Zeller’s Rule: Day on any date in the calendar

Vineet Patawari Calendars & Clocks, Speedy Calculation

Zeller’s Rule : With this technique named after its founder Zeller, you can solve any ‘Dates and Calendars’ problems.

Zellers rule can be used to find the day on any particular date in the calendar in the history. All you have to know is the formula given below and how to use it.

Zeller’s Rule Formula:

F = K + [(13xM – 1)/5] + D + [D/4] + [C/4] – 2C

K = Date => for 25/3/2009, we take 25
In Zeller’s rule months start from march.
M = Month no. => Starts from March.
March = 1, April = 2, May = 3
Nov. = 9, Dec = 10, Jan = 11
Feb. = 12
D = Last two digits of the year, using previous year for January and February. Thus for 2009 = 09
C = The first two digits of century => for 2009 = 20

Example: 25/03/2009

F = 25 + [{(13 x1)- 1}/5] + 09 + 09/4 + 20/4 – (2 x 20)
= 25 + 12/5 + 09 + 09/4 + 20/4  – 2×20
=25+2+09+2+5-40 [ We will just consider the integral value and ignore the value after decimal]
= 43 – 40 =

Replace the number with the day using the information given below.
1 = Monday
2 = Tuesday
3 = Wednesday
4 = Thursday
5 = Friday
6 = Saturday
7 or 0 = Sunday

So it’s Wednesday on 25th march, 2009.

If the number is more than 7, divide the no. by 7. The remainder will give you the day.

I shall be very grateful if anyone of you can provide me the java script for this formula, so that I can post it here for everyone’s convenience.

Thanks in advance. I hope you will find the above method very useful.

you may find another post named, cyclicity very interesting.

Vineet Patawari – PGDM, ACA, B.COM (H)

Please follow and like us:

You may also like:

  1. Find Day of the Week on Any Date
  2. Speed Multiplication by 111 : Vedic Maths
  3. Vedic Multiplication by 9, 99, 999 and so on

Vineet Patawari

Hi, I'm Vineet Patawari. I fell in love with numbers after being scared of them for quite some time. Now, I'm here to make you feel comfortable with numbers and help you get rid of Math Phobia!

72 thoughts to “Zeller’s Rule: Day on any date in the calendar”

    1. Yes because you might have taken D=18 as last two digits of year but in this case it should be D=18-1=17 because we are considering that our year starts from march and ends in february ..So if we are given jan/feb months than in those cases D =last two digits-1 . I hope It works.

      Reply

  8. Sir,
    I want to know short cut to find date.
    Here the question is: what will be the date after 73days from 23rd September 2017

    Reply

    1. Add 7 days to 23rd September……now remaining days 66…..subt 31 days for oct and 30 for nov respectively …..now left days 5 . So it will be 5th dec

      Reply

    1. Answer for 14 jan 2001

      Put this in zeller’s rule ….we get

      14 + (13*11-1)/5 + 00 + 00/4 + 5 – 40
      = 14+28+5-40
      =7 now divide by 7
      =0 i.e. sunday

      Reply

    1. Answer for 14 jan 2001

      Put this in zeller’s rule ….we get

      14 + (13*11-1)/5 + 00 + 00/4 + 5 – 40
      = 14+28+5-40
      =7 now divide by 7
      =0 i.e. sunday

      Reply

    2. For 1st jan 2000 ….put 1999 instead 2000 u will get ur answer…..bcoz u hv to subtract 1 for jan & feb month from the given year

