20^th June, 1837 means 1836 complete years + first 5 months of the year 1837 + 20 days of June
1600 years give no odd days
200 years give 3 odd days
36 years give (36 + 9 ) or 3 odd days
Thus 1836 years give 6 odd days
From 1^st January to 20^th June there are 3 odd days
Odd days: January 3, February 0, March 3, April 2, May 3, June 6 = 17
Thus the total number of odd days = 6 + 3 or 2 odd days
This means that the 20th of June fell on 2nd day commencing from Monday.
The required day was Tuesday.

Number of days from 26-Jan-1996 to 15-May-1996 (both days included)
= 6(Jan) + 29(Feb) + 31 (Mar) + 30(Apr)+ 15(May)
= 111

94 days = (13 × 7) + 3 = 3 odd days.
The required day is 3 days beyond Thursday i.e., Sunday

Counting the years 1600 + 300 + 74
In 1600 years, there are zero odd days
In 300 years, there is one odd day
In 74 years, there are 18 leap years and 56 normal years, so the odd days are:
18(2) + 56(1) = 36 + 56 = 92,
Which is 13 weeks and 1 odd day
In 25 days of January, 1975, there are 3 weeks and 4 odd days
Total odd days = 0 + 1 + 1 + 4
six odd days, so it was a Saturday.

Given that seventh day of a month is three days earlier than Friday
⇒ Seventh day is Tuesday
⇒ 14^th is Tuesday
⇒ 19^th is Sunday

This is a leap year. So, none of the next 3 years will be leap years. Each year will give one odd day so the day of the week will be 3 odd days beyond Monday i.e. it will be Thursday.

Given that 1st October is Sunday
Number of days in October = 31
31 days = 3 odd days
(As we can reduce multiples of 7 from odd days which will not change anything)
Hence 1^st November = (Sunday + 3 odd days) = Wednesday