Orthodox Easter and public chronology guides

In the article “On the Reform of the Orthodox Calendar”, published in the “Church Bulletin” (No. 6 of 1892, pp. 86–89), the true foundations of Christian chronology were indicated and the reasons for which we consider the Russian style to be more appropriate for the time being - Old or Julian, since only with it alone is the canonical requirement of the Apostolic Rules and the Council of Nicea satisfied, that Easter is Christian should neither warn the Jewish one nor coincide with it in the day, meanwhile in the West both happen. Not everyone was pleased with our comments, and some understood them in the sense that we were arguing against scientific accuracy, when it was only about the fact that the latter should not expel and destroy the canons, which have at least equal importance in the question of the Christian calendar. However, it is also important that even the most ardent defender of the “calendar reformation,” Mr. Filippov, recognized the legality of the formulation of the matter indicated by us and promised to adapt his system to it (see the newspaper Day, 1892, No. 1327:1347). However, even if a project appears that is impeccable both in church and scientific respects, a change in the now accepted calendar cannot happen soon; at the same time, in any case, the Paschal, essentially, should be preserved in its original form. Therefore, for practical purposes, it is not useless to indicate some of the simplest methods for finding the most important chronological points according to our old style. For our immediate purposes, we will first try to explain the table attached to the church monthly books entitled “The Key in Brief”, where all calculations are based on Sunday letters or vruceles. The first question in chronology is to find a rule for determining the day of the week on a given day of the month, assuming that the creation of the world was on Friday, the first of March, 5508 years before the birth of Christ, as adopted by the Easter calendar. For this purpose, all days of the week are designated by Slavic letters in the following order: G - Friday, B - Saturday, A - Sunday, Z - Monday, S - Tuesday, E - Wednesday, D - Thursday. In the first year, all its days will correspond to these letters and Sunday will always coincide with A; but since the number of weeks in a year is not complete (52 weeks + 1 day or, on a leap year, 52 weeks + 2 days), then the Sunday letters will not be the same in different years and will not fall on the same days. Sunday letters, obviously, must change, which continues for 28 years (which period is called the circle of the sun), after which they again take the same order for the next 28 years. From here it is clear that if we know the birthday for a given year, then we already know on what day of the week its beginning or March 1st fell. Therefore, to determine the Sunday letter of a given year, you need to find its position in the 28-year period. To do this, the chosen year is added to 5508 from the creation of the world and divided by 28; the quotient (0=28) will show which one he will be in the circle of the sun. To simplify the work, you can simply apply only 20 (for 20 years before Christ the new 197th circle of the sun began). Based on the remainder in the table, we find the key letter, for example, for 12 - A, for 14 - G, for 26 - D. After this, it is no longer difficult to assign each day of this year to the corresponding letter. To do this, you need to count the number of days from March 1 to the given number, subtract three (since March 1 is equal to Friday) and divide by seven; the quotient will be equal in order to one of the following letters: Z, S, E, D, G, B, A, with 0=7 or A. Now you can find the day of the week for any day of any year. Let’s take at least the 15th day of the month of May 1892. 