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how we get the week day if given date and year
We assume that we need to find the day of Aug 15, 1947 (India's Independence Day).

Step 1: Find leap year or not.
Using the above said conditions, find whether your year is leap year or not. To know whether a number is divisible by 4, you can see here
* 1947 is not a leap year as it is not divisible by 4

Step 2: Find out total odd days on months.
Method 1: Add all the days in the months before your month (that is august) and the date (that is 15) and find the remainder when the sum is divided by 7
* 31 + 28 (since, not a leap year) +31+30+31+30+31+ 15 = 227
* 227%7 = 3

Method 2: If adding like above is difficult, add all the odd days the months before august and add 15 % 7 = 1 to it. And then find the remainder when the sum is divided by 7.
* 3 +0(since not leap year) +3+2+3+2+3+1 = 17
* 17%7 = 3

Consider 1601 - 1700
Every year we have one odd day (meaning 365 % 7 = 1)
Every leap year have two odd day (meaning 366 % 7 = 2)
In 100 years (say 1601 - 1700) we have 24 leap years, 76 normal years.
That implies no of odd days in every 100 year = (2 x 24 + 1 x 76 ) % 7 = 124 % 7 = 5

so far 1601 - 1700 - 5 odd days
for 1701-1800 - 5 odd days
for 1801-1900 - 5 odd days
for 1901-2000 - 5 + 1 odd days
Since 2000 is a leap year where as 1700 and others are not, we have added 1 in the last statement.

So far a set of 400, we have (5+5+5+6 )%7 = 21 % 7 = 0 odd days

This implies, if 1-1-1601 is some day, then 1-1-2001 is also the same day, as the diff between them is 400.

So we can reduce multiples of 400 from the year we have.
that is 1947-1600= 347


Step 3: Find out the total odd days on year.
Step 3(i): (only if greater than or equal to 100 in the above calculation)

In 347 we have, 3 hundreds

Therefore we have
(no of odd days per hundred x no of hundreds) % 7 = (5 x 3) % 7 = 15 % 7 = 1

Now we can reduce multiple of 100 from the year we have.
347-300=47

Step 3 (ii) : Only if year greater than 1 in last calculation
We have already found the odd days in the 1947th year in step 2
Excluding the 47th year, there are 11 leap years and 35 normal years from 1901-1946
so (11x2 + 35x1) % 7 = 57 % 7 = 1

Step 4: (step 2 + step 3(i) + step 3(ii)) % 7
(3 + 1 + 1 ) % 7 = 5 % 7

Jan 1, 2001 is a Monday, depending on odd day, we have
1 is Monday
2 is Tuesday
3 is Wednesday
4 is Thursday
5 is Friday
6 is Saturday
0 or 7 is Sunday

so 5 is Friday , which is the day of Aug 15th, 1947.


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