristov


Although the CALENDAR seems mysterious in its workings, most of this article should be simple enough for Form 1 pupils to understand. So, share the ideas presented here with your brothers, sisters, and friends. By memorizing a few things, you will be able to make easy calendar calculations in your head.

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 2021, 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).
Mon
1
8
15
22
29
Tue
2
9
16
23
30
Wed
3
10
17
24
31
Thu
4
11
18
25
Fri
5
12
19
26
Sat
6
13
20
27
Sun
7
14
21
28

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 some number 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) (\(\ m \equiv n\) (mod p) means that the difference m-n is divisible by p, i.e. m and n have the same remainder when divided by p.). For example 26 November 2021 is Friday because 26 \(\equiv\) 5 (mod 7). This is a short expression for \(26 = 3\times 7+5\ \) or \(\ \displaystyle{26\over 7} = 3 +{5 \over 7}\).

Now we want to use the same method for other months in 2021. Let us start from March, in which there are 31 days, and note that 31 \(\equiv\) 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 2021, it is Tuesday, because 20+3 = 23 \(\equiv\) 2 (mod 7). Let us call the number 3 the correction for April 2021 and we can take the number 0 as the correction for February, March, and November 2021.

What is the correction for May 2021? Using April as a basis, there are 2 extra days in it after excluding the full weeks (30 \(\equiv\) 2 (mod 7)), so we have to add 2 to the April correction 3 and therefore the correction for May 2021 is 5. Then you can check that 15 May 2021 is Saturday (15+5 \(\equiv\) 6 (mod 7)).

In this way, you can continue calculating consecutively the corrections for all months until the end of 2021 - 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 since 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):

(144 = 122, 025 = 52, 036 = 62, 146 = 122+2)

  
J
F
M
A
M
J
J
A
S
O
N
D
normal
1
4
any year
4
0
2
5
0
3
6
1
4
6
leap
0
3

Now, let us adapt this fixed table for 2021. 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 2021. For example, November has correction 4 in the fixed table. If we add 3 for 2021, then the sum 7 can be reduced to 0 by subtracting 7 (we need the remainder), which corresponds to the correction 0 for November 2021 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 we found 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 \(\equiv\) 1 (mod 7), i.e. we remove the 52 full weeks and then equal dates in the months from March to December differ by one day of the week. For January and February the upper corrections 1 and 4 in the table should be incresed by 1.

If the next year is leap, we can add 2 for the months from March to December, because 366 \(\equiv\) 2 (mod 7), but then the special smaller lower corrections 0 and 3 for January and February from the fixed table should be incresed by 1.

Let us write the corrections for consecutive years, starting with an artificial 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

0
1
2
3
4
5
6
7
8
9
10
11
12
13
0
1
2
3
5
6
0
1
3
4
5
6
1
2
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
3
4
6
0
1
2
4
5
6
0
2
3
4
5
0

Where is the year 2021 in this cycle? It has correction 3 and is a normal year immediately after a leap year (the position is important!). You can check that it corresponds to the year numbered 25 in our sequence, so the initial year 0 must be 1996 ( = 2021 - 25 ). With this initial data, let us find when the new century has started. (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 \(\equiv\) 1 (mod 7), so 1 January 2001 has been 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 was the day of the week 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 \(\equiv\) 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 the leap years need additional increment of the correction by 1 after every 4 years. Therefore

$$ YC \equiv N+\left[{N\over 4}\right]\,({\rm mod\ } 7), $$ where the second term \(\displaystyle\left[{p \over q}\right]\) denotes the whole part of the quotient \(\displaystyle{p \over q}\). Note that \(\displaystyle\left[{N \over 4}\right]\) is 0 if  N = 0,1,2,3; it is 1 if  N = 4,5,6,7; etc. You can memorize the initial years of the cycle and the following easy formula for N which uses them:
N
=
Year (between 1901 and 2099) -
-1884, 1912, 1940, 1968, 1996, 2024, 2052, 2080,
 where 0 \(\le\) N \(\le\) 27.

(A short expression for the formula is N \(\equiv\) 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.

OLD STYLE:

(Last date October 4, 1582) (In Bulgaria: last date March 31, 1916)

\[DW\equiv 6+D+2\cdot M+\left[{3\cdot(M+1)\over 5}\right]+Y_2+\left[{Y_2\over 4}\right] +6\cdot Y_1\quad \mbox{(mod 7)}\]

NEW STYLE:

(First date October 15, 1582) (In Bulgaria: first date April 14, 1916)

\[DW\equiv 1+D+2\cdot M+\left[{3\cdot(M+1)\over 5}\right]+Y_2+\left[{Y_2\over 4}\right] +5\cdot Y_1+\left[{Y_1\over 4}\right]\quad \mbox{(mod 7)}\]


\[\begin{array}{ll} D\mbox{ - Date (1-31)}&\mbox{Example: } 27.01.2021 = 27.13.2020\cr M\mbox{ - Month (3-14)}&D = 27,\ M = 13,\ Y1 = 20,\ Y2 = 20.\cr Y_2\mbox{ - Year in the century}& DW\equiv 1+27+2\cdot 13+8+20+5+5\cdot 20+5\equiv\cr Y_1\mbox{ - Full centuries} \left( \left[ {\mbox{year}\over\mbox{100}}\right]\right) &\hskip 21pt\equiv 1+6+2\cdot 6+1+6+5+5\cdot 6+5=\cr &\hskip 21pt =66\equiv 3\ (\mbox{mod 7})\mbox{ - Wednesday} \cr &DW\mbox{ - Day of the week (0-6)}\cr &\hskip 20pt 0\mbox{ - Sunday}\cr &\hskip 20pt 1\mbox{ - Monday}\cr &\hskip 20pt 2\mbox{ - Tuesday}\cr \left[ a \right] \mbox{ is the whole part of }a &\hskip 20pt 3\mbox{ - Wednesday}\cr &\hskip 20pt 4\mbox{ - Thursday}\cr c\equiv d\ \ (\mbox{mod 7})\mbox{ means }c\mbox{ and }d \mbox{ have} &\hskip 20pt 5\mbox{ - Friday}\cr \mbox{equal remainders divided by 7}& \hskip 20pt 6\hbox{ - Saturday}\cr \end{array}\]

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 \(\ \displaystyle Y_2+\left[{Y_2\over 4}\right]\) 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 \(\ \displaystyle 5\cdot Y_1+\left[{Y_1\over 4}\right]\) (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.


Footnote:

1+3+29 \(\equiv\) 5 (mod 7)

Dr Valentin Z. Hristov


My Stuff


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