20+1892=1912=68*28+8 gives the remainder 8, i.e. his birthday G. Until May 15, 76–3 = 73:7 days elapsed from March 1 (31+30+15): 7 gives a remainder of 3; therefore, for May 15th the letter of this year will be E. Counting G as Sunday, according to the first order of letters we will see that E will fall on Friday, which is May 15th in 1892. It goes without saying that for the current civil style, the key letter for January and February must be taken from the previous year, because in the Easter calendar, from March 1, these months constitute the end of the old year, and not the beginning of the new. The same result can be achieved even easier using the formula of Prof. D. Perevoshchikova. Divide the previous year by 4 and find the number of days from January 1 to the chosen date; add up all these numbers and divide the sum by 7: - the remainder will be shown by the day, if you consider Sunday = 1, etc., Saturday = 7 or 0. So, for May 15, 1892 we have: 1891+(1891=472*4+3=)472+(31+29+31+30+14)=2498=356*7+6 gives a remainder of 6, equal to Friday. But for Orthodox chronology it is even more important to know what date Easter will fall in a given year, since it is this holiday that underlies it. It is done: a) after the first spring full moon, if it falls on the day or after the spring equinox; b) after the second (i.e. April), when March will be earlier than the specified dates; c) on the first Sunday or the next if the Easter full moon falls on Friday, Saturday and Sunday. This means that in this case you need to find the date of the Easter full moon and the corresponding day of the week. Since 12 lunar months, from full moon to full moon, are almost 11 days shorter than the Julian year, the phases of the moon will be different for each year; in other words: not every year the full moon will coincide with its beginning. It has been noticed that the phases of the moon repeat after 19 years (Metonian cycle): this is the circle of the moon, in relation to which it is necessary to determine what place a given year occupies in it. To do this, you need to calculate how many full lunar circles have passed; the remainder, called the golden number, will show the desired value. In church timekeeping, the beginning of the circles of the moon is taken to be 5508 from the creation of the world, and from (5508=19*289+17) 17 BC. a new (290th) circle begins. But it turned out that the new moon of the Christian era coincided with January 1, a year before Christ. and, therefore, the church calendar, which places this point in (5508+2=) 5510. instead of 5507, it lags behind in the circles of the moon by 3 years; therefore, adding 17 to the chosen year and dividing the sum by 19, we get only the circle of the moon, and to find the true golden number we must add 3 more and from the sum, if it is more than 19, subtract the figure 19. Multiplying the golden number by 11 and dividing by 30, we find in the quotient - the number of complete circles for this period and in the remainder (in “epact” astronomy, in in church timekeeping the basis) is its actual position in March of a given year (for January and February constitute two full lunations). The new moon occurs after 30 days, therefore it = 30 - the foundation; the full moon occurs after 15 days and, therefore, is equal to 30-base + 15. It is recognized as Easter if it is greater than 19 or if it is equal to the 19th day of March; otherwise, it will not be such, and the true Easter full moon will occur in April, for which we add the March one with the 30th and subtract the 31st day of March. So for 1892: circle of the moon = =(1892+17=1909=19*100+9)=9, golden number=9+3=12; base = =(12*11=30*4+12)=12; new moon = (30–12) = 18, full moon (18+15) = March 33 or (33–31) April 2. The day of the week is found as above. Vrucelny anniversary for 1892; 33 days passed from March 1 to April 2; therefore, it will correspond to (33–3=7*4+2)=2 or the letter S, which will fall on Thursday. Easter will be next Sunday or in 3 days, i.e. April 5th 1. Having determined all the reasons for a 19-year period, we find that Easter can only occur within the boundaries between March 22 and April 25, that is, on one of the 35 days between them. Each of them is designated by letters of the Slavic alphabet in order (A = 1, S (zelo) = 8, Z = 9, Ѿ (ot) = 24, Ѫ = 34, Ѧ (YUS small) = 35), called “key” or “border keys”; they indicate how many days a given Easter is removed from March 21st. So, this year 2 it is (31+5=36–21) = 15 days away, and the key letter for it will be N (our). Using these key letters, all holidays and fasts are found as follows: 1) The half-life is equal to the key letter + April 14, i.e. N or April 15+14 = April 29; 2) Ascension N +29 Apr.