Total extra days=12 ( Which is greater than 7, so again must divide by 7 to get a total extra day)

First, write down DBY then Y, TD, T, DAT, 1, 2, 3  serially and then count from Thursday to 3.  3 is Friday. Hence the third day after the day after tomorrow is Friday

First of all, we have to find the day on 01.03.2005

01.03.2005 = (2004 years + Period from 01.01.2005 to 01.03.2005)

Number of odd days in 2000 years = 0 —-(1)
Number of odd days in 4 years = 1 leap year + 3 ordinary year
= 1X2 + 3X1 = 5 days —-(2)

From 01.01.2005 to 01.03.2005, we have,
Jan-31days + Feb-28days + Mar-1day = 60days
= 8weeks + 4days = 4 odd days —-(3)

Adding (1),(2)&(3), we get,total no.of odd days = 0+5+4
= 9 days = 1 week + 2days = 2 odd days.
2 odd days corresponding to Tuesday.So 01.03.2005 was Tuesday. Therefore, Friday lies on 04.03.2005.

Hence, the dates of March,2015 on which Fridays fell are 4,11,18&25

16th July,1776 = (1775 years + Period from 01.01.1776 to 16.07.1776)

1775 = 1600 + 100 + 75

No.of odd days in 1600 years = 0 —-(1)
No.of odd days in 100 years = 5 —-(2)
No.of odd days in 75 years = 18 leap years + 57 ordinary years = 18X2+57X1=93days=13weeks+2days=2 odd days —-(3)

From 01.01.1776 to 16.07.1776, we have,

Jan-31days+Feb-29days+Mar-31days+Apr-30days+May-31days+Jun-30days+July-16days = 198days = 28weeks + 2 days = 2 odd days —-(4)

Adding (1),(2),(3) & (4), we get,
Total no.of odd days = 0+5+2+2 = 9days = 1 week + 2days = 2 odd days. 2 odd days corresponding to Tuesday

Hence, the day of the week on 16th July,1776 is Tuesday

For example,if our answer is 2018, we must have the same day on 01.01.07 & 01.01.18 and number of odd days from 01.01.07 to 31.12.17 must be zero. Let us check the same thing for each of the given options.

Option (a)2015: No.of odd days from 01.01.2007 to 31.12.2014 = 2 leap years + 6 ordinary years = 2X2+6X1 = 10days = 1week + 3days = 3 odd days[not correct]

Option (b)2016: No.of odd days from 01.01.2007 to 31.12.2015 = 2 leap years + 7 ordinary years = 2X2+7X1 = 11 days = 1 week + 4 days = 4 odd days [not correct]

Option (c)2017: No.of odd days from 01.01.2007 to 31.12.2016 = 3 leap years + 7 ordinary years = 3X2+7X1 = 13 days = 2 weeks + 1 day = 1 odd day [not correct]

Option (d)2018: No.of odd days from 01.01.2007 to 31.12.2017 = 3 leap years + 8 ordinary years = 3X2+8X1 = 14 days = 2 weeks + 0 day = 0 odd day [correct]

Number of days in “x” weeks = 7x
Number of days in “x” days = x

Number of days in “x”weeks”x”days = 7x+x = 8x

Hence, the number of days in “x”weeks”x”days = 8x days

From the notes given on this web page, we have

Number of odd days in 100 years (1st century) = 5
So, the last day of 1st century is Friday

Number of odd days in 200 years (2nd century) = 3
So, the last day of 2nd century is Wednesday

Number of odd days in 300 years (3rd century) = 1
So, the last day of 3rd century is Monday

Number of odd days in 400 years (4th century) = 0
So, the last day of 4th century is Sunday

This pattern is repeated

Therefore, the last day of a century can not be Tuesday or Thursday or Saturday.

1988 april 4th to 1987 november 3rd we have 27+31+31+29+31+4 = 153 = 153/7 = 21 weeks + 6 odd days.
rnHere is where everyone got confused,As per rules we take
rn[0 -sun 1 -mon 2 -tues 3 -wed 4 – thurs 5 – fri 6 -sat]
rnAs no.of odd days = 6 you may think the answer is saturday…
rnBut here later date is given and former date is asked,so the solution should be something like this..
rnwe have six odd days which means that we are short of one day to get the complete week i.e.., november 2nd would be monday and hence november 3rd is tuesday

This being a leap year none of the next 3 years is a leap year. So, the day of the week will be 3 days beyond Monday ie, it will be Thursday.

→ 4th June, 2002 = (2001 years + Period from 1.1.2002 to 4.6.2002)

→ Odd days in 1600 years = 0

→ Odd days in 400 years = 0

→ Odd days in 1 ordinary year = 1

→ Odd days in 2001 years = (0+0+1) = 1

→ Jan(31) Feb(28) March(31) April(30) May(31) June(4)

→ (31 + 28 + 31 + 30 + 31 + 4) = 155 days

→ 22 weeks + 1day ≡ 1 odd day

→ Total number of odd days = (1+1) = 2

→ Required day is Tuesday

61 days = 8 weeks 5 days = 5 odd days

Hence if today is Monday, After 61 days, it will be = (Monday + 5 odd days)
= Saturday

17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March. April. May. June.
(31 + 28 + 31 + 30 + 31 + 17) = 168 days
168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) = 6 odd days
(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 dayss = (21 weeks + 1 day) = 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 = 0 odd days.
Given day is Sunday

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 = 4 odd days.

