First method

Let Sunday be $1$, Monday be $2$, ..., Saturday be $7$.

1. Divide the last 2 digits of the year by $4$ and discard the fraction, call it $a_1$.
2. Add the date of the month to $a_1$, call the result $a_2$.
3. Add to $a_2$ the month's key value as given by the table below and call the result $a_3$.

   Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
   1	4	4	0	2	5	0	3	6	1	4	6

4. Subtract $1$ from $a_3$ if the month is a Jan or Feb of a leap year. Call the result $a_4$.
5. To $a_4$, add
   $$\left\{ \begin{tabular}{cc} 0,& if the year was in the $1900$'s,\... ... in the $1700$'s,\\ 2,& if the year was in the $1800$'s, \end{tabular}\right.$$
   and if the year is not above then add a multiple of $400$ to convert to one of the above cases. Call the result $a_5$.
6. Add last 2 digits of the year to $a_5$ and call the result $a_6$.
7. Let $1\leq a_7\leq 7$ satisfy $a_6\equiv a_7\, ({\rm mod}\ 7)$.

This is the desired day of the week.

Notation: Let $[x]$ denote the integer part of a real number $x$.

Example 1.7.28   Find the day of the week of $9-14-99$ using the above algorithm.

(1) $[99/4]=24$, (2) $24+14=38$, (3) $38+6=44$, (4) $44+0=44$, (5) $44+99=143$, (6) $143\equiv 3\ ({\rm mod}\ 7)$.

The third day is Tuesday.