Calendar Test-2

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Sum = 14 odd days ≡0 odd days.
∴Calendar for the year 2018 will be the same as for the year 2007.

The century divisible by 400 is a leap year.
∴The year 700 is not a leap year.

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.
∴8th Dec, 2006 is Friday.

The year 2008 is a leap year. So, it has 2 odd days.
1^st  day of the year 2008 is Tuesday (Given)
So, 1^st  day of the year 2009 is 2 days beyond Tuesday.
Hence, it will be Thursday.

The year 2007 is an ordinary year. So, it has 1 odd day.
1^st  day of the year 2007 was Monday.
1^st  day of the year 2008 will be 1 day beyond Monday.
Hence, it will be Tuesday.

22 Apr 2222 i.e.  2221 years + period from 1-Jan-2222 to 22-Apr-2222

We know that number of odd days in 400 years = 0
Hence the number of odd days in 2000 years = 0 (Since 2000 is a perfect multiple of 400)

Number of odd days in the period 2001-2200 = Number of odd days in 200 years = 5 x 2 = 10 = 3

As we can reduce perfect multiples of 7 from odd days without affecting anything

Number of odd days in the period 2201-2221 = 16 normal years + 5 leap years = 16 x 1 + 5 x 2 = 16 + 10 = 26 = 5 odd days
Number of days from 1-Jan-2222 to 22 Apr 2222 = 31 (Jan) + 28 (Feb) + 31 (Mar) + 22(Apr) = 112

112 days = 0 odd day
Total number of odd days = (0 + 3 + 5 + 0) = 8 = 1 odd day  = Monday

Hence 22 Apr 2222 is Monday

Each day of the week is repeated after 7 days. So, after 63 days, it will be on Monday. After 61 days, it will be on Saturday.

350 days, 350/7  = 50, no odd days, so it will be a Monday.

number of odd days in 338 days = 338 / 7 = 248 complete weeks + 2odd days. 2nd day after Saturday is Monday.

Formula : (Date + Month code + No.of years + No.of leap year + Century code)/7
= 43/7 =1
Thus, the day of the week on 24th July 2011will be Sunday.