On 14 ^th Feb the valentine day, 2009 it was Saturday. What was the day of the week on 14 ^th Feb, 2008?

January and August or October depends on leap year or non leap year. But if we find the number of odd days between March and November we will get number of odd days is 0 hence they will have same calendar.

From Zeller's Formula we can find that 1^st March 2009 is Sunday so well have 5 Saturdays and 5 Sundays in total 10 weekends.

Here we have to find the number of odd days between, 5^th march and 5^th November,
Number of days in March is 26 or 5 odd days
(Here we have not included 5^th march)
Number of days in April is 30 or 2 odd days
Number of days in May is 31 or 3 odd days
Number of days in June is 30 or 2 odd days
Number of days in July is 31 or 3 odd days
Number of days in August is 31 or 3 odd days
Number of days in September is 30 or 2 odd days
Number of days in October is 31 or 3 odd days
Number of days in November is 5 or 5 odd days
(Here 5^th November is included)
So total number of odd days = 5 + 2 + 3 + 2 + 3 + 3 +2 + 3 + 5 = 28 when divided by 7 gives remainder 0 hence 5^th November will be same as that of 5^th march.

From Zeller's Formula
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 18 (since 18^th October)
Month m = 8 (As march = 1, April = 2, May = 3, October = 8)
D is the last two digit of year here D = 50 (As year is 2050)
C is the 1^st two digit of century here C = 20 (As year is 1950)
f = 18 + [13 x 8 - 1 / 5] + 50 + [50/4] + [20/4] - 2 x 20.
f = 18 + [103/5] + 50 + [12.5] + [5] - 40.
f = 18 + 20 + 50 + 12 + 5 - 40 = 65.
When divided by 7 we will get remainder 2, hence number of odd days is 2,
So 18^th October 2050 is 2 days more than Sunday, i.e Tuesday.

From Zeller's Formula:
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 15 (since 15^th August)
Month m = 6 (As march = 1, April = 2, May = 3, August = 6)
D is the last two digit of year here D = 47 (As year is 1947)
C is the 1^st two digit of century here C = 19 (As year is 1947)
f = 15 + [13 x 6 - 1 / 5] + 47 + [47/4] + [19/4] - 2 x 19.
f = 15 + [77/5] + 47 + [11.75] + [4.75] - 38.
f = 15 + 15 + 47 + 11 + 4 - 38 = 54.
When divided by 7 we will get remainder 5, hence number of odd days is 3,
A remainder of 0 corresponds to Sunday, 1 means Monday,
So 15^th August 1947 is 5 days more than Sunday, i.e Friday.

Number of days in K weeks is 7K hence total number of days is 7K + K = 8k or number of days must be a multiple of 8.

Number of days in K weeks is 7K hence total number of days is 7K + K = 8k
Similarly number of days in 2Kth day of the 2Kth week is 2k x 7 + 2k = 16k
Required number of days is 16K - 8K = 8K or number of days must be a multiple of 8.

At 6 : 30 A.M. from the formula here P = 6 and Q = 30 so required angle is 11 x 30/2 - 30 x 6 = 165 - 180 = - 15 or ignoring the negative sign the required angle is 15°

The last day of century cannot be either Tuesday, Thursday or Saturday.

Total number of odd days
30 September, 1997-98 = 1
30 September, 1998-99 = 1
30 September, 1999-2000 = 2
30 September, 2000-02 = 1
30 September, 2001-02 = 1
30 September, 2002-03 = 1/7
? Tuesday + 7 = Tuesday
So, the next Tuesday will come on the Mrs Susheela's wedding anniversary in 30th September, 2003