January 16, 1997 was a Thursday. What day of the week will it be on January 4, 2000?

1600 years contains 0 odd day; 300 years conains 1 odd day. Also, 83 years contains 20 leap years and 63 ordinary years and therefore, (40 + 0) odd days i.e., 5 odd days.

? 1983 years contain (0 + 1 + 5) i.e., 6 odd days
Number of days from Jan. 1984 to 31st Oct 1984 = (31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31) = 305 days = 4 odd days
? Total number of odd days = 6 + 4 = 3 odd days
So, 31st Oct 1984 was wednesday

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

We can say that ,
August 1, 1988 means = 1987 years + 7 months
Number of odd days in 1987 years = 1600 years have 0 odd days + 300 years have 1 odd day + 87 years have 21 leap years and 66 ordinary years.
So, there are [ ( 21 × 2 ) + ( 66 × 1 ) ] = 108 days, = 15 weeks and 3 odd days.
Number of days between January 1, 1988 to August 1, 1989 = January + February + March + April + May + June + July + August = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 1 = 241 days = 30 weeks and 4 odd days.
Total number of odd days = 0 + 1 + 3 + 4 = 8 odd days = ( 1 x 7 + 1 ) = 1 odd day.
Thus, Friday falls on 5th, 12th, 19th and 26th in August 1988.

According to question , we know that
The hands of a clock point towards each other 11 times in every 12 hours. (because between 5 and 7, at 6 O'clock only they point towards each other)
So, in a day the hands point towards each other 22 times.
Hence , correct answer will be 22 times.

Try yourself. It was Wednesday.

Time from 10 am to 3 pm of the next day = 29 h
Now, 24 h 10 min of the clock in question = 24 h of the correct clock
i.e., 145 h of the clock in question = 24 h of the correct clock
29 h of the clock in question = 24 x (6/145) x 29) h of correct clock = 28 h 48 min correct clock
So, the correct time is 28 h 48 min after 10 am i.,e 48 min past 2 pm.

Count the number of odd days from the year 2008 onward to get the sum equal to 0 odd day.

Let us see
2008 = 2
2009 = 1
2010 = 1
2011 = 1
2012 = 2
2013 = 1
2014 = 1
2015 = 1
2016 = 2
2017 = 1
2018 = 1

Sum = 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 = 14 days = 2 weeks = 0 odd days
? Calendar for the year 2019 will be the same as for the year 2008

Given that n = 5 and n + 1 = 6
The hands will make right angle at
(5n ± 15) x 12/11 min past 5 .
? (5 x 5 ± 15) x 12/11 min past 5.
? (25 + 15) x 12/11 min past 5 and (25 - 15) x 12/11 min past 5.
? 40 x 12/11 min past 5 and 10 x 12/11 min past 5.
? 4801/11 min past 5 and 120/11 min past 5.
? 43⁷/₁₁ min past 5 and 10¹⁰/₁₁ min past 5.

Time from 7 am on Sunday to 7 pm on following Sunday
= 7 days 12 h = 180 h
? Watch gains (5 + 5⁴/₅) min or 54/5 in 180 h.
Now, 54/5 min are gained in 180 h.
? 5 min gained in (180 x 5/54 x 5) h = 83 h 20 min.
= 3 days 11 h and 20 min

? Watch is correct after 3 days 11 h and 20 min after 7 am of Sunday.
? It will be correct at 20 min past 6 pm on Wednesday.