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| TABLE OF MONTHS | TABLE OF DAYS | ||
| Sunday = 1 | |||
| January = 1 (0 in leap yr) | July = 0 | Monday = 2 | |
| February = 4 (3 in leap yr) | August = 3 | Tuesday = 3 | |
| March = 4 | September = 6 | Wednesday = 4 | |
| April = 0 | October = 1 | Thursday = 5 | |
| May = 2 | November = 4 | Friday = 6 | |
| June = 5 | December = 6 | Saturday = 0 |
From Arthur Overton comes an enhanced version of this that will allow you to seek out the day of the week for ANY A.D. year, not just 20th century ones...thank you kindly Arthur!
Step 1) Write down the last two digits of
the year. Call it A.
Step 2) Divide that number by four dropping the remainder call this answer B.
Step 3) Find the month number corresponding to the month from the table below.
Call it C.
Step 4) The day call it D.
Step 5) Take the century number of the year and divide it by four.
If the remainder is 0 then E = 6
If the remainder is 1 then E = 4
If the remainder is 2 then E = 2
If the remainder is 3 then E = 0
Step 6) Now add A+B+C+D+E.
Divide this sum by 7. The remainder you get is the key to the day of the week.
Step 7) In the table of days below, match the remainder with the day of week.
| TABLE OF MONTHS | TABLE OF DAYS** | LEAP YEARS | ||
| Sunday = 1 | If year divides 400, yes | |||
| Jan = 1 (0 in leap year) | July = 0 | Monday = 2 | If year divides 100, no | |
| Feb = 4 (3 in leap year) | Aug = 3 | Tuesday = 3 | If year divides 4, yes | |
| Mar = 4 | Sept = 6 | Wednesday = 4 | If year does not divide 4, no | |
| April = 0 | Oct = 1 | Thursday = 5 | ||
| May = 2 | Nov = 4 | Friday = 6 | ||
| June = 5 | Dec = 6 | Saturday = 0 |
**From Alison comes this comment, "In the enhanced Overton method, for determining which day of the week based on birthdate, it seems the table relating the remainder after division by 7 with the days of the week is offset by one day. I believe it should read Sunday = 0, Monday = 1, Tuesday = 2, etc." THANK YOU, ALISON!
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