=44–30=14 May; 3) Trinity Day N +9 May=24 May; 4) meat emptying or fasting before Peter's fast N +16 May = 31 May (next fast from June 1); 5) Lent: in a simple year the key is +January 24 and in a leap year the key is +January 25 (for 1892: January 15 + 25) = meat empty (February 40–31–9), and after 7 days there will be a raw food or complete fasting (for 1892: February 9 + 7 = February 16); 6) The continuation of meat eating in a simple year is equal to the key letter +31 and in a leap year it is equal to the key letter 32; for 1892 we have 15+32=47 or 6 weeks and five days. To this we should add an explanation of two terms. An indict is a fifteen-year period, which is why, adding this year with 5508 from the creation of the world and dividing by 15, we get the number of past indicts and the remainder - the year of a new indict. For 1892: 1892+5508=7400=15*493+5 (remainder 5), i.e. this year is the 5th in the 494th index. For brevity, the very remainders are called indicts (for 1892 there will be indict 5th or D). It is distinguished from the "indiction" or "great Easter circle", embracing 532 (the 19-year circle of the moon multiplied by the 28-year circle of the sun), after which the Easter numbers and moving holidays return to their previous order. If such an indication exists, you just need to find the position of a given year in it, and then the corresponding table will indicate all the necessary holidays. The year 1892 will be the 484th in the 14th indiction (1892+5508=7400=532*13+484:484 in the remainder). Up to 7000 from the creation of the world (or 1492 A.D.) inclusive, the year in Rus' was counted from March 1, and from 7001 to 7207 from the creation of the world (1669 A.D.) - from September 1; from 7208 (or 1700 A.D.) from 1 January. When calculating ancient dates and converting them to our civil (January) calendar in March or Easter years for the months of January and February, you need to subtract 5507 years from them; in September or church years for the months of September, November and December, you need to subtract 5509 years; when the September year is converted to March, we subtract 5509 years from it for the months from September to February; on the contrary, we subtract 5507 years for the same months. We have explained all the elements of the church calendar and indicated how to use them for various calculations. Based on them, you can also check historical data. So, in the “Pskov Chronicle” it says: “Indicta 1, in the summer of 6497, the key of boundaries is 100, the circle of the sun is 28, the circle of the sun is 7, and the circle of the moon is 17, and for the Jews Easter is April 5 on Friday, and for Christians - Easter is April 8.” 6497 divided by 15 gives 433 with a remainder of 2, i.e. the second indict. It is clear that here the counting is based on the September year; therefore, 5509 years are already subtracted from it. The result is 988. Added to 5508 and divided by 28, it gives the 28th circle of the sun. 988+17:19 3 gives the 17th circle of the moon. In the 28th circle of the sun, the birthday is 3 or 7. Golden number (17+3=20; 20–19)=1; the base is 1*11=11, therefore the new moon is 30–11=March 19. Full moon 19+15=34–31=April 3. 31+3–3:7 gives the remainder 3 or the letter E, which, when the birthday is S (green), will correspond to Tuesday. From here, April 5 will be equal to Thursday, but not Friday (which means there is an error in the chronicle either in the day or the number), and Orthodox Easter will fall on April 8 next Sunday. 39 days have passed from March 1 to this number, therefore, from March 21 it is 39–21 or 18, which letter in the keys falls on P, and it means 100. From the analysis we have presented, it is clear, however, that, despite the simplicity, calculation using the indicated methods requires great dexterity, dexterity and “purity”, and even some tables. Therefore, various methods have long been invented to make it easier to find, at least, the day of Easter. Thus, in the ancient book “The Hand of the Theologian” and the manuscript of the Imperial Public Library No. 199 there is a “Hand Easter”, where all Easter terms are associated with the knuckles. This is visual, but it doesn’t make things easier, because it requires