I am going to tell you one of the fastest method to find to day of any date from the year \(1900 to 1999\) just by giving out 4 inputs. For this method you must know the code of the day and months.

These are the codes for the month:-

Jan=1

Feb=4

Mar=4

Apr=0

May=2

Jun=5

Jul=0

Aug=3

Sept=6

Oct=1

Nov=4

Dec=6

These are the codes for the days:-

Sun=1

Mon=2

Tue=3

Wed=4

Thu=5

Fri=6

Sat=0

Let's take an example:-

Find which day was \(19 July 1967\)

Let's start solving by giving the four inputs:-

--> Date = \(19\)

--> Last 2 digits of the year = \(67\)

--> Number of leap years occurred in that century = 67/4 = \(16\)

Note that for this input JUST TAKE THE QUOTIENT NOT THE REMAINDER.

--> Code of the month = \(0\)

By adding all the inputs we get the answer \(102\)

Now, divide the result by 7 and ONLY TAKE THE REMAINDER.

= \(102 / 7 \) = 14 and the remainder is 4.

And 4 is the code which day you have to find.

And 4 is the code of \(Wednesday\)

So, the answer is \( 19 July 1967 was Wednesday.\)

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