Doomsday rule

Doomsday for the current year in the Gregorian calendar (2026) is Saturday. Simple methods for finding the doomsday of a year exist.

One can find the day of the week of a given calendar date by using a nearby doomsday as a reference point. To help with this, the following is a list of easy-to-remember dates for each month that always land on the doomsday.

The last day of February is always a doomsday. For January, January 3 is a doomsday during common years and January 4 a doomsday during leap years, which can be remembered as "the 3rd during 3 years in 4, and the 4th in the 4th year". For March, one can remember either Pi Day or "March 0", the latter referring to the day before March 1, i.e. the last day of February.

For the months April through December, the even numbered months are covered by the double dates 4/4, 6/6, 8/8, 10/10, and 12/12, all of which fall on the doomsday. The odd numbered months can be remembered with the mnemonic "I work from 9 to 5 at the 7–11", i.e., 9/5, 7/11, and also 5/9 and 11/7, are all doomsdays (this is true for both the day/month and month/day conventions).^[8]

John Conway wrote that: "Summary: 'Last' in Jan and Feb, otherwise nth in even months, (n ± 4) in odd ones".^[1] He clarified that: "The sign is + for long odd months (31 days), and − for short ones (30 days)".^[1]

Several well-known dates, such as Independence Day in United States, Boxing Day, Halloween and Valentine's Day in common years, also fall on doomsdays every year.

Since the doomsday for a particular year is directly related to weekdays of dates in the period from March through February of the next year, common years and leap years have to be distinguished for January and February of the same year.

To find which day of the week Christmas Day of 2027 is, proceed as follows: in the year 2027, doomsday is on Sunday. Since December 12 is a doomsday, December 25, being thirteen days afterwards (two weeks less a day), will fall on a Saturday. Christmas Day is always the day of the week before doomsday. In addition, July 4 (U.S. Independence Day) is always on the same day of the week as a doomsday, as are Halloween (October 31), Pi Day (March 14), and December 26 (Boxing Day).

Since this algorithm involves treating days of the week like numbers modulo 7, John Conway suggested thinking of the days of the week as "Noneday" or "Sansday" (for Sunday), "Oneday", "Twosday", "Treblesday", "Foursday", "Fiveday", and "Six-a-day" to recall the number-weekday relation without needing to count them out in one's head.^[11]

There are some languages, such as Slavic languages, Chinese, Estonian, Greek, Portuguese, Galician and Hebrew, that base some of the names of the week days in their positional order. The Slavic, Chinese, and Estonian agree with the table above; the other languages mentioned count from Sunday as day one.

First take the anchor day for the century. For the purposes of the doomsday rule, a century starts with '00 and ends with '99. The following table shows the anchor day of centuries 1600–1699, 1700–1799, 1800–1899, 1900–1999, 2000–2099, 2100–2199 and 2200–2299.

For the Gregorian calendar:

Mathematical formula

5 × (c mod 4) mod 7 + Tuesday = anchor.

Algorithmic

Let r = c mod 4

if r = 0 then anchor = Tuesday

if r = 1 then anchor = Sunday

if r = 2 then anchor = Friday

if r = 3 then anchor = Wednesday.

For the Julian calendar:

6c mod 7 + Sunday = anchor.

Note: ${\displaystyle c={\biggl \lfloor }{{\text{year}} \over 100}{\biggr \rfloor }}$.

Next, find the year's anchor day. To accomplish that according to Conway:^[12]

1. Divide the year's last two digits (call this y) by 12 and let a be the floor of the quotient.
2. Let b be the remainder of the same quotient.
3. Divide that remainder by 4 and let c be the floor of the quotient.
4. Let d be the sum of the three numbers (d = a + b + c). (It is again possible here to divide by seven and take the remainder. This number is equivalent, as it must be, to y plus the floor of y divided by four.)
5. Count forward the specified number of days (d or the remainder of ⁠d/7⁠) from the anchor day to get the year's one.