a lot of memory. The only correct method here remains a purely mathematical one. The formula of Prof. is well known among us. A.N. Savich, recognized academician. Bunyakovsky is very witty, but “conditions” or “reservations” play an important role in it, complicating the process. Without a doubt, it is accurate, but still somewhat complex and, if not carefully considered, can lead to incorrect conclusions. And we personally know how one mathematics teacher calculated Easter on Wednesday using it. In terms of simplicity, the Gauss formula remains the best, by which you can find both Orthodox and Western Easter - according to the Gregorian calendar. Here the elements M and N come first. For the old style they are constant: M =15, N =6; but in the new one they change: according to Delambert’s table for the years from 1800 to 1899 - M = 23, N = 4, from 1900 to 1999 - M = 24, N = 5; from 2000 to 2099 – M =24, N =5. Then the Gaussian system is as follows: 1) The selected year must be divided by 19, 4 and 7; we get the remainders: a, b and c. 2) The sum 19* a + M divided by 30; – the remainder will be d. 3) Divide the sum 2* b +4* c +6* d + N by 7; – the remainder will be e. 4) Easter will be this year either 22+ d + e March, or (d + e –9) April. From here we can determine Orthodox and Western Easter for 1892. In the first case we have: a = (1892 = 99*19+11) = 11; b=(1892=473*4+0)=0; c=(1892=270*7+2)=2; d=(19*11+15=7*30+14)=14; e=(2*0+4*2+6*14+6=14*7+0)=0. Easter will be 22+14+0=March 36 or (36–31)=April 5. For the new style d = (19* a +23 or 19*11+23=232=7*30+22)=22; e =(2* b +4* c +6* d +4 or 0+8+132+4=144=20*7+4)=4; therefore, this year Easter in the West will be 22+22+4=March 48 or April 17, i.e. on the same day as the Orthodox one, since the “Gregorians” count numbers 12 days 4 ahead against ours. However, Gauss’s formula, for all its merits, cannot be considered the simplest. In addition to the additional elements M and N, it requires remembering the order of actions and in the calculations itself gives too large numbers, in which errors and oversights are more likely to occur. Therefore, it is not without interest to report on a new attempt on this subject by Mr. Splendorov, published in issue 6 (March 15) of “News on the Kazan Diocese” for 1892 (pp. 133–138); it is alien to these shortcomings and is visual, and its analysis opens up the possibility of calculating the day of Easter without any formulas using four simple arithmetic operations. This is the essence of the author's considerations. Easter is celebrated on one of the dates from March 22 to April 25 inclusive, 5, i.e., for 35 days, or seven weeks. These form a table (No. 3) with five vertical and seven horizontal columns 6. According to this row, a second one is compiled, also in five vertical and seven horizontal rows, from numbers from 0 to 27 inclusive, but in such a way that every four digits there remains one empty space - (No. II). In addition to this, another table is needed (No. I), which, like the first two, consists of five vertical and seven horizontal rows. It is obtained in the following way. In the last column, in the last column is written 1; each further figure must be greater than the previous one by 11 (that is, the second will be equal to 12, the third - 23), but if it exceeds 29, then 30 must be subtracted from it and continued in the same way until all thirty-five places are filled. After this we will have the following diagram: Now the chosen year must be divided by 19 and 28 and the remainder of the division must be found in Tables I and II; in this case, the last five (underlined) numbers of the fifth vertical column of table I should not be taken into account, because they were encountered earlier, so for the remainders 15; 4; 23; The years 12 and 1 will be in the first, not the fifth, column. If the places of the remainders from dividing the year by 19 and 28 in tables I and II fall on the same horizontal line, then the number of Easter in table III will be the one that is on the same horizontal line in the vertical column corresponding to the column of the first remainder in table I. If the remainders do not fall on the same horizontal line, then in the first table you need to take from those following the remainder the number that will be in relation to the second in exactly this position, and then, according to it, find in Table III required date. Thus, from the division of 1892, the remainders are 11 and 16 on one horizontal line, with 11 being in the third column of the table. I; therefore Easter will be on April 5th. 1893 gives remainders 12 and 17 on more than one horizontal line; after 12 the nearest number in the table. I on the same line will be 9, and for him in the table. III equals March 28. 