days-of-the-week formula 117

The numbers in parentheses pertain to leap years.

Thus, for example, July 1, 2000, fell on day 6 +0=6,

Saturday; and December 1, 2000, fell on day 6 + 6 ≡5,

Friday.

Now it is a matter of determining on which week-

day the first day of any given year fell. Assume, for the

sake of the mathematical argument, that the Gregorian

calendar has been in use for two millennia, and that

January 1 in the year 0 was day d. Consider the day the

weekday of New Year’s Day Nyears later.

As each ordinary year contains 365 (≡1) days, the

day on which January 1 falls advances one weekday

each year. For each leap year, it advances an additional

day. We need to determine the number of leap years

over a period of Nyears.

In general, a leap year occurs every 4 years, yielding

possible occurrences of a leap year, including year

zero. (Here we are making use of the

CEILING FUNC

-

TION

.) However, no leap year occurs on a year value

that is a multiple of 100—and this occurs

times—except if Nis a multiple of 1,000, which occurs

times. (The year 1900, for instance, was not a

leap year, but the year 2000 was.) The total number of

leap years L that occur in a period of Nyears from year

zero is thus given by:

For instance, February 29 appeared, in theory,

times

before the date of January 1, 2000. Thus, the weekday

of January 1, year N, is given by:

d+N+L

(January 1 year zero, was day d. There is an advance

for each of the Nyears, and an advance of an addi-

tional day for each of the Lleap years.)

Knowing that New Year’s day, 2000, was day 6, we

deduce then that the appropriate value of dis given by:

d+ 2000 + 482 = 6

That is, working with remainders upon division by 7,

d+5+6=6, yielding d= 2. Thus, in our theory, Jan-

uary 1 in the year zero was a Tuesday.

We now have the following

ALGORITHM

for com-

puting the weekday of any given date. Assume we wish

to compute the weekday of the Dth day, of month M,

in year N.

1. Consider the year number Nand compute

its remainder upon division by 7.

2. Compute and its

remainder upon division by 7.

3. Sum the answers of the previous two steps

and add 2. Compute the remainder of this

number, if necessary, when divided by 7.

This is the weekday number of January 1 of

year N.

4. To this weekday number add the day D,

subtract 1, and add the appropriate month

number from the table above. Look at the

remainder upon division by 7, if necessary.

This final result is the weekday number of

the desired day.

For example, for the date of March 15, 2091, N= 2091

≡5, L=523–21+3=505≡1, yielding January 1 of

that year to be day 2 +5+1≡1, a Monday. To this we

add 14 days (the number of days later is 1 less than the

date D) with a month adjustment of value 3 (this is not

a leap year). Thus March 15, 2091, will fall on day

1+14+3≡4, a Thursday. (Warning: as the Gregorian

calendar was not used before October 15, 1582, this

algorithm cannot be applied to dates earlier than this. )

Simplifying the Procedure

This method can be simplified to some extent. Write

the year number as mcyy, with mfor millennia, cfor

century, and yy as the two-digit year number. More

precisely, we mean:

N = 1000m + 100c + yy

with 0 ≤c ≤9 and 0 ≤yy ≤99. For example, the year

3261 will be written:

N = 1000 × 3 + 100 × 2 + 61

Notice that