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

**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 show the calculation for 3/3/89…

thanks

sirra formula.great formula

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.”

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The formula does not work for the dates Sept 3, 1752 – Sept 13, 1752. This is when the Gregorian calendar was changed. Please explain.

The Gregorian calendar was instituted on AD 1582 Oct 15th.

and 1752 Sep 13 was on wed day.

The formula is not working for Sept 3 – Sept 13. These dates were eliminated during that calendar year. Explain please.

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

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Its not working for 16/01/1988. I am getting it Wednesday!!

(28+16+87+21+4-38)mod 7=6. It is Saturday.

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.

This is very informative though some are difficult.

thank you

31/08/1990 it is friday but i’m getting wrong answer………. once try..

(31+15+90+22+4-38)mod 7+5. This is Friday

not workin for

26 jan 1952

itz saturday on tht day…..0 or 7 shud come..

bt

1is comin.

( REPLY )

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!).

bro… its not C

its D=51 (since its jan)

C will remain same i.e 19

its working man..

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.

and if u have some time please visit http://www.splmaths.worpress.com

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;

}

}

}

it just a logical works

18+{(13*12-1/5)}+91+91/4+19/4-38

=18+35+91+22+4-38=132

132/7 now remainder=6 implies thd day is saturday

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.

if remainder is 0 than we can take remainder =7

how to calculate dates between 2000 to 2004

very nice trick But I want to find out days between 2000 to 2009

sir its coming wrong for 18/2/1991 please give the corrected formula

not workin for

28/02/2010

itz sunday on tht day…..0 or 7 shud come..

bt

1 is comin.

according to Zellers rule,the last two digits of the year should be 1 less than that of the year i for any dates on the months january and february.

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.

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!

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.

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

it will be really great if u help us in explaining the logic behind this formula…

Very well organized formula ..

especially, the leap year part, I Liked the most…

Thanks Abhimanyu

informative but little difficult