1897 gives remainders 16 and 21 not on the same line, and after 16 in table I, position 21 of table II will correspond to 13, which in table III indicates April 13. For 1907, the remainders will be 7 and 3 not on the same horizontal line, and after 7 in the fifth column of Table I, the position of the second remainder will correspond to the number 4, for which in Table III we have 22. Therefore, in 1907 Easter will be April 22. From the above, it is clear that in the stated system the main role is played by the position of the remainders from dividing the year by 19 and 28 in the horizontal and vertical rows in their mutual relation and, accordingly, to the numbers from March 22 and April 25. Is there some general law here? Considering the proposed tables from this side, we notice that if the given remainder from dividing the year by 19 is increased by as many tens as there are in the remainder of units from dividing it by 3, and subtracting the sum from 46, we obtain its vertical and horizontal positions in the table. I. Thus 1894 divided by 19 leaves a remainder of 13; after dividing it by 3, we get a remainder of 1 and therefore increase it by one ten. 13+10 or 23 subtracted from 46 gives 23. Since there are seven places in each column, for 1894 it will be the second in the fourth column, for 23:7 gives 3 full rows and a remainder of 2, i.e. the second place of the fourth row. In Table II there is an empty space after every fourth number; therefore, to the second remainder you need to add as many units as the number of times it contains 4, and divide the sum by seven: the quotient will show that it is in the next vertical row, and the remainder will show its place in the last one. 1894 divided by 28 leaves a remainder of 18, which contains four fours (18=4*4+2). Therefore 18+4=22=3*7+1; that is, for this year in the second table there will be place 1 (or second, since there is a zero at the bottom of the first column) in the fourth row. After this, it is necessary to determine whether these places coincide in the horizontal direction, and if not, then which vertical row of the table. I will correspond to the year taken. The first, obviously, will happen only when the sum of their places is equal to 7 (for example, for 1892 with remainders 11 and 16 we have: 11=3*3+2; 11+20=31; 46–31=15=2*7+1; 16=4*4+0, therefore 16+4=20:7=2*7+6; 1+6=7); If this is not the case, it means that in the horizontal rows the remains of this year will be separated from each other by as much as the sum of their places in the vertical rows is short of 7, by which difference the number of the place of the first remainder must be moved forward. For 1894 we have 2 and 1; therefore 2+1=3; 7–3=4; 23+4=27. It is clear that for this year Easter has moved forward from March 22 by 27 days, of which 10 days (from March 22 to 31) fall on March; therefore, subtracting the number 10 from 27, we will have April 17 for Easter this year. For all these reasons, Easter can be calculated without any tables using just four arithmetic operations. The whole operation will be as follows (method one): A) The chosen year must be divided by 19 and 28 and note the remainders obtained from the division. B) Increase the first remainder (add to it) by as many tens as the remainder of units from dividing it by 3. For 1895, the remainder is 14; 14=4*3+2; therefore, we get 14+20=34. If the remainder of dividing the year by 19 is equal to zero or 1 and 2, which are not divisible by three by an integer with a remainder, then you need to add as many tens as units such a remainder shows. Thus, for the years from 1900 to 1902 the remainders are: 0; 1 and 2; this means that for them we have 0; 11 and 22. If the remainder is completely divisible by three, then no additions need to be made to