## Mentally compute the day of the week (the perpetual calendar made easy)

**October 29, 2008 is… wednesday!**

Being able to know the day of the week for any particular date is a handy ability. In our times, a cellphone or computer can answer the question in seconds, but it is more handy to just know.

Above all, being able to do it is *impressive*, which is of course the main reason for learning how to do it.

All methods involve the simple notion of counting by sevens from a date where the day of the week is known. The issue is complicated because calendar dates are not just consecutive numbers but a series of month days which alternate between 30 and 31, except for February. Then you have the added complication of leap years.

To make the method practical people have devised lists of numbers to assign to years and lists of numbers to assign to months. With a few rules and 4 mental computations the day of the week can be determined.

Over the years I have tried several variations on the methods, and found them all to be too cumbersome to be practical. So I created my own method.

### The Bellon method

*in 2 steps*

If you can remember you will be able to tell the day of the week for any date in just seconds. The number sequence 033614625035 represents each month of the year (a month per digit).

- First you take your date, add the month digit, and add 2 (for the year 2008, explained below).
- Now all you do is obtain the remainder of your total divided by 7.

Once I got the month digits memorized, it takes me 2-3 seconds to come up with the answer. It’s really easy.

Examples:

**What day of the week was the 12 of November of 2008?**

- Take 12 and add 3 (november digit) add 2 (year digit) = 12 + 3 + 2 = 17.
- Now divide 17 by 7, and you have a remainder of 3. The 3rd day of the week is
**Wednesday**.

**What day of the week was the 25 of May 2008?**

- 25 + 1 (month) + 2 (year) = 28
- 28 / 7 = 4 with a remainder of 0. The zero or seventh day of the week is
**Sunday!**

Reference:

Months (This is the only real memorization)

- January = 0
- February = 3
- March = 3
- April = 6
- May = 1
- June = 4
- July = 6
- August = 2
- September = 5
- October = 0
- November = 3
- December = 5

Days (Just in sequential order, where monday is 1 and sunday 7)

- Sunday = 0
- Monday = 1
- Tuesday = 2
- Wednesday = 3
- Thursday = 4
- Friday = 5
- Saturday = 6
- Sunday = 7

Years (Just remember the current year number and go backward or forward as needed. Once it reaches 7, it resets to zero)

- 2006 = -1
- 2007 = 0 (I started using my system on this year so I set it so that 2007 was 0)
- 2008 before Feb 29 = 1
- 2008 = 2
- 2009 = 3
- 2010 = 4
- …

Hello, I was trying the Bellon method.

QUOTE

” What day of the week was the 12 of November of 2008?

Take 12 and add 3 (november digit) add 2 (year digit) = 12 + 3 + 2 = 17.

Now divide 17 by 7, and you have a remainder of 3. The 2nd day of the week is Wednesday.” END QUOTE

here it says the 2nd day of the week is Wednesday.

QUOTE

“Days (Just in sequential order, where monday is 1 and sunday 7)

Sunday = 0

Monday = 1

Tuesday = 2

Wednesday = 3

Thursday = 4

Friday = 5

Saturday = 6

Sunday = 7 ” END QUOTE

Here it says the 3rd day of the week is Wednesday.

Is this a mistake or intentional?

Good catch, it was a mistake. It should read “3rd”, not 2nd. It has been corrected now.

Thank you!

I am unclear with two things about the number assigned for the year. One, if I want to determine the year number for, say 1790, do I count backwards from 2007 and come up with “6″? If so, that is quite a mental computation. The other part I am unclear is the leap year 2008 having 2 numbers assigned. Does every leap year get assigned two numbers, one for Jan. 1 – Feb. 29 dates and another for March 1 – Dec. 31 dates? In that case, that backwards counting to 1790 got even harder. In that case, 1790 gets “4″.

wow really cool. im an idiot and even i got this down pat ^^. post some more algorithms!

Larry,

Yes you would have to count backwards. If you want to compute any date of ANY year it may be more practical to have a list of years with their appropriate 1 to 7 values. The way I present the method is more suitable for computing dates during the current year or a few back or forward.

(You are also correct on the leap years, you have two numbers per year)

My aim was to simplify the known methods for the most useful everyday life situation: knowing the weekday of the present year.

This is very interesting, and it works quickly!

I was having problems with the year value when I was trying to figure out days from 1994. Thanks to Larry’s question about this, I now know what the problem was (that leap years have 2 numbers).

[QUOTE]

* 2006 = -1

* 2007 = 0 (I started using my system on this year so I set it so that 2007 was 0)

* 2008 before Feb 29 = 1

* 2008 = 2

* 2009 = 3

* 2010 = 4

* …

[/QUOTE]

Above, it says that the value for 2006 is -1. After thinking about it, I’m sure that both x and 7-x would work as the year value, right? So if the year value is 5, you could subtract 2 instead and it still works.

Once again, it’s a great resource.

How do you mentally calculate 1790 to be 4 or 2029 to be 0?

Sorry. Clarification. Following your method above I know how to calculate 1790 to be 4 or 2029 to be 0 on a piece of paper.

But the question is what formula / mental calculation do you use to come with the correct Year Digit?

This is fantastic!

But I have just one question.

[QUOTE

Years (Just remember the current year number and go backward or forward as needed. Once it reaches 7, it resets to zero)

2006 = -1

2007 = 0 (I started using my system on this year so I set it so that 2007 was 0)

2008 before Feb 29 = 1

2008 = 2

2009 = 3

2010 = 4

…

QUOTE]

The number for 2006 is -1, so does that mean that if you want say 11 January 2006 you would do

11+0+(-1)= 10

10/7=1 r3,

so 11 January 2006 would be a Wednesday?

Is this right?