      Reply

  11. “””implementation of Zeller’s Equation to calculate day of week for a given date using python.
    Its modified for convenience of people.You can take jan=1,feb=2…so on.It can take input in any of the following forms :
    13-Jan-2015 or 13-01-2015
    13/Jan/2015 or 13/1/2015
    13,Jan,2015 or 13,1,2015
    Note:for any month use 3 letter short-form with initial letter capitalised. e.g. Jan, Apr etc.
    Implementation is in Python 3
    ————————————————————————–
    “””
    from enum import Enum
    class days(Enum):
    Sunday=0
    Monday=1
    Tuesday=2
    Wednesday=3
    Thursday=4
    Friday=5
    Saturday=6
    class months(Enum):
    Jan=1
    Feb=2
    Mar=3
    Apr=4
    May=5
    Jun=6
    Jul=7
    Aug=8
    Sep=9
    Oct=10
    Nov=11
    Dec=12
    def findDayOfWeek(day,month,year):
    if(month.isalpha()):
    month=months[month].value
    K=int(day)
    if(int(month)-2<=0):
    M=int(month)+10
    else:
    M=int(month)-2
    if(M==11 or M==12):
    year=str(int(year)-1)
    C=int(year[0:2])
    D=int(year[2:])
    return (K+int((13*M-1)/5)+D+int(D/4)+int(C/4)-2*C)%7
    date=input().strip()
    if "/" in date:splitter="/"
    elif "-" in date:splitter="-"
    else:splitter=","
    (day,month,year)=date.split(splitter)
    x=findDayOfWeek(day,month,year)
    print(days(x).name)
    #Code ends here.
    """
    —————————————————-
    """

    Reply

    1. THIS IS A BETTER CODE FOR ZELLER RULE USING PYTHON
      THIS IS MADE BY A 14 YEAR OLD CODER WITH BASIC KNOWLEDGE OF PYTHON
      I HAVE LEARNT IT IN THE 2ND MODULE OF PYTHON LEARNING ON SOLOLEARN APP
      REPLY TO PRESENT YOUR VIEWS
      —————————————————
      i=0
      while i==0:
      print(“DATE TO DAY CONVERTER”)
      date = int(input(“ENTER A DATE (DD) “))
      month = int(input(“ENTER THE MONTH NUMBER(MM) “))
      yearcent = int(input(“ENTER THE 1ST 2 DIGITS OF THE YEAR (YY–) “))
      yearaft = int(input(“ENTER THE LAST 2 DIGITS OF THE YEAR (–YY) “))
      RISHIMA=[11,12,1,2,3,4,5,6,7,8,9,0]
      monthno=0
      monthno=int(RISHIMA[month-1])
      if month==1 or month==2:
      if yearaft==00:
      yearaft=99
      else:
      yearaft=yearaft-1
      monthfunc=int((13*monthno-1)/5)
      day=date+monthfunc+yearaft+int(yearaft/4)+int(yearcent/4)-(2*yearcent)
      while day=7:
      day=day%7
      PRAGYANSH=[“SUNDAY”,”MONDAY”,”TUESDAY”,”WEDNESDAY”,”THURSDAY”,”FRIDAY”,”SATURDAY”]
      print (PRAGYANSH[day])
      x=input(“DO YOU WANT TO QUIT (y/n) “)
      if x==”y”:
      print(“THANKS FOR USING ME ;)”)
      break
      else:
      print(“LETS DO IT ALL AGAIN”)
      ———————————————————–
      CODE ENDED
      ITS IN AN OPTIONAL LOOP

      Reply

      1. DATE TO DAY CONVERTER V2
        NOW INPUTS DATA IN DD/MM/YY, DD.MM.YY, DD,MM,YY AND ANY FORMAT
        ————————————————————————————
        i=0
        while i==0:
        print(“DATE TO DAY CONVERTER”)
        mum=input(“INPUT A DATE IN DD/MM/YYYY FORMAT “)
        date= int(mum[0]+mum[1])
        month=int(mum[3]+mum[4])
        yearcent=int(mum[6]+mum[7])
        yearaft=int(mum[8]+mum[9])
        RISHIMA=[11,12,1,2,3,4,5,6,7,8,9,0]
        monthno=0
        monthno=int(RISHIMA[month-1])
        if month==1 or month==2:
        if yearaft==00:
        yearaft=99
        else:
        yearaft=yearaft-1
        monthfunc=int((13*monthno-1)/5)
        day=date+monthfunc+yearaft+int(yearaft/4)+int(yearcent/4)-(2*yearcent)
        while day=7:
        day=day%7
        PRAGYANSH=[“SUNDAY”,”MONDAY”,”TUESDAY”,”WEDNESDAY”,”THURSDAY”,”FRIDAY”,”SATURDAY”]
        print (PRAGYANSH[day])
        x=input(“DO YOU WANT TO QUIT (y/n) “)
        if x==”y”:
        print(“THANKS FOR USING ME ;)”)
        break
        else:
        print(“LETS DO IT ALL AGAIN”)
        ———————————————————————–
        END OF CODE
        ENJOY THE VERSION 2 CODE