it. Therefore, for example, 1893=99*19+12; 12=3*4+0; therefore we simply take 12. For 1903 we take 3, because 3=3*1+0. C) We now subtract the found number from 46 and notice this new remainder; for 1895 we get 46–34=12. If the subtraction results in a number greater than 30, then subtract another 30 from it. For 1903, we have 1903=19*100+3; 46–3=43; 43–30=13. D) Divide the last number B by seven and remember the remainder. For 1903, this will be equal to 13=7*1+6. If this division by 7 results in zero in the quotient, then this means that the year taken in the column of the first table occupies, in its position, the last or seventh place from the top; therefore, such zeros must be considered equal to seven. For 1899, with a remainder of 18 (from dividing it by 19), we have 18=6*3+0; 46–18=28=7*4+0, which is what we count as 7. If the remainder is less than 7, then, obviously, the chosen year occupies in the first vertical column on the left of Table I the place indicated by the remainder figure; It is clear that in such cases there is no need to divide by 7, but you should simply remember for G the very remainder B. For 1904, with a remainder of 4 (from division by 19), it has 4 + 10 = 14; 46–14=32; 32–30=2; this last digit (2) will serve for both B and G. D) The remainder of dividing the year by 28 is increased by as many units as the number of times it contains 4. 1895: 28=67*28+19; 19=4*4+3. Therefore 19+4=23. If this remainder is zero or less than 5 (no more than 4), then no additions are made, because the remainder itself already shows the position of the given year in the first column from the bottom of the second table. For the years from 1904 to 1908, dividing by 28 gives 0; 1; 2; 3; 4. E) Divide the found number by 7 and note the remainder. For 1895 we get 23=3*7+2. For 1911 we have 1911=28*68+7; 7=4*1+3; 7+1=8; 8=7*1+1. If the number D is less than 7, then division is not performed, because for such a year the position in the first column of the second table will be the place that the number D indicates. For the years from 1904 to 1908 we have 0; 1; 2; 3; 4. For 1909 we get 1909=28*68+5; 5=4*1+1; therefore 5+1=6. If dividing this number by 7 results in 0, then the year taken is equal to the seventh place in the column of the second table; because 0=7. So, 1899=28*67+23; 23=4*5+3; 23+5=28=7*4+0; hence 0=7. G) Numbers G and E are added and subtracted from 7. For 1894, G=(1894=19*99+13; 13=4*3+1; 13+10=23; 46–23=23=7*3+2)=2; and E=(1894=28*67+18; 18=4*4+2; 18+4=22=7*3+1)=1. Therefore F=(2+1=3; 7–3)=4. If the sum of G and E is greater than 7, then it should be subtracted from 14. For example, for 1890 G and E = 0 and 3 or 7 + 3 = 10; 14–10=4. H) We apply this number to number B and, if the sum is greater than 10, subtract 10 from it: then the number for Easter will be April. For 1895, B=12, and W=(1895=28*67+19; 19=4*4+3; 19+4=23=7*3+2; 5+2=7; 7–7)=0; 12+0=12; 12–10=2. Easter will be on the second of April. For 1890, F equals 4, and Z = (7 + 4) = 11; Easter was (11–10)=April 1st. If the sum of F and W is less than 10, then it is clear that Easter from the first point of the Easter limit has not moved beyond the border of March (21 + 10 = thirty-one days of March); in this case, you need to add 21 to this amount and the result will indicate the corresponding March date for Easter. Let's take 1877. B=(1877=19*98+15; 46–15=31; 31–30)=1; Г=1; E=(1877=28*67+1)=1; F=(1+1=2; 7–2)=5; F+B=(5+1)=6; 6+21=27. Easter in 1877 was March 27th. Let us now give, for illustration, an example for each case. 1) The case is the most correct: 1894. Its remainders are 13 and 18. B=(13=3*4+1; 13+10=23; 46–23)=23; Г=(23=7*3+2)=2; E=(18=4*4+2; 18+4=22=7*3+1)=1; F=(2+1=3; 7–3)=4; 23+4=27; 27–10=April 17. 2) 1900. Remains 0 and 24. B= [46=16+30]=16; Г=(16=7*2+2)=2; E=(24=4*6+0; 24+6=30=7*4+2)=2; F=(2+2=4; 7–4)=3; Easter=(16+3=19; 19–10)=April 9. 1901 Residues 1 and 25. B= [46–(1+10)=35; 35–30]=5; G=5; E=(25=4*6+1; 25+6=31=7*4+3)=3; F=(5+3=8; 14–8)=6; Easter will be=(5+6=11; 11–10)=April 1st. 1902 Residues 2 and 26. B= [46–(2+20)]=24; Г=(24=7*3+3)=3; E=(26=4*6+2; 26+6=32=7*4+4)=4; F=(3+4=7; 7–7)=0; Easter will be (24–10)=April 14. 