HOW'S IT CALCULATED?

Ever wonder how the infamous "They" go about generating perpetual calendars. Well, wonder no more. To calculate the day on which a date falls, use the following method:
 

Compute the values a ,m , and y as follows--

Then for the Julian calendar (used from 1 January 45 B.C. until 4 October 1582 B.C.E) calculate the following--

Note that mod 7 represents a modulo 7 calculation. The modulo of a number is generated by dividing the number by the modulo base (in this case 7) and reporting what's left over (the remainder of the division). So the result of a

mod 7 (50) = mod (50/7) = 1 (since 50/7 = 7 with a remainder of 1)

Then for the Gregorian Calendar (used from 15 October 1582 A.D. * )--

NOTE:

• All divisions are integer divisions; remainders are thrown out
• The value of d represents the days of the week as follows:
   

*11 days were added to the year in 1582, so 4 October was followed by 15 October.

Example. If you want to know what day of the week June 4, 1998 fell on, you'd calculate it as follows:

GIVEN:

month = 6
day = 4
year = 1998

CALCULATIONS:
 

• a = (14 - 6)/12 would normally equal 8/12, but since this is an integer division, the result of the division is 0 with a remainder of 8 . . . and since in integer division we throw out remainders,a = 0
• m =6 + 12(0) - 2 = 4
• y = 1998 - 0 = 1998
• d = mod 7 [ 4 + 1998 + 1998/4 - 1998/100 + 1998/400 + {31(4)}/12 ]
• d = mod 7 [ 4 + 1998 + 499 - 19 + 4 + 10 ]    Remember: all divisions are integer divisions
• d = mod 7 [ 2501 - 19 + 14 ]
• d = mod 7 [ 2501 - 4 ]
• d = mod 7 [ 2496 ]
• 2496/7 = 356 with a remainder of 4.  So . . .
• d = 4 which we see from the table means Thursday

If you check with a 1998 calendar, you'll see that this is correct.
 

Return to Top  

Back to Science/Math Page