${\displaystyle {\begin{matrix}\left({\left\lfloor {\frac {y}{12}}\right\rfloor +y{\bmod {1}}2+\left\lfloor {\frac {y{\bmod {1}}2}{4}}\right\rfloor }\right){\bmod {7}}+{\rm {{anchor}={\rm {Doomsday}}}}\end{matrix}}}$

For the twentieth-century year 1966, for example:

${\displaystyle {\begin{matrix}\left({\left\lfloor {\frac {66}{12}}\right\rfloor +66{\bmod {1}}2+\left\lfloor {\frac {66{\bmod {1}}2}{4}}\right\rfloor }\right){\bmod {7}}+{\rm {Wednesday}}&=&\left(5+6+1\right){\bmod {7}}+{\rm {Wednesday}}\\\ &=&{\rm {Monday}}\end{matrix}}}$

As described in bullet 4, above, this is equivalent to:

${\displaystyle {\begin{matrix}\left({66+\left\lfloor {\frac {66}{4}}\right\rfloor }\right){\bmod {7}}+{\rm {Wednesday}}&=&\left(66+16\right){\bmod {7}}+{\rm {Wednesday}}\\\ &=&{\rm {Monday}}\end{matrix}}}$

So doomsday in 1966 fell on Monday.

Similarly, doomsday in 2005 is on a Monday:

${\displaystyle \left({\left\lfloor {\frac {5}{12}}\right\rfloor +5{\bmod {1}}2+\left\lfloor {\frac {5{\bmod {1}}2}{4}}\right\rfloor }\right){\bmod {7}}+{\rm {{Tuesday}={\rm {Monday}}}}}$

The doomsday's anchor day calculation is effectively calculating the number of days between any given date in the base year and the same date in the current year, then taking the remainder modulo 7. When both dates come after the leap day (if any), the difference is just 365y + ⁠y/4⁠ (rounded down). But 365 equals 52 × 7 + 1, so after taking the remainder we get just

${\displaystyle \left(y+\left\lfloor {\frac {y}{4}}\right\rfloor \right){\bmod {7}}.}$

This gives a simpler formula if one is comfortable dividing large values of y by both 4 and 7. For example, we can compute

${\displaystyle \left(66+\left\lfloor {\frac {66}{4}}\right\rfloor \right){\bmod {7}}=(66+16){\bmod {7}}=82{\bmod {7}}=5}$

which gives the same answer as in the example above.

Where 12 comes in is that the pattern of ${\displaystyle {\bigl (}y+{\bigl \lfloor }{\tfrac {y}{4}}{\bigr \rfloor }{\bigr )}{\bmod {7}}}$ almost repeats every 12 years. After 12 years, we get ${\displaystyle {\bigl (}12+{\tfrac {12}{4}}{\bigr )}{\bmod {7}}=15{\bmod {7}}=1}$. If we replace y by y mod 12, we are throwing this extra day away; but adding back in ${\displaystyle {\bigl \lfloor }{\tfrac {y}{12}}{\bigr \rfloor }}$ compensates for this error, giving the final formula.

For calculating the Gregorian anchor day of a century: three “common centuries” (each having 24 leap years) are followed by a “leap century” (having 25 leap years). A common century moves the doomsday forward by

${\displaystyle (100+24){\bmod {7}}=2+3=5}$

days (equivalent to two days back). A leap century moves the doomsday forward by 6 days (equivalent to one day back).

So c centuries move the doomsday forward by

${\displaystyle \left(5c+{\biggl \lfloor }{c \over 4}{\biggr \rfloor }\right){\bmod {7}}}$,

but this is equivalent to

${\displaystyle (5(c{\bmod {4}})){\bmod {7}}}$.

Four centuries move the doomsday forward by

${\displaystyle -2-2-2-1=-7,\qquad -7\equiv 0\quad {\pmod {7}}}$;

so four centuries form a cycle that leaves the doomsday unchanged (and hence the “mod 4” in the century formula).

