Calendars Test - 1

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.

• Every year has one odd day and a leap year has 2 odd days.
• Though 1984 is a leap year, we don't have Feb 29 in the required period.
• So, we get only one odd day and as we are moving back we get Tuesday as the answer.

• 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.

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.

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

We shall first calculate the number of odd days till 31 December 1946:

• Number of odd days in the first 1600 years = 0 odd day
• Number of odd days in the next 300 years = 1 odd day
• Now, 46 years had 11 leap years and 35 ordinary years.
• The number of odd days in 46 years = (2 × 11) + (1 × 35) = 22 + 35 = 57 = 8 weeks and 1 odd day.

Now, we shall calculate the number of odd days in 1947 till 15 August:
Month (Days):

• January(31)
• February(28)
• March(31)
• April(30)
• May(31)
• June(30)
• July(31)
• August(15)

Days = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 i.e. 32 weeks and 3 odd days
So, the total number of odd days till 15 August 1947 = 0 + 1 + 1 + 3 = 5
On counting five days from Monday, we get Friday.
Therefore, 15 August 1947 was a Friday.

Given that the month begins on a Friday and has 31 days

• Sundays = 3^rd, 10^th, 17^th, 24^th, 31^st
  ⇒ Total Sundays = 5
  Every second & fourth Saturday is holiday.
• 2^nd & 4^th Saturday in every month = 2
• Total days in the month = 31
• Total working days = 31 - (5 + 2) = 24 days

• 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.

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

• March, April, May, June, July, Aug, Sept, Oct , i.e. 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 241 days = 35 weeks = 0 odd day.
• Number of odd days during this period = 0.

Thus, 5th March of a year will be the same day as 5th November of that year.