are the last two digits of the year, using the previous year for January and February for some information about the history of the months, including why the year began in March and why their names and lengths are strange, see Formulas can be used proleptically, but “year 0” is actually the year 1 BC (see numbering of astronomical years). The Julian calendar is actually proleptic until March 1, 4 A.D. due to mismanagement in Rome (but not egypt) in the period since the calendar was set on January 1, 45 B.C. Effective date (which was not a leap year). In addition, the Modulo operator can cut integers in the wrong direction (ceiling instead of floor). To take this into account, one can add a sufficient multiple of 400 Gregorian years or 700 Julian years. k is the day of the month. m is the number of the month. D are the last two digits of the year. This is the first of the two digits of the year. But according to the rule of the Gregorian calendar, each year that is exactly divisible by four is not a leap year.
We must exclude years that are precisely divisible by 100, but not by 400. That is, we must exclude years that end in 00, but not those that are divisible by 400. We celebrated a new year with some research on the data: the fact that 2020 is a new decade (sort of) and a leap year (definitely), and now some details about how the entire Gregorian calendar works. Some of it is collected in the Ask Dr. Math FAQ about calendar and days of the week. Hello I can find out how can I solve if the last two dogmaits of the year are less than 4, i.e. 2002, 2003, etc. in the method of the key value, because comes from the previous year for February (it is 2009). This leads to , so the appointment was Wednesday. Now, of course, simple questions arise, the key value method uses codes for different months and years to calculate the day of the week.
It would be easier if you could remember the codes that are very easy to learn. The reason why the formula of the Julian calendar differs is that this calendar has no separate rule for leap centuries and is balanced from the Gregorian calendar by a fixed number of days per century. Here we do not have to worry about leap years, as we have already taken them into account in the previous sections. The number of cumulative days is counted at the beginning of the month, so if we divide it by 7, the rest shows how many days of the week the beginning of the month is from the start day for “the year”. Note the pattern in excess: 3,2,3,2,3 repeats every five months, and the accumulation reaches 13 in this period. So every 5 months we want to add 13 days. This can be modeled by the equation of the form Note that if the result for f is negative, caution should be exercised when calculating the correct rest. Suppose f = -17.
If we divide by 7, we must follow the same rules as for the largest integer function; We find the largest multiple of 7 less than -17, so the rest will be positive (or zero). -21 is the largest multiple of 7 less than -17, so the rest is 4, since -21 + 4 = -17. Alternatively, we can say that -7 goes to -17 twice, creating -14 and leaving a remnant of -3, and then adding 7 since the rest is negative, so that -3 + 7 is again a remnant of 4. Normally, 1 year is equivalent to 365 days. Since each normal year advances the day of the week by 1, so nothing more to handle! Just enter the year in the formula. In addition, the years that end in 00 but are not divisible by 400 are those that are divisible by 4, so we must add the semester days. Therefore, the day of the week goes back 2 days. Each century can be represented as the first two digits of the year (the 17th century is the collection of years that begin with two digits 16, the 20th century is the collection of years that begin with two digits 19). So we need to change our decimal arithmetic formula used in Zeller and found it convenient to use J and K in the representation of the year. But if you use a computer, it is easier to manage the modified year Y and month m, which in January and February are Y – 1 and m + 3: each century contains 100 years, 24 of which are usually leap years (without those with an end 00). Since the Gregorian calendar has been adopted at different times in different parts of the world, the location of an event is important in determining the right day of the week for a date that occurred during this transition period.