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sci.math FAQ: Day of Week

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Archive-Name: sci-math-faq/dayWeek
Last-modified: December 8, 1994
Version: 6.2

How to determine the day of the week, given the month, day and year

   First a brief explanation: In the Gregorian Calendar, over a period of
   four hundred years, there are 97 leap years and 303 normal years. Each
   normal year, the day of January 1 advances by one; for each leap year
   it advances by two.

   303 + 97 + 97 = 497 = 7 * 71

   As a result, January 1 year N occurs on the same day of the week as
   January 1 year N + 400 . Because the leap year pattern also recurs
   with a four hundred year cycle, a simple table of four hundred
   elements, and single modulus, suffices to determine the day of the
   week (in the Gregorian Calendar), and does it much faster than all the
   other algorithms proposed. Also, each element takes (in principle)
   only three bits; the entire table thus takes only 1200 bits, or 300
   bytes; on many computers this will be less than the instructions to do
   all the complicated calculations proposed for the other algorithms.

   Incidental note: Because 7 does not divide 400, January 1 occurs more
   frequently on some days than others! Trick your friends! In a cycle of
   400 years, January 1 and March 1 occur on the following days with the
   following frequencies:

           Sun      Mon     Tue     Wed     Thu     Fri     Sat
    Jan 1: 58       56      58      57      57      58      56
    Mar 1: 58       56      58      56      58      57      57

   Of interest is that (contrary to most initial guesses) the occurrence
   is not maximally flat.

   The Gregorian calendar was introduced in 1582 in parts of Europe; it
   was adopted in 1752 in Great Britain and its colonies, and on various
   dates in other countries. It replaced the Julian Calendar which has a
   four-year cycle of leap years; after four years January 1 has advanced
   by five days. Since 5 is relatively prime to 7, a table of 4 * 7 = 28
   elements is necessary for the Julian Calendar.

   There is still a 3 day over 10,000 years error which the Gregorian
   calendar does not take into account. At some time such a correction
   will have to be done but your software will probably not last that

   Here is a standard method suitable for mental computation:

    1. Take the last two digits of the year.
    2. Divide by 4, discarding any fraction.
    3. Add the day of the month.
    4. Add the month's key value: JFM AMJ JAS OND 144 025 036 146
    5. Subtract 1 for January or February of a leap year.
    6. For a Gregorian date, add 0 for 1900's, 6 for 2000's, 4 for
       1700's, 2 for 1800's; for other years, add or subtract multiples
       of 400.
    7. For a Julian date, add 1 for 1700's, and 1 for every additional
       century you go back.
    8. Add the last two digits of the year.
    9. Divide by 7 and take the remainder.

   Now 1 is Sunday, the first day of the week, 2 is Monday, and so on.

   The following formula, which is for the Gregorian calendar only, may
   be more convenient for computer programming. Note that in some
   programming languages the remainder operation can yield a negative
   result if given a negative operand, so mod 7 may not translate to a
   simple remainder. W = (k + floor(2.6m - 0.2) - 2C + Y + floor(Y/4) +
   floor(C/4)) mod 7 where floor() denotes the integer floor function,
   k is day (1 to 31)
   m is month (1 = March, ..., 10 = December, 11 = Jan, 12 = Feb) Treat
   Jan &Feb as months of the preceding year
   C is century (1987 has C = 19)
   Y is year (1987 has Y = 87 except Y = 86 for Jan &Feb)
   W is week day (0 = Sunday, ..., 6 = Saturday)
   Here the century and 400 year corrections are built into the formula.
   The floor(2.6m - 0.2) term relates to the repetitive pattern that the
   30-day months show when March is taken as the first month.

   The following short C program works for a restricted range


   The program appeared was posted by (Tomohiko
   Sakamoto) on comp.lang.c on March 10th, 1993.

   A good mnemonic rule to help on the computation of the day of the week
   is as follows. In any given year the following days come on the same
   day of the week:


   to remember the next four, remember that I work from 9-5 at a 7-11 so


   and the last day of Feb.

   "In 1995 they come on Tuesday. Every year this advances one other than
   leap-years which advance 2. Therefore for 1996 the day will be
   Thursday, and for 1997 it will be Friday. Therefore ordinarily every 4
   years it advances 5 days. There is a minor correction for the century
   since the century is a leap year iff the century is divisible by 4.
   Therefore 2000 is a leap year, but 1900, 1800, and 1700 were not."

   Even ignoring the pattern over for a period of years this is still
   useful since you can generally figure out what day of the week a given
   date is on faster than someone else can look it up with a calender if
   the calender is not right there. (A useful skill that.)


   Winning Ways for your mathematical plays. Guy Conway and Elwyn
   Berlekamp London ; Toronto : Academic Press, 1982.

   Mathematical Carnival. Martin Gardner. New York : Knopf, c1975.

   Elementary Number Theory and its applications. Kenneth Rosen. Reading,
   Mass. ; Don Mills, Ont. : Addison-Wesley Pub. Co., c1993. p. 156.

   Michael Keith and Tom Craver. The Ultimate Perpetual Calendar? Journal
   of Recreational Mathematics, 22:4, pp. 280-282, 19

    Tue Apr 04 17:26:57 EDT 1995

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