A simpler method for finding the year's anchor day was discovered in 2010 by Chamberlain Fong and Michael K. Walters.^[13] Called the "odd + 11" method, it is equivalent^[13] to computing

${\displaystyle \left(y+\left\lfloor {\frac {y}{4}}\right\rfloor \right){\bmod {7}}}$.

It is well suited to mental calculation, because it requires no division by 4 (or 12), and the procedure is easy to remember because of its repeated use of the "odd + 11" rule. Furthermore, addition by 11 is very easy to perform mentally in base-10 arithmetic.

Extending this to get the anchor day, the procedure is often described as accumulating a running total T in six steps, as follows:

1. Let T be the year's last two digits.
2. If T is odd, add 11.
3. Now let T = ⁠T/2⁠.
4. If T is odd, add 11.
5. Now let T = 7 − (T mod 7).
6. Count forward T days from the century's anchor day to get the year's anchor day.

Applying this method to the year 2005, for example, the steps as outlined would be:

1. T = 5
2. T = 5 + 11 = 16 (adding 11 because T is odd)
3. T = ⁠16/2⁠ = 8
4. T = 8 (do nothing since T is even)
5. T = 7 − (8 mod 7) = 7 − 1 = 6
6. Doomsday for 2005 = 6 + Tuesday = Monday The explicit formula for the odd+11 method is:

${\displaystyle 7-\left[{\frac {y+11(y\,{\bmod {2}})}{2}}+11\left({\frac {y+11(y\,{\bmod {2}})}{2}}{\bmod {2}}\right)\right]{\bmod {7}}}$.

Although this expression looks daunting and complicated, it is actually simple^[13] because of a common subexpression ⁠y + 11(y mod 2)/2⁠ that only needs to be calculated once.

Anytime adding 11 is needed, subtracting 17 yields equivalent results. While subtracting 17 may seem more difficult to mentally perform than adding 11, there are cases where subtracting 17 is easier, especially when the number is a two-digit number that ends in 7 (such as 17, 27, 37, ..., 77, 87, and 97).

Another method for calculating the doomsday was proposed by H. Nakai in 2023.^[14]

As above, let the year number n be expressed as ${\displaystyle n=100c+y}$, where ${\displaystyle c}$ and ${\displaystyle y}$ represent the century and the last two digits of the year, respectively. If ${\displaystyle c_{2}}$ and ${\displaystyle y_{2}}$ denote the remainders when ${\displaystyle c}$ and ${\displaystyle y}$ are divided by 4, respectively, then the number representing the day of the week for the doomsday is given by the remainder ${\displaystyle 5(c_{2}+y_{2}-1)+10y\quad {\text{mod}}\;(7)}$.

(August 7, 1966) The remainder on dividing ${\displaystyle (c,y)=(19,66)}$ by 4 is ${\displaystyle (c_{2},y_{2})=(3,2)}$, which gives ${\displaystyle 5\cdot (3+2-1)=20\equiv 6}$; 10 times ${\displaystyle y}$ is ${\displaystyle 10\cdot 66=660\equiv {2}}$, so doomsday for 1966 is ${\displaystyle 6+2=8\equiv 1}$, that is, Monday. The difference between 7 and the doomsday in August (namely 8) is ${\displaystyle 7-8=-1\equiv 6}$, so the answer is ${\displaystyle 1+6=7\equiv 0}$, Sunday.^[15]

Doomsday is related to the dominical letter of the year as follows.

Since in the Gregorian calendar there are 146,097 days, or exactly 20,871 seven-day weeks, in 400 years, the anchor day repeats every four centuries. For example, the anchor day of 1700–1799 is the same as the anchor day of 2100–2199, i.e. Sunday.

The full 400-year cycle of doomsdays is given in the adjacent table. The centuries are for the Gregorian and proleptic Gregorian calendar, unless marked with a J for Julian. The Gregorian leap years are highlighted.

Negative years use astronomical year numbering. Year 25BC is −24, shown in the column of −100J (proleptic Julian) or −100 (proleptic Gregorian), at the row 76.

