for finding the day of the week for any given date

Take the given date in 4 portions, viz. the number of centuries, the number of years over, the month, the day of the month.Compute the following 4 items, adding each, when found, to the total of the previous items. When an item or total exceeds 7, divide by 7, and keep the remainder only.

The Century-itemFor Old Style (which ended September 2, 1752) subtract from 18. For New Style (which began September 14) divide by 4, take overplus from 3, multiply remainder by 2.

The Year-itemAdd together the number of dozens, the overplus, and the number of 4s in the overplus.

The Month-itemIf it begins or ends with a vowel, subtract the number, denoting its place in the year, from 10. This, plus its number of days, gives the item for the following month. The item for January is "0"; for February or March, "3"; for December, "12".

The Day-itemThe total, thus reached, must be corrected, by deducting "1" (first adding 7, if the total be "0"), if the date be January or February in a leap year: remembering that every year, divisible by 4, is a Leap Year, excepting only the century-years, in New Style, when the number of centuries isnotso divisible (e.g. 1800).The final result gives the day of the week, "0" meaning Sunday, "1" Monday, and so on.

Examples1783, September 18

17, divided by 4, leaves "1" over; 1 from 3 gives "2"; twice 2 is "4".

83 is 6 dozen and 11, giving 17; plus 2 gives 19, i.e. (dividing by 7) "5". Total 9, i.e. "2"

The item for August is "8 from 10", i.e. "2"; so, for September, it is "2 plus 31", i.e. "5" Total 7, i.e. "0", which goes out.

18 gives "4". Answer, "Thursday".

1676, February 23 16 from 18 gives "2"

76 is 6 dozen and 4, giving 10; plus 1 gives 11, i.e. "4".

Total "6"

The item for February is "3". Total 9, i.e. "2"

23 gives "2". Total "4"

Correction for Leap Year gives "3". Answer, "Wednesday"