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

Consider a set of every four years (So as to include the leap year with Feb 29).
So in 3 years, 29 will occur 11 times, and once it would accur 12 times.
So in 4 years, there will be 3*11 + 12 = 45 days in which you will have 29th.
So in 400 years, it would be 4500 times.

Every year day increase by 1 and if leap year come then by 2

=> leap year increase on day extra

2001 – 1971 = 30 years

so increase of 30 days

Leap years 1972 , 1976  , 1980 , 1984 , 1988 , 1992 , 1996 ( divisible by 4)  , 2000 (divisible by 400)

8 Leap years

Total days increase = 30 + 8 = 38 days

38 = 5*7 + 3

3 days increased from 09/12/1971 to  09/12/1971

so it would be Thursday on 09/12/1971

Read more on Brainly.in – https://brainly.in/question/6133112#readmore

In 400 consecutive years there are 97 leap years. Hence, in 400 consecutive years February has the 29th day 97 times and the remaining eleven months have the 29th day 400 × 11 or 4400 times.
Therefore, the 29th day of the month occurs (4400 + 97) or 4497 times.

1 – May – 2007
=(1 + 2 + 7 + 1 + 6)/7 = 17/7 = 3 = Tuesday
= May 1st –> Tuesday + 5 days = Saturday = 5th may
5th may + 7 days = Saturday = 12th may
12th may + 7 days = Saturday = 19th may
19th may + 7 days = Saturday = 26th may
= Answer = 26th may

x weeks x days = (7x + x) days = 8x days

Let us do this iteratively. Feb 29^th 2012 = Wednesday => Feb 28^th 2012 = Tuesday

Feb 28^th 2013 = Thursday (because 2012 is a leap year, there will be 2 odd days)

Feb 28^th 2014 = Friday, Feb 28^th 2015 = Saturday, Feb 28^th 2016 = Sunday, Feb 29^th 2016 = Monday

Or, Feb 29^th to Feb 29^th after 4 years, we have 5 odd days.

So, every subsequent birthday, would come after 5 odd days.
2016 birthday – 5 odd days
2020 birthday – 10 odd days = 3 odd days
2024 birthday – 8 odd days = 1 odd day
2028 birthday – 6 odd days
2032 birthday – 11 odd days = 4 odd days
2036 birthday – 9 odd days = 2 odd days
2040 birthday – 7 odd days = 0 odd days. So, after 28 years he would have a birthday on Wednesday

The next birthday on Wednesday would be on 2068 (further 28 years later), the one after that would be on 2096. His 84th birthday would again be a leap year.

Now, there is a twist again, as 2100 is not a leap year. So, he does not have a birthday in 2100. His next birthday in 2104 would be after 9 odd days since 2096, or 2 odd days since 2096, or on a Thursday.

From now on the same pattern continues. 2108 would be 2 + 5 odd days later = 7 odd days later. Or, 2108 Feb 29^th would be a Wednesday.

So, there are 4 occurrences of birthday falling on Wednesday – 2040, 2068 and 2096, 2108.

Correct Answer: 4.

Given year 1908, when divided by 4 leaves a remainder of 1.

NOTE: When remainder is 1, 6 is added to the given year to get the result.

So, 1897 + 6 = 1903

Today is Friday.

Lets solve this problem is reverse order

Then we have day after tomorrow = Sunday.

The before day of the day after tomorrow = Saturday

Three days after the before day of the day after tomorrow = Tuesday

Hence Tuesday is the correct answer

Radha: May 8, Thursday
Geeta: May 10, Thursday
Revathi: June 8, Friday

We know that April 1^st = Tuesday

⇒ May 1^st = Tuesday + Remainder of 30/7 = Tuesday + 2days = Thursday.
Also, May 1^st = Thursday
⇒ May 8^th = Thursday
Also, June 8^th = Thursday + Remainder of 31/7 = Thursday + 3 days = Sunday
⇒ Revathi is wrong.
Also, May 10^th = May 8 + 2 days = Thursday + 2 days = Saturday
⇒ Geeta is also wrong.
Hence, only Radha is correct.

Hence, correct date is May 8, Thursday.

59 days = 8 weeks 3 days
Every day of the week is repeated after 7 days hence after 56 days it
would be Thursday again .And after 59 day it would be Sunday

We will show that the number of odd days between last day of February and last day of October is zero. .

March April May June July Aug. Sept. Oct.

31 + 30 + 31 + 30 + 31 + 31 + 30 + 31

= 241 days = 35 weeks = 0 odd day. ,Number of odd days during this period = 0.

Thus, 1st March of an year will be the same day as 1st November of that year. Hence, the result follows

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Year    : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd day : 1    2    1    1    1    2    1    1    1    2    1    

Sum = 14 odd days  0 odd days.

 Calendar for the year 2018 will be the same as for the year 2007.

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

Here, 300%4 = 75.

But, as 100,200 and 300 are not leap years => 75 – 3= 72 leap years.

This is a leap year.

So none of the next 3 years will be leap years.

Each ordinary year has one odd day, so there are 3 odd days in next 3 years.

So the day of the week will be 3 odd days beyond Wednesday i.e. it will be Saturday

April 4th 2001 =(2000 years+Period of 1.1.2001 to 4.4.2001)

Odd days in 2000 years = 0 days

From January to April 4

Total no.of.days: 31+28+31+4=94 days =(13 weeks +3 days)=3 odd days

So the day will be Wednesday

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

Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
∴ After 61 days, it will be Saturday.

On 31st December it was Friday.Number of Odd days between 2006 and 2008 = (1+1)=2days

Then December 31st 2007 will be Monday then 1st Jan 2008 will be Tuesday.

We first find the day of the week on 1^st April 2009.

       1^st April 2009 means, 2008 years 3 months and 1 day

           =  2000 years + 8 years  + January + February + March + 1^st April

          = 0 odd days  + 2 leap years  + 6 ordinary years + 31 days + 28 days +31 days + 1 day

           = 0+ 4 odd days + 6 odd days + 3 odd days + 0 odd days +3 odd days + 1 odd day

          =17 odd days = 3 odd days

  From the date in the above problem, when number of days is 3, the day of the week becomes Wednesday.

Thus, first Thursday falls on 2^nd April.

In that month, Thursday falls on 2nd, 9th, 16^th, 23^rd, and 30^th.

Formula:-(Date + Month code + No.of years + No.of leap year + Century code)/7
= (26 + 1 + 50 + 12 + 0)/7 = 89/7 = 12
= Thursday

Day Codes:-
Sat = 0
Sun = 1
Mon = 2
Tue = 3
Wed = 4
Thu = 5
Fri = 6
Month Codes:-
Jan = 1
Feb = 4
Mar = 4
Apr = 0
May = 2
Jun = 5
Jul = 0
Aug = 3
Sep = 6
Oct = 1
Nov = 4
Dec = 6
Century Codes:-
1500 – 1599 —> 0
1600 – 1699 —> 6
1700 – 1799 —> 4
1800 – 1899 —> 2
1900 – 1999 —> 0
2000 – 2099 —> 6
2100 – 2199 —> 4
2200 – 2299 —> 2
2300 – 2399 —> 0

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

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

100 years contain 5 odd days.

 Last day of 1^st century is Friday.

200 years contain (5 x 2)  3 odd days.

 Last day of 2^nd century is Wednesday.

300 years contain (5 x 3) = 15  1 odd day.

 Last day of 3^rd century is Monday.

400 years contain 0 odd day.

 Last day of 4^th century is Sunday.

This cycle is repeated.

 Last day of a century cannot be Tuesday or Thursday or Saturday.