3) 1903. Residues 3 and 27. B=(46–3=43; 43–30)=13; Г=(13+7*1+6)=6; E=(27=4*6+3; 27+6=33=7*4+5)=5; F=(6+5=11; 14–11)=3; 13+3=16; 16–10=6; those. Easter April 6th. 4) 1899. Residues 18 and 23. B=(18=3*6; 46–18)=28. Г=(28=7*4+0)=0 or 7; E=(23=4*5+3; 23+5=28=7*4+0)=0 or 7. F=(7+7=14; 14–14)=0; 28–10=April 18, when Easter will be. 5) 1904. The remainders are 4 and 0. B=(4=3*1+1; 4+10=14; 46–14=32; 32–30)=2; G=2; E=0; F=(2+0=2; 7–2)=5; 2+5=7; 7+21=March 28, on which Easter will fall. 6) 1909. Residues 9 and 5. B=(9=3*3+0; 46–9=37; 37–30)=7; Г=(7*1+0)=0 or 7; E=(5=4*1+1; 5+1)=6; F=(7+6=13; 14–13)=1; 7+1=8; 9+21=March 29, when Easter falls. 7) 1915. Residues 15 and 11. B=(46–15=31; 31–30)=1; Г=1; E=(11=4*2+3; 11+2=13=7*1+6)=6; F=(1+6=7; 7–7)=0; 1+0=1; 1+21=22; Easter will be on March 22. 1885 Residues 4 and 9. B=(4=3*1+1; 4+10=14; 46–14=32; 32–30)=2; G=2; E=(9=4*2+1; 9+2=11=7*1+4)=4; F=(2+4=6; 7–6)=1; 2+1=3; 3+21=24; Easter fell on March 24th. The entire procedure, as outlined above, can be slightly modified and, perhaps, simplified. Then we get the second method for finding the day of Easter celebration in any year. All actions will be arranged in the following order. α) Divide the selected year by 28, and add the number of fours contained to the resulting remainder; then divide this amount by 7 and subtract the remainder (whatever it may be, even if zero) from seven and remember the difference. For example, 1916. 1916=28*68+12; 12=4*3+0; 12+3=15; 15=7*2+1; 7–1=6. If the remainder of dividing a year by 28 is less than 4, then you need to directly subtract it from seven and notice this difference. For 1904 it will be equal to (1904=28*68+0; 7–0)=7. For 1907 it is equal to (1907=28*68+3; 7–3)=4. If the remainder of dividing a year by 28, increased by the number of fours it contains, is less than 7, then it must be directly subtracted from seven. For 1908 it will have (1908=28*68+4; 4:4=1; 4+1=5; 7–5)=2. For 1909 it turns out (1909=28*68+5; 5=4*1+1; 5+1=6; 7–6)=1. If the remainder of dividing the year by 28, increased by the corresponding number of fours, after dividing it by 7, gives zero, then subtract this zero from 7. For 1910 we have (1910=28*68+6; 6=4*1+2; 6+1=7; 7=7*1+0; 7–0)=7. β) Divide the same chosen year by 19 and increase the remainder by as many tens as there are units in the remainder of dividing it by 3, and subtract the sum from 46. If this remainder by three is divisible by an integer or less than three (i.e. 0, 1 and 2), then you need to subtract it directly; if after subtraction the difference is equal to 30 or more than 30, then subtract another 30. For 1916 we have (1916=19*100+16; 16=3*5+1; 16+10=26; 46–26)=20. For 1918 we have (1918=19*100+18; 18=3*6+0; 46–18)=28. For 1902 we have (1902=19*100+2; 46–2=44; 44–30)=14. For 1904 it turns out (1904=19*100+4; 4=3*1+1; 4+10=14; 46–14=32; 32–30)=2. γ) If the last difference is less than 7 and the difference α or equal to it, then add the last one (α) to 21 - and in total we get the corresponding March date for Easter. For 1942 α=(1942=28*69+10; (10)=4*2+2; 10+2=12; 12=7*1+5; 7–5)=2; γ=(1942=19*102+4; 4=3*1+1; 4+10=14; 46–14=32; 32–30)=2. γ=2 <7 and=α or 2; therefore 21+2=23. Easter in 1942 will be on March 23. δ) If the difference β is greater than α, then as many sevens must be added to α as the number of times 7 is contained in γ; add the amount to 21. If the number is less than 31, it will indicate the date of March for Easter; if it is more than 31, subtract 31 from it, and we get the number of April for Easter. For 1909 we have: α=(1909=28*68+5; 5=4*1+1; 5+1=6; 7–6)=1, β=(1909=19*100+9; 9=3*3+0; 36–9=37; 37–30=7; 7=7*1+0; 7*1)=7; α+ β=1+7=8; 21+8=29; Easter will be on March 29th. For 1905 we get: α=(1905=28*68+1; 7–1)=6; β=(1905=19*100+5; 5=3*1+2; 5+20=25; 46–25=21=7*3+0; 7*3)=21; 6+21=27; 27+21=48; 48–31=17; Easter in 1905 will be April 17th. In conclusion, we offer an analysis of two examples for the deadlines when Easter generally occurs, i.e. for March 22 and for April 25. For the first case, let's take 1915. For it we get: α=(1915=28*68+11; 11=4*2+3; 11+2=13; 7=1+6; 7–6)=1; β=(1915=19*100+15; 15=3*5+0; 46–15=31; 31–30)=1; 21+1=22; Easter in 1915 will occur on March 22. For the second case, let's take 2800 g. For it we have: α=(2800=28*100+0; 7–0)=7; β=(2800=19*147+7; 7=3*2+1; 7+10=17; 46–17=29=7*4+1; 7*4)=28; 7+28=35; 21+35=57; 57–31=25; this shows