A: Coming soon.
Q: What can you tell me about the formula involved with this technique, so I can learn it more quickly and efficiently.
A: Besides the description of the technique at the basic site, the basic formula can be expressed in terms of the below three steps. These steps are addressed below in these FAQs.
- (Value for Month) + Day = Result#1; Result#1 - 7X = Result#1 (as adjusted)
- Year - 28X = Year (adjusted to 0-27); Year (adjusted) + INT [Year adjusted / 4] + (Value for Century) - 1 (only if Month = January/February and Year is a leap year) = Result#2
- Result#1 + Result#2 = Result#3; Result#3 - 7X = Day-of-Week (between 0-6)
FAQs for Step 1:
Q: Why so we add a value from a table for the month and not just calculate it each time?
A: It would precious add time to performing the formula. Besides, the simplest calculation would also involve having to look up in a table the lengths of the various months. Memorizing a simple one-digit number is easy than remembering the two digit lengths of the months.
Q: What does the Month Table mean and how is it derived?
A: The values in the Month Table are the number of days of the week away from the first day of the week of the year that the first day of that month is in a non-leap year. Take October, for which the value is zero. That means that January's and October's calendars are exactly the same. The table is derived by taking the excess days (i.e., number of days beyond four full weeks of 28 days) in a month and adding them to the cumulative figure for the prior month, and then reducing the result to the base of 7, that is, a number between 0 and 6.
Q: Why do you reduce the Result in step 1 by the highest multiple of 7 contained in it? (7X means the largest multiple of 7 in the previous number).
A: Because the weekly calendar repeats itself every 7 days, there is no need to carry around a large number, especially since we are going to have to do this at the end of the formula. Also, it is a lot easier to hold a number between 0 and 6 (say, on you hands) until step 3 instead of some number as large as 37.
FAQs for Step 2:
Q: Why do we reduce the Year (i.e., the last two digits of the year, that is) by 28X, the largest multiple of 28 in it?
A: We are going to have to do it anyone at the end so this is a good place to do it since we can then work with a smaller number in step 2. As for the 28X, since there are 7 days in a week and 4 years in a leap year cycle, the calendar repeats itself every 28 years within a century (except for the first two months of '00 years not divisible by 400), so the calendar for 1910 is the same as for 1938, 1966 and 1994, and so on. We call this result the "adjusted year."
Q: What is the INT (Adjusted Year/4) thing and why do we do that?
A: That means the integer (round down or drop any decimal) that you get when you divided the adjusted year by 4. The reason we use 4 is that we are trying to find how many leap years occurred since the base year because we have to add one extra day to the calendar for every leap year after February 29th. (The calendar moves forward 1 day each regular year anyway.)
Q: What is that Value for the Century and what can you tell me about it?
A: When we talk about the century, we refer to the cardinal meaning (1900s and not 20th century) and refer to the first two digits of the full year. The Century Value from the table means the value of the day of the week before that century started. For instance, the 1 for the 1900s means that that century started on a day with a value of 2, a Monday, which it did. (In pivotal century years, i.e., ’00 years evenly divisible by 400) it means the day the century started.) Again, we want a simple table to use here instead of having to waste time by calculating the value each and every time. Note that there are only 4 entries in this table ( 0, 5, 3, 1). This means that the cycle of those four values repeats itself every 400 years, so 1600's value is 2000's and 2400's etc. Note also that the distance in days of the week from 1600 to 1700 is 5 days (0 to 5), from 1700 to 1800 also 5 (5 to 10, 10 because 3+7 equals 10), and 1800 to 1900 also 5 (3 to 8, 8 because 3+5 equals 8), but from 1900 to 2000 (1 to 0 or 7) 6 days (6 because 7 minus 1 equals 6). Why is this? The reason is there is no leap year in-turn-of century years unless that are evenly divisible by 400, which 1600 and 2000 are. So one extra day in the century value of 0 is for that extra day in those special turn-of-century years.
Q: Why do you subtract 1 form the adjusted year if the month is January or February and the year is a leap year?
A: The formula adds in the 1 extra day from February 29th in a leap year for the whole year, so it must be removed it if that extra day has not occurred yet.
FAQs for Step 3:
Q: Why do you reduce the result in step 3 by 7X, the largest multiple of 7 in it?
A: If we did not go through this reduction process, the largest number we would have would be for, say, December 31, 1799, the unreduced value for which would be 164 (5+31+5+99+24). Why memorize a table of days of the week from 0 to 164 with the days repeating themselves every 7 days if you can memorize a simple table of 7 numbers between 0 and 6?