A leap year with Monday as doomsday means that Sunday is one of 97 days skipped in the 400-year sequence. Thus the total number of years with Sunday as doomsday is 71 minus the number of leap years with Monday as doomsday, etc. Since Monday as doomsday is skipped across February 29, 2000, and the pattern of leap days is symmetric about that leap day, the frequencies of doomsdays per weekday (adding common and leap years) are symmetric about Monday. The frequencies of doomsdays of leap years per weekday are symmetric about the doomsday of 2000, Tuesday.

The frequency of a particular date being on a particular weekday can easily be derived from the above (for a date from January 1 – February 28, relate it to the doomsday of the previous year).

For example, February 28 is one day after doomsday of the previous year, so it is 58 times each on Tuesday, Thursday and Sunday, etc. February 29 is doomsday of a leap year, so it is 15 times each on Monday and Wednesday, etc.

Regarding the frequency of doomsdays in a Julian 28-year cycle, there are 1 leap year and 3 common years for every weekday, the latter 6, 17 and 23 years after the former (so with intervals of 6, 11, 6, and 5 years; not evenly distributed because after 12 years the day is skipped in the sequence of doomsdays).^{[citation needed]} The same cycle applies for any given date from March 1 falling on a particular weekday.

For any given date up to February 28 falling on a particular weekday, the 3 common years are 5, 11, and 22 years after the leap year, so with intervals of 5, 6, 11, and 6 years. Thus the cycle is the same, but with the 5-year interval after instead of before the leap year.

Thus, for any date except February 29, the intervals between common years falling on a particular weekday are 6, 11, 11. See e.g. at the bottom of the page Common year starting on Monday the years in the range 1906–2091.

For February 29 falling on a particular weekday, there is just one in every 28 years, and it is of course a leap year.

The Gregorian calendar is currently accurately lining up with astronomical events such as solstices. In 1582 this modification of the Julian calendar was first instituted. To correct for calendar drift, 10 days were skipped, so doomsday moved back 10 days (i.e. 3 weekdays): Thursday, October 4 (Julian, doomsday is Wednesday) was followed by Friday, October 15 (Gregorian, doomsday is Sunday). The table includes Julian calendar years, but the algorithm is for the Gregorian and proleptic Gregorian calendar only.

Note that the Gregorian calendar was not adopted simultaneously in all countries, so for many centuries, different regions used different dates for the same day.

Suppose we want to know the day of the week of September 18, 1985. We begin with the century's anchor day, Wednesday. To this, add a, b, and c above:

• a is the floor of ⁠85/12⁠, which is 7.
• b is 85 mod 12, which is 1.
• c is the floor of ⁠b/4⁠, which is 0.

This yields a + b + c = 8. Counting 8 days from Wednesday, we reach Thursday, which is the doomsday in 1985. (Using numbers: In modulo 7 arithmetic, 8 is congruent to 1. Because the century's anchor day is Wednesday (index 3), and 3 + 1 = 4, doomsday in 1985 was Thursday (index 4).) We now compare September 18 to a nearby doomsday, September 5. We see that the 18th is 13 past a doomsday, i.e. one day less than two weeks. Hence, the 18th was a Wednesday (the day preceding Thursday). (Using numbers: In modulo 7 arithmetic, 13 is congruent to 6 or, more succinctly, −1. Thus, we take one away from the doomsday, Thursday, to find that September 18, 1985, was a Wednesday.)

Suppose that we want to find the day of week that the American Civil War broke out at Fort Sumter, which was April 12, 1861. The anchor day for the century was 94 days after Tuesday, or, in other words, Friday (calculated as 18 × 5 + ⌊⁠18/4⁠⌋; or just look at the chart, above, which lists the century's anchor days). The digits 61 gave a displacement of six days so doomsday was Thursday. Therefore, April 4 was Thursday so April 12, eight days later, was a Friday.

• Ordinal date
• Computus – Gauss algorithm for Easter date calculation
• Zeller's congruence – An algorithm (1882) to calculate the day of the week for any Julian or Gregorian calendar date.
• Mental calculation