Test: Calendars- 1 - UPSC

Remember the leap year rule:

• Every year divisible by 4 is a leap year, if it is not a century.
• Every 4th century is a leap year, but no other century is a leap year.
• 800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc).

Hence, 800,1200 and 2000 are leap years.

• 100 years contain 5 odd days.
  ∴ Last day of 1st century is Friday.
• 200 years contain (5 x 2) ≡ 3 odd days.
  ∴ Last day of 2nd century is Wednesday.
• 300 years contain (5 x 3) = 15 ≡ 1 odd day.
  ∴ Last day of 3rd century is Monday.
• 400 years contain 0 odd day.
  ∴ Last day of 4th century is Sunday.

This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.

1895 is not a leap year. So it will have 1 odd day.

Since 1896 is a leap year, it will add 2 odd days.

Similarly 1987, 1898, 1899, 1900 will add 1,1,1,1 odd days.

Now the total number of odd days add up to 7.

So the next year 1901 will have the same calendar as 1895.

• We know that in 1600 years, there will be 0 odd days. And in the next 300 years, there will be 1 odd day.
• From 1901 to 1978 we have 19 leap years and 59 non-leap years.
• So, the total number of odd days up to 31st Dec. 1978 is 19 x 2 + 59 = 97. On dividing 97 by 7 we get 6 as the remainder, which is the total number of odd days in these years.
• So, till 31st Dec. 1978 we have 1 + 6 = 7 odd days, which forms one complete week. Now, in 1979, we have 3 odd days in January, and 2 odd days in the month of February (up to 9th Feb).
• So, the total odd days are 3 + 2 = 5.
• Hence, 9th February 1979 was a Friday.

30 years. The number of leap years is 8 (1972,1976,1980,1984,1988,1992,1996,2000).

So, the total number of days = 22*365 + 8*366 = 10958

10958 mod 7 = 3

Since 9/12/2001 is a Sunday, 9/12/1971 should be a Thursday.

Since it has been mentioned that Mohan was not born in February, so he can’t be born on 29th Feb.

Hence He will celebrate his next birthday on a Wednesday in the year for which the sum of the odd days becomes 5 or a multiple of 5.

By his birthday in 2017, there will be 1 odd day.

By his birthday in 2018, there will be 2 odd days.

By his birthday in 2019, there will be 3 odd days.

By his birthday in 2020, there will be 5 odd days, as 2020 is a leap year.

So in 2020 He will celebrate his birthday on Wednesday.

We are given that 15th March 1816 was a Friday.

Now we know that 100 years have 5 odd days. So till 15th March 1916, we will be having 5 odd days. 

So if we move from 15th March 1816 to 15th March 1916, we will encounter 5 odd days.

Now from 15th March 1916 to 15th April 1916 there would be 3 odd days.

So total number of odd days = 5 + 3 = 8

8 mod 7 = 1

So 15th April 1916 would be Friday + 1 = Saturday

For every 28 years, the calendars will same,
so the years 2008,2036 have the same calendar as 1980.

In this question, the reference point is May 10, 1997 and we need to find the number of odd days from May 10, 1997 up to Oct 10, 2001.

► Now, from May 11, 1997 - May 10, 1998 = 1 odd day
► May 11, 1998 - May 10, 1999 = 1 odd day
► May 11, 1999 - May 10, 2000 = 2 odd days (2000 was leap year)
► May 11, 2000 - May 10, 2001 = 1 odd day
► Thus, the total number of odd days up to May 10, 2001 = 5.
► The remaining 21 days of May will give 0 odd days.
► In June, we have 2 odd days; in July, 3 odd days; in August, 3 odd days; in September,2 odd days and up to 10th October, we have 3 odd days. Hence, total number of odd days = 18 i.e. 4 odd days.
Since, May 10, 1997 was a Monday, and then 4 days after Monday would be Friday. So, Oct 10, 2001 was Friday.

It is given that 28th August 1946 was Wednesday.

From 28th August 1946 to 28th August 1961, we have 4 leap years and 11 normal years.

So the number of odd days would be 11*1 + 4*2 = 19

Now the date which is asked is 31 Aug 1961. So if we move from 28th August to 31st August, we will have 3 more odd days.

So total number of odd days = 5 + 3 = 8

Now 8 mod 7 = 1 .

So 31st August 1961 would be Wednesday + 1 = Thursday.