How to remember the last days in dec month of a year?

Question

Hi all,

I came across a few questions like

How to solve these kind of questions whic is related to the days of a year?? any formulas??

“I'm a greater believer in luck, and I find the harder I work the more I have of it”

jumsumtak · Answer

Not going to solve it. Giving general directions:

if today is 16 July, 2013 - Tuesday and I ask you what day will be 24 July, 2013. Either you count on your hand or see that 23 July is going to be Tuesday as well and hence, 24 will be Wednesday. 
in essence, what you did was (24-16)mod7=1. So, you add 1 to Tuesday and make it Wednesday.

what day will be 16 July 2014?
(16July 2014-16July 2013) = 365 days. 
365mod7= 1

Therefore, it will be a Wed again.

You need to track backwards for the questions you mentioned. (keep in mind the leap years)

CMcAboy · Answer

You can do the knuckle trick to figure out how many days are in a month. https://en.wikipedia.org/wiki/File:Month_-_Knuckles_(en).svg

RnH · Answer

This is a very nice and interesting piece from Wikipedia:
(Search for: Determination_of_the_day_of_the_week; Content #: 3.1 Basic method for mental calculation)
https://en.wikipedia.org/wiki/Determinat ... alculation

Please see the Wikipedia entry to understand this completely. There are a few tables on the page that I am unable to insert here

Basic method for mental calculation

This method is valid for both the Gregorian calendar and the Julian calendar. Britain and its colonies started using the Gregorian calendar on Thursday, September 14, 1752 (the previous day was Wednesday, September 2, 1752 (Old Style). The areas now forming the United States adopted the calendar at different times depending on the colonial power: Spain and France had been using it since 1582, while Russia was still using the Julian calendar when Alaska was purchased from it in 1867.

The formula is (d + m + y + [y/4] + c ) mod 7, where:

d is the day of the month,
m is the month's number in the months table,
y is the last two digits of the year, and
[y/4] is the Quotient of the result of y/4
c is the century number. For a Gregorian date, this is 6 if the first two digits of the year are evenly divisible by 4, and subsequent centuries are 4-2-0 (so the century numbers for 2000, 2100, 2200, and 2300 are respectively 6, 4, 2, and 0). For a Julian date, this is 6 for 1200, and subsequent centuries subtract 1 until 0, when the next century is 6 (so 1300 is 5, and 1100 is 0).

Add the aforementioned 5 numbers - d, m, y, y/4, c - and divide the result by 7 and get the remainder. If the remainder is 0, the date was a Sunday; if 1 it was a Monday, and so on through the week until 6 = Saturday.

Months Table:

RnH · Answer

I found another one. This one was on Yahoo! Answers: https://uk.answers.yahoo.com/question/i ... 026AAgSsXC

It does take a little work, but it isn't impossible. With practice you can do it in your head.

Take the two digit year. (Example '08 in 2008).
Divide the year by 4 --> 2
Divide the year by 7 and take the remainder --> 1
Add these --> 3

Remember this table (it goes down by 2 each century) and repeats:
1600, 2000 = +6
1700, 2100 = +4
1800, 2200 = +2
1900, 2300 = 0

2000 has an offset of +6
3 + 6 --> 9

At any point you can take your number and divide it by 7 and only keep the remainder. So 9 is the same as 2.

Now remember this table of offsets for the months
033 - 614 - 625 - 035

September is the 9th month, so add 5.
2 + 5 = 7 (again you can divide by 7 and take the remainder) --> 0

Now add the day (30th) --> 0 + 30 = 30 (divide by 7, keep the remainder) --> 2

Here's the final lookup table:
Sunday = 0
Monday = 1
Tuesday = 2
Wednesday = 3
Thursday = 4
Friday = 5
Saturday = 6

So today (Sept 30, 2008) is a Tuesday.

Leap Year Rule:
The other rule is if it is a leap year and it is Jan. or Feb. you have to subtract 1. Remember that century years like 1700, 1800, 1900 are *not* leap years, unless they are divisible by 400 (e.g. 2000 was a leap year, 2400 is a leap year).

Here's another example:
12/07/1941

41 / 4 --> 10 (divide by 7, keep remainder ) --> 3
41 / 7 --> remainder 6
3 + 6 = 9 --> remainder 2

Century offset for 1900
--> +0 --> still 2

Month number for December is 5
2 + 5 = 7 --> remainder 0

Day = 7
0 + 7 = 7 --> remainder 0

12/07/1941 was a Sunday

Edit: Yes, it does work for dates before 2000. Another example:
July 4, 1776:
76 / 4 = 19 --> div 7 = remainder 5
76 / 7 --> remainder 6
5 + 6 = 11 --> remainder 4

1700 has a +4 offset for the century.
4 + 4 = 8 --> remainder 1

July (7th month) has an offset of 6
1 + 6 = 7 --> remainder 0

Add the day of the month (4)
0 + 4 = 4

July 4th, 1776 was a Thursday

One final note, the current Gregorian calendar started in 1583, so you can't go backward before then for dates (without adjusting for the Julian calendar).

Try it with your birthday and see if you can do it. If you want some help, let me know.

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How to remember the last days in dec month of a year?

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