      2. DATE TO DAY CONVERTER V3
        NOW WITH ERRORCHECKS
        ————————————————————-
        i=0
        while i==0:
        print(“DATE TO DAY CONVERTER V3”)
        print(“VERSION UPDATES :”)
        print(“18/02/2017 16:57 V3 : ADDED ERRORCHECKS”)
        print(“18/02/2017 16:40 V2 : ADDED DD/MM/YY INPUT”)
        mum=input(“INPUT A DATE IN DD/MM/YYYY FORMAT “)
        if len(mum)==10:
        date= int(mum[0]+mum[1])
        month=int(mum[3]+mum[4])
        yearcent=int(mum[6]+mum[7])
        yearaft=int(mum[8]+mum[9])
        RISHIMA=[11,12,1,2,3,4,5,6,7,8,9,0]
        monthno=0
        monthno=int(RISHIMA[month-1])
        if month==1 or month==2:
        if yearaft==00:
        yearaft=99
        else:
        yearaft=yearaft-1
        monthfunc=int((13*monthno-1)/5)
        day=date+monthfunc+yearaft+int(yearaft/4)+int(yearcent/4)-(2*yearcent)
        while day=7:
        day=day%7
        PRAGYANSH=[“SUNDAY”,”MONDAY”,”TUESDAY”,”WEDNESDAY”,”THURSDAY”,”FRIDAY”,”SATURDAY”]
        print (PRAGYANSH[day])
        x=input(“DO YOU WANT TO QUIT (y/n) “)
        if x==”y”:
        print(“THANKS FOR USING ME ;)”)
        break
        elif x==”n”:
        print(“LETS DO IT ALL AGAIN”)
        else:
        print(“INVALID INPUT…….”)
        print(“CONTINUING THE PROGRAM…….”)
        else:
        print(“INVALID INPUT”)
        ———————————————————————
        CODE ENDED
        ITS IN AN OPTIONAL LOOP
        MADE BY A 14 YEAR OLD CODER

      1. For jan and feb …. subtract 1 from the given year…..and putthe values in zeller’s formula.
        You’llget ur answer

    1. 3+(13*1-1)/5+89+89/4+19/4-38
      ==> 3+2.4+89+22.25+4.75-38
      =3+2+89+22+4-38
      =82/7
      =5 remaining days. I.e friday

      Reply

  15. Hi,
    In response to “I shall be very grateful if anyone of you can provide me the java script for this formula, so that I can post it here for everyone’s convenience.”

    Today I just read the Zeller’s Rule from ur blog and have build a js code for u.
    I need ur email Id.

    Reply

  16. I see a lot of interesting articles on your blog.
    You have to spend a lot of time writing, i know how to save you a lot of time, there is a tool that creates unique,
    SEO friendly articles in couple of minutes,
    just type in google – laranita’s free content source

    Reply

    1. The Gregorian calendar was instituted on AD 1582 Oct 15th.
      and 1752 Sep 13 was on wed day.

      Reply

  19. at last i have got solution. really it is a awesome work. it should be given a thanks to admin for the demonstration

    Reply

  20. hello!,I like your writing so a lot! share we be in contact extra about your post on AOL? I need a specialist on this house to resolve my problem. May be that is you! Looking ahead to peer you.

    Reply

    2. 1988 was leap year .and Jan 1st was on Friday. And Jan 8th and Jan 15th also on Friday. So Jan 16th was on Satur day.

