94 when divided by 7 gives remainder 3 hence today it must be Thursday, after 3 more we will get a Sunday, next Sunday will be after 3 + 7 = 10 days and so on so we will get Sunday after a day 7K + 3 days.
Counting the number of days after 3rd November. 1994 we have :
Number of Days in November = 27 or 6 odd days.
Number of Days in December = 31 or 3 odd days.
Number of Days in January = 31 or 3 odd days.
Number of Days in February = 28 or 0 odd days.
Number of Days in March = 20 or 6 odd days.
So total number of odd days = 6 + 3 + 3 + 0 + 6 = 18 when divided by 7 gives remainder 4.
Number of odd days = 4
The day on 3rd November, 1994 is (7 - 4) days beyond the day on 20th March, 1995.
So, the required day is Thursday.
If month ends with Thursday then next month will start with Friday and it may have 5 Friday otherwise it may have 4 Fridays.
Let us take two cases
Case (i) : When year start with Sunday then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays
Case (ii) : When year start with Sunday and then we have 53 Sundays that means year is a leap year then next 4 years will always have 52 Sundays hence total number of Sundays are 53 + 3 x 52 = 209 Sundays.
In a year we have 53 Sundays only when year start with Sunday (For non leap year) and either with Saturday of Sunday (For leap year).
From zeller's formula 1st January 2001 was Monday
From number of odd days 1st January 2006 will start with Sunday so it had 53 Sunday
Similar 1stJanuary 2009 will start with Thursday.
Number of odd days between 2013 and 2015 is 2, so we have to find the year which will have 2 odd days between 1977 and required year.
Consider options one by one -
(a) Number of odd days between 1977 and 1981 is 1 + 1 + 1 + 2 = 5 odd days
(b) Number of odd days between 1977 and 1985 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 10 odd days or 3 odd days
(c) Number of odd days between 1977 and 1990 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 = 16 odd days or 2 odd days, hence calendar of 1990 is the answer.
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 4rd century is Sunday.
This cycle is repeated that means last day of century is Friday, Wednesday, Monday, or Sunday and it repeats the cycle.
Here we have two cases :
Case (i) : If a month start with Thursday the month will have 5 Saturdays and 4 Sundays i.e total 9 weekends. In this case 31st December will be Sunday hence 1st January will be either Sunday (For non leap year) or Saturday ( For leap year) then January will have 9 or 10 weekends.
Case (ii) : If a month start with Sunday the month will have 4 Saturdays and 5 Sundays i.e total 9 weekends.
In this case 31st December will be Wednesday hence 1st January will be either Wednesday (For non leap year) or Tuesday ( For leap year) then January will have 8 or 9 weekends.
Hence number of weekends in January will be 8 or 9 or 10.
If there are 53 Saturday and Sunday that means year is a leap year and last day of year and last day of year is Sunday so 1st day of January must be Saturday then January will have 10 weekends.
In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap leap year, as 1900 was not a leap year)
Number of odd days in a period of 100 years is 100 + 23 or 100 + 24 or 123/124 odd days.
Number of odd days in 100 years is when 123/124 divided by 7, remainder is 4 or 5. So next century should start with Monday or Tuesday.
In a period of 100 years there are 23 or 24 leap (as for century year it might be or might not be a leap leap year, as 1900 was not a leap year)
Number of odd days in a period of 100 years is 100 + 23 or 100 + 24 or 123/124 odd days.
Number of odd days in 100 years is when 123/124 divided by 7, remainder is 4 or 5. So next century should start with Friday or Saturday.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.