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

Repetition of non leap year

Step 1 : Add +11 to the Given Year. If 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 non leap 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

Therfore, 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.

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.

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.

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.