Jan. Feb. Mar.  Apr.  May.  Jun.  Jul.  Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 =  0 odd days.

Given day is Sunday

First we find the day on 1.3.2005

→ 1.3.2005 = (2004 years + Period from 1.1.2005 to 1.3.2005)

→ Odd days in 1600 years = 0

→ Odd days in 400 years = 0

→ 4 years = (1 leap year + 3 ordinary years)

→ (1×2+3×1) odd days = 5 odd days

→ Jan(31) Feb(28) March(1)

→ (31 + 28 + 1) = 60 days = (8 weeks + 4 days) ≡ 4 odd days

→ Total number of odd days = (0+0+5+4) = 9 ≡ 2 odd days

→ 1.3.2005 was Tuesday. So, Friday lies on 4.3.2005

→ Hence, Friday lies on 4th, 11th, 18th, and 25th of March, 2005

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days

1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day.

Jan   Feb   Mar   Apr   May   Jun   Jul

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days)

Total number of odd days = (0 + 2) = 2.

Required day was ‘Tuesday’.

On 31st December, 2005 it was Saturday.

Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.

On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

8th Feb, 2005 was Tuesday

The year 2004 is a leap year. It has 2 odd days.

The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.

Hence, this day is Sunday

Firstly in terms of years, the year 1911 to 1912 would give us 2 odd days and 1913, 1914, 1915 would give 1, 1 and 1 odd day respectively.
Now shift the focus on months. If you move one month ahead i.e. from 11th April to 11th May, the month ending in between is April, which gives you 2 days. Now after that the month of May, June, July, and August gives you 3, 2, 3, and 3 odd days respectively.
With this you reach on 11th September 1915. After this, there are 6 more September days (from 11th to 17th September).
The total number of odd days is 2 + 1 + 1 + 1 + 2 + 3 + 2 + 3 + 3 + 6 = 24.
Subtracting 21 (3 full weeks) from this the odd number of days left is 3.
Adding three days to the day given i.e. Tuesday, the answer becomes Friday.

29th Feb, 2016 = Monday => 28th Feb, 2012 = Sunday
28th Feb, 2017 = Tuesday (because 2016 is a leap year, there will be 2 odd days)
Therefore» Feb 28th 2018 (Wednesday), Feb 28th 2019 (Thursday), Feb 28th 2020 (Friday), Feb 29th 2020 (Saturday)
Or, Feb 29th to Feb 29th after 4 years, we have 5 odd days.
So, every subsequent birthday, would come after 5 odd days.
2020 birthday – 5 odd days
2024 birthday – 10 odd days = 3 odd days
2028 birthday – 8 odd days = 1 odd day
2032 birthday – 6 odd days
2036 birthday – 11 odd days = 4 odd days
2040 birthday – 9 odd days = 2 odd days
2044 birthday – 7 odd days = 0 odd days. So, after 28 years, his birthday would fall on Monday.
The next birthday on Monday would be in year 2072 (further 28 years later), the one after that would be in year 2100. But we are told that she lives upto year 2099.
So, there are 2 occurrences of his birthday falling on Monday – 2044 & 2072.

Repetition of leap year ===> Add +28 to the Given Year.

Repetition of nonleap year

Step 1 : Add +11 to the Given Year. If the Result is a leap year, Go to step 2.

Step 2:  Add +6 to the Given Year.

Solution :

Given Year is 1897, Which is a nonleap year.

Step 1 : Add +11 to the given year (i.e 1897 + 11) = 1908, Which is a leap year.

Step 2 : Add +6 to the given year (i.e 1897 + 6) = 1903

Therefore, the calendar of the year 1897 can be used again in the year 1903

We know that, After every 400 years, the same day occurs.

Thus, if 9th August 2016 is Saturday, before 400 years i.e., on 9th August 1616 has to be Saturday.

350 days = (350/7=50 weeks) i.e  No odd days,

So it will be a Saturday.

Given year is divided by 4, and the quotient gives the number of leap years.

Here, 100%4 = 25.

But, as 100 is not a leap year => 25 – 1= 24 leap years.

The year 2006 is an ordinary year. So, it has 1 odd day.

So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.

But, 8th Dec, 2007 is Saturday

S0, 8th Dec, 2006 is Friday.

Calculate the number of odd days from starting till 8th June 2215

Step 1:

2000 = 0 odd days

200 = 3 odd days

14 years = 11 ordinary years + 3 leap years = 11 + 6 = 17 odd days = 3 odd days

Jan till June = 4 odd days

8 days = 8 odd days

Step 2:

Add all the odd days

3 + 3 + 4 + 8 = 18 odd days = 4 odd days

Hence, 4 odd days indicates a Thursday (According to reference chart)

Therefore, 8th June 2215 was a Thursday.