Calendar Reasoning

Hint

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th Aug., 1947)
Counting of odd days :
1600 years have 0 odd day. 300 years have 1 odd day.
47 years = (11 leap years + 36 ordinary years)
= [(11 × 2) + (36 × 1)] odd days = 58 odd days
⇒ 2 odd days.

= 227 days = (32 weeks + 3 days) = 3 odd days.
Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.
Hence, the required day was ‘Friday’.

Hint

26th Jan., 1950 = (1949 years + Period from 1st Jan., 1950 to 26th Jan., 1950)
1600 years have 0 odd day. 300 years have 1 odd day.
49 years = (12 leap years + 37 ordinary years)
= [(12 × 2) + (37 × 1)] odd days = 61 odd days = 5 odd days.
Number of days from 1st Jan. to 26th Jan. = 26 = 5 odd days
Total number of odd days = (0 + 1 + 5 + 5) = 11 = 4 odd days
∴ The required days was ‘Thursday’

Hint

2000 years have 0 odd days.

= 11 odd days (11-7) ⇒ 4 odd days.

1st January, 2010 has 1 odd day.

Total number of odd days = (0 + 4 + 1) = 5.

∴ 5th day of week is Friday.

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

Hint

16th July, 1776 mean (1775 years + 6 months + 16 days)
Now, 1600 years have 0 odd days.
100 years have 5 odd days
75 years contain 18 leap years and 57 ordinary years and therefore (36 + 57) or 93 or 2 odd days.
∴1775 years given 0 + 5 + 2 = 7 and so 0 odd days.
Also number of days from 1st Jan. 1776 to 16th July, 1776
Jan. Feb. March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 16
= 198 days = 28 weeks + 2 days = 2 odd days
∴Total number of odd days = 0 + 2 = 2.
Hence the day on 16th July, 1776 was ‘Tuesday’.

Hint

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

= 7 or 0 odd day.
∴ Calender for the year 2005 is the same as that for the year 2011.

Hint

We go on counting the odd days from 1991 onwards till the sum is divisible by 7. The number of such days are 14 upto the year 2001. So, the calender for 1991 will be repeated in the year 2002.

Hint

09/12/2001—— Sunday
No. of days between 9/ 12/ 71 & 9 / 12/ 2001
we know every year has 1 odd day and leap year has 2 odd days
Here, No. of normal years = 22And no. of leap years = 8
So odd days = 22 + 16 = 38 i.e., 3odd days
(remainder when 38 is divided by 7, i.e. 3)
Hence it was a Thursday

Hint

100 years contain 5 odd days. So, last day of 1st century is ‘Friday’
200 years contain (5 × 2) = 10 odd days = 3 odd days.
So, last day of 2nd century is ‘Wednesday’.
300 years contain (5 × 3) = 15 odd days = 1 odd day.
∴ Last day of 3rd century is ‘Monday’.
400 years contain 0 odd day.
∴ Last day of 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 1980 is one day beyond on the day on 12th January, 1979.
But, January 12, 1980 being Saturday.
∴January 12, 1979 was Friday.

Hint

If the first working day happened to be Tuesday then 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