      Reply

    1. It is working. Above the extra rule is forgotten: If you have to determine a date in January (M = 11) or February (M = 12) it belongs to the prior year.
      So in the above case C = 51 (and not 52!).

      Reply

    3. 1988 was leap year .and Jan 1st was on Friday. And Jan 8th and Jan 15th also on Friday. So Jan 16th was on Satur day.

      I prepared such a book named CALENDAR R FOR EVER form which you can find out any day of any date since the Gregorian Calendar was institued.

      Reply

  26. Hey buddy i wrote the java code but oit is not working for the jan feb and march remaining all r fine i’m sending u the code lets c wat u can say abt the code and do reply me on raj.tvss@gmail.com or u can message me on 7204974366 i’m resedent of bangalore and here is the code

    CODE:

    import java.util.Scanner;
    import java.*;

    public class Date{
    public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    int d,c,day;

    //
    // Read year
    //
    System.out.print(“Year: “);
    int y = scanner.nextInt();

    // Read date
    System.out.print(“Date: “);
    int k = scanner.nextInt();

    //
    // Read month
    //
    System.out.print(“Month in numeric formate: “);
    int m = scanner.nextInt();

    switch (m)
    {
    case 1: m=11;
    break;
    case 2: m=12;
    break;
    case 3: m=1;
    break;
    case 4: m=2;
    break;
    case 5: m=3;
    break;
    case 6: m=4;
    break;
    case 7: m=5;
    break;
    case 8: m=6;
    break;
    case 9: m=7;
    break;
    case 10: m=8;
    break;
    case 11: m=9;
    break;
    case 12: m=10;
    break;
    }

    d=y%100;
    c=y/100;
    day = k*((13*(m-1))/5)+d+(d/4)+(c/4)-2*c;
    //System.out.println(“Day is”+day);

    if (day<0)
    {
    int day1 = -(day);
    int div = day1/7;
    int rem = 7*(div+1)+day;
    day=rem;
    }

    else

    {
    day=day%7;
    }

    switch (day)
    {
    case 0: System.out.print("The Given Date is a "+"Sunday");
    break;
    case 1: System.out.print("The Given Date is a "+"Monday");
    break;
    case 2: System.out.print("The Given Date is a "+"Tuesday");
    break;
    case 3: System.out.print("The Given Date is a "+"Wednesday");
    break;
    case 4: System.out.print("The Given Date is a "+"Thursday");
    break;
    case 5: System.out.print("The Given Date is a "+"Friday");
    break;
    case 6: System.out.print("The Given Date is a "+"Satday");
    break;

    }
    }

    }

    Reply

    1. USE PYTHON CODE——————————-
      BY A 14 YEAR OLD———————-
      ——————————————————-
      THIS IS A BETTER CODE FOR ZELLER RULE USING PYTHON
      THIS IS MADE BY A 14 YEAR OLD CODER WITH BASIC KNOWLEDGE OF PYTHON
      I HAVE LEARNT IT IN THE 2ND MODULE OF PYTHON LEARNING ON SOLOLEARN APP
      REPLY TO PRESENT YOUR VIEWS
      —————————————————
      i=0
      while i==0:
      print(“DATE TO DAY CONVERTER”)
      date = int(input(“ENTER A DATE (DD) “))
      month = int(input(“ENTER THE MONTH NUMBER(MM) “))
      yearcent = int(input(“ENTER THE 1ST 2 DIGITS OF THE YEAR (YY–) “))
      yearaft = int(input(“ENTER THE LAST 2 DIGITS OF THE YEAR (–YY) “))
      RISHIMA=[11,12,1,2,3,4,5,6,7,8,9,0]
      monthno=0
      monthno=int(RISHIMA[month-1])
      if month==1 or month==2:
      if yearaft==00:
      yearaft=99
      else:
      yearaft=yearaft-1
      monthfunc=int((13*monthno-1)/5)
      day=date+monthfunc+yearaft+int(yearaft/4)+int(yearcent/4)-(2*yearcent)
      while day=7:
      day=day%7
      PRAGYANSH=[“SUNDAY”,”MONDAY”,”TUESDAY”,”WEDNESDAY”,”THURSDAY”,”FRIDAY”,”SATURDAY”]
      print (PRAGYANSH[day])
      x=input(“DO YOU WANT TO QUIT (y/n) “)
      if x==”y”:
      print(“THANKS FOR USING ME ;)”)
      break
      else:
      print(“LETS DO IT ALL AGAIN”)
      ———————————————————–
      CODE ENDED
      ITS IN AN OPTIONAL LOOP

