I have been asked the following question in an exam in Mathematics Phd Programme as a level of IQ Test. However I don't know how to figure it out.
November 9,1994 was a Wednesday. Then what day will be November 9,1965 and November 9,1970?
One way would be to find the number of days in between the two dates and divide it by 7 which is very long and time consuming. Is there a specific way to work this out?
A regular year is $365=7\cdot52+1$ days, so the day of the week of a given calendar date advances by one day each regular year and by two days each leap year. From $1965$ to $1994$ is $29$ years, of which those that are multiples of $4$ are leap years. The multiples of $4$ start with $1968$ and end with $1992$, so there are $7$ of them. $29+7\equiv1\pmod7$, so the day of the week advances one place from $9$ November $1965$ to $9$ November $1994$, and $9$ November $1965$ must have been a Tuesday. $1970$ is $5$ years later, and one of those years is a leap year, so $9$ November occurs $5+1=6$ days later in the week in $1970$, on Monday.
Rule 1. The day advances by 1 every normal year, and by 2 every leap year.
Rule 2. The calendar repeats after a span of 28 years with 7 leap years.
Thus by Rule 1, Nov 9, 1993 (-1 from Wed.) will be Tuesday, and by Rule 2, so will Nov 9, 1965
Also, by rule 1, Nov 9, 1970 will be 1+1+2+1+1 = 6 days beyond Tuesday, i.e. Monday.
Find how many leap years there are(which is easy), multiply by $365$ days for each year and add $1$ for each leap year.
Once you have found one of them, there is necessarily only one leap year in there, multiply the years by $365$ add $1$, convert to modulo arithmetic. Done.
Note that $365=7\times52+1$, so you just need to count years (and adding $1$ for each February 29th in between)
