The last day of century can be which of the following day?

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 1^st January 2001 was Monday
From number of odd days 1^st January 2006 will start with Sunday so it had 53 Sunday
Similar 1^stJanuary 2009 will start with Thursday.

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 31^st December will be Sunday hence 1^st 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 31^st December will be Wednesday hence 1^st 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 1^st 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.

Last day of century can be Friday, Wednesday, Monday, or Sunday and it repeats the cycle. So 1^st day of next century must be either Saturday, Thursday, Tuesday, or Monday and so it must not be Wednesday, friday & Sunday