      Reply

  29. Dear Quicker Maths,

    How to Calculate the days of the week for a following date:

    1. 7th Dec, 1941
    2. 16th Apr, 2000
    3. 1st Jan, 2004
    4. 8th March, 2005
    5. 26th Jan, 1950
    etc

    how to calculate if the remainder is zero.

    Reply

  32. In case of negative f, simply add a big multiple of 7 (say 70) to f and check for the remainder.
    For ex: f= -17
    -17 + 70 = 53
    Divide 53 by 7 to get remainder of 4.

    Reply

  33. Dear, friends thanks for the great discussion and comments!

    @Bharat: Thanks for explaining the whole thing in your own words. From our side we tried to keep it as simple as possible. Thanks for including the explanation for negative ‘f’.

    Just to remove any error in your explanation I would like to point out this, “so -3 + 6 is again a remainder of 4.” which seems wrong. Please confirm.

    Thanks anyway for your contribution!

    Reply

  34. its actually not explained proprrlyy hav a luk below 2 get the corrrect idea n mahesh ur query will also be solved …

    The following formula is named Zeller’s Rule after a Reverend Zeller. [x] means the greatest integer that is smaller than or equal to x. You can find this number by just dropping everything after the decimal point. For example, [3.79] is 3. Here’s the formula:

    f = k + [(13*m-1)/5] + D + [D/4] + [C/4] – 2*C.

    k is the day of the month. Let’s use January 29, 2064 as an example. For this date, k = 29.
    m is the month number. Months have to be counted specially for Zeller’s Rule: March is 1, April is 2, and so on to February, which is 12. (This makes the formula simpler, because on leap years February 29 is counted as the last day of the year.) Because of this rule, January and February are always counted as the 11th and 12th months of the previous year. In our example, m = 11.
    D is the last two digits of the year. Because in our example we are using January (see previous bullet) D = 63 even though we are using a date from 2064.
    C stands for century: it’s the first two digits of the year. In our case, C = 20.
    Now let’s substitute our example numbers into the formula.

    f = k + [(13*m-1)/5] + D + [D/4] + [C/4] – 2*C
    = 29 + [(13*11-1)/5] + 63 + [63/4] + [20/4] – 2*20
    = 29 + [28.4] + 63 + [15.75] + [5] – 40
    = 29 + 28 + 63 + 15 + 5 – 40
    = 100.

    Once we have found f, we divide it by 7 and take the remainder. Note that if the result for f is negative, care must be taken in calculating the proper remainder. Suppose f = -17. When we divide by 7, we have to follow the same rules as for the greatest integer function; namely we find the greatest multiple of 7 less than -17, so the remainder will be positive (or zero). -21 is the greatest multiple of 7 less than -17, so the remainder is 4 since -21 + 4 = -17. Alternatively, we can say that -7 goes into -17 twice, making -14 and leaving a remainder of -3, then add 6 since the remainder is negative, so -3 + 6 is again a remainder of 4.

    A remainder of 0 corresponds to Sunday, 1 means Monday, etc. For our example, 100 / 7 = 14, remainder 2, so January 29, 2064 will be a Tuesday.

    Reply

  35. Sir,
    How to find the days by using Zeller’s Rule Formula for the year 2010 ,This Zeller’s Rule is applicable only for the year 2009,but the answer is not coming for the other years…

    plz send the correct solution to find the day for other years..

    Thank u for finding the answer for year 2009

    Reply

Leave a Reply

Your email address will not be published. Required fields are marked *