Calendar Reasoning

Hint

15th Aug 1947 ⇒ (1946 years + Period from 1st Jan 1947 to 15th Aug 1947)

Counting of odd days:
1600 years have 0 odd days.
300 years have 1 odd day.

47 years contain 11 leap years and 36 ordinary years.
⇒ (11×2)+(36×1) odd days
⇒ 58 odd days
⇒ 2 odd days.

No. of the odd day from 1st Jan 1947 to 15th Aug 1947:
31+28+31+30+31+30+31+16=227 days
⇒ 227 days =(32 weeks+3 days)
⇒ 3 odd days.

Total number of odd days =(0+1+2+3) =6
Hence, the 6th day of the week is Friday.

Hint

26th Jan 1950 ⇒ (1949 years + Period from 1st Jan 1950 to 26th Jan 1950)

Counting of odd days:
1600 years have 0 odd days.
300 years have 1 odd day.

49 years contain 12 leap years and 37 ordinary years.
⇒ (12×2) + (37×1) odd days
⇒ 61 odd days
⇒ 5 odd days.

Number of odd days from 1st Jan to 26th Jan (26 days)
⇒ 5 odd days
Total number of odd days =(0 + 1 + 5 + 5) = 11 = 4
Hence, the 4th day of the week is Thursday.

Hint

1st January 2030 ⇒ (2030 years + 1st Jan)

Counting of odd days:
2000 years have 0 odd days.
29 years contain 7 leap years and 22 ordinary years.
⇒ (7×2)+(22×1)=36 odd days
⇒ 1 odd day
1st January 2030 has 1 odd day.

Total number of odd days =(0+1+1)=2.
Hence, 2nd day of the week is Tuesday.

Remember: Number of odd days in 400 years =(5×4+1)=0
Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years, etc. has 0 (zero) odd days.

Hint

16th July 1776 ⇒ (1775 years + 6 months + 16 days)

Counting of odd days:
Now, 1600 years have 0 odd days.
100 years have 5 odd days.

75 years contain 18 leap years and 57 ordinary years.
⇒ (18×2)+(57×1)=93 days
⇒ 2 odd day

No. of the odd day from 1st Jan 1976 to 16th Jul 1976:
31+29+31+30+31+30+16=198 days
= 28 weeks + 2 days
⇒ 2 odd days

Total number of odd days
= (0+5+2+2)=9.
⇒ 2 i.e., 2nd day of the week
Hence the day on 16th July 1776 was Tuesday.

Hint

Count the number of days from 2005 onwards to get 0 odd days.

Year–odd day: 2005–1, 2006–1, 2007–1, 2008–2, 2009–1, 2010–1
= 7 or 0 odd days.

Thus, the calendar for the year 2005 is the same as that for the year 2011.

Hint

We go on counting the odd days from 1990 onwards till the sum is divisible by 7.

The number of such days are 14 up to the year 2000.
Year–odd day: 1990–1, 1991–1, 1992–2, 1993–1, 1994–1, 1995–1, 1996–2, 1997–1, 1998–1, 1999–1, 2000–2

The year next to 1991 having the same calendar as that of 1990 is 2001.

Hint

No. of years from 9/12/71 to 9/12/2001=30
We know every ordinary year has 1 odd day and a leap year has 2 odd days.

Here, no. of ordinary years are 22, and no. of leap years are 8.
So total counting of odd days =(22×1)+(8×2)=38
i.e., 3 odd days (3 is remainder when 38 is divided by 7)

So, the day on 09/12/1971 is 3 days before the day on 09/12/2001.
But, 09/12/2001 is Sunday.
Thus, 09/12/1971 will be 3 days before Sunday.
Hence, it was Thursday.

Hint

100 years contain 5 odd days.
So, the last day of the 1st century is Friday.
200 years contain (5×2=10)⇒3 odd days.
So, the last day of the 2nd century is Wednesday.
300 years contain (5×3=15)⇒1 odd day.
So, the last day of the 3rd century is Monday.
400 years contain 0 odd days.
So, the last day of the 4th century is Sunday.

Since the order is continually kept in successive cycles, we see that the last day of a century cannot be Tuesday, Thursday or Saturday.

Hint

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.

Hint

The year 1979 being an ordinary year, it has 1 odd day.
So, the day on 12th January 1979 is one day before the day on 12th January 1980.

But, 12 January 1980 is Saturday.
Thus, 12 January 1979 was Friday.

Hint

If the first working day happened to be Tuesday then the 8th, 15th, 22nd and 29th of the month will be Tuesday. Hence, the last day of the month will be Wednesday (since number of days in the month is 30). Thus, the next casual leave will be on Thursday.

Hint