that Easter in 2800 will be on April 25th. But it goes without saying that all the methods we have discussed do not yet provide everything that may be required in chronological research for practical and scientific purposes. Therefore, we add a review of publicly available manuals on chronology in Russian literature. A good guide to this subject is provided by the book by Mr. G.M. “Time calculation among ancient and new peoples” (Kazan, 1884, II+96 pp. Price 75 kopecks). Here the author gives a condensed overview of calendars and eras among different peoples, and then (Chapter III) carefully outlines the “Russian chronology,” which is not in the translated work (1867) by Lalosh, “Time Calculation of the Christian and Pagan World.” The last (IV) chapter is devoted to detailed teaching about the “Orthodox Easter” and provides all the necessary - scientific, historical and church - information. In the appendices we find “Easter circle with the designation of the years XI, XIII and XIV (current) indictions”, “Table of the lunar course”, “Easter sighted by key letters” with a schedule of movable holidays, “Index of the days of the week on which 1 falls; 8; 15; On the 22nd and 29th of each month in March, September and January common and leap years, on all dates on which Easter occurs, i.e. from March 22 to April 25”, “Table for finding the days of the week corresponding to the given numbers of months in the current and last centuries, covering the period of Russian history” and “Translation of March and September years from January ones”. After studying this book, everyone will be able to do all the calculations they need without much difficulty. But it goes without saying that this requires some knowledge and mathematical experience, meanwhile, not everyone has such qualities, and sometimes chronological information is required quickly. For such persons, we can recommend the “Reference tabular calendar with Easter for all years according to the Orthodox chronology (old style)”, compiled according to the Orthodox Easter by the priest of the Vilna Palace Church Kapiton Petrov (Vilna 1887; Price 25 kopecks). This is a small book of 14 leaves with two indexes and XIV tables, which, for ease of use, can be pasted on cardboard or canvas: - then the “calendar” will be wall-mounted. To use it you only need the ability to divide the chosen year by 28 or 19 and make the simplest subtraction, and then it will show a) the date of Easter in any year and the holidays and fasts that depend on it (weeks preparatory to fasting, Palm Sunday, Ascension, Pentecost, the duration of Peter’s Fast, etc.) and b) in all years from the Creation of the World to the Nativity of Christ and after it - 1) the day of the week when the year, month and day are known, 2) the day of the month, when we know the year, month and day and 3) month, when the year, date and day are given. This book is very practical and suitable “for historical, legal and family references.” The whole procedure takes no more than two or three minutes, but its readings are certainly correct. For those who are embarrassed even by the first arithmetic operations or are burdened to turn to the help of a pencil, the “Public Perpetual Calendar” (Kharkov, 1891, price including postage 40 kopecks) is very useful. It consists of three tables. Table I - “for determining the days of the week corresponding to the numbers of months” in the years from the creation of the world and after the Nativity of Christ, and for Russian history - in the years of March (ending with the 7000th year from the creation of the world) and in September (from 7001 to 7203 from the creation of the world). From this table you can find out: 1) day of the week by year and date, 2) according to the same data, the distribution of the numbers of all months by day of the week and 3) in what years does such and such a day of the month coincide with a known day, for example, April 1 with Sunday, February 1 with Friday, etc. Table II - “for finding the days of Easter celebration” - presents the Easter full moons themselves, but the holiday of the Resurrection of Christ is found easily and unmistakably. Table III shows the distribution of fasts, meat-eaters and moving holidays according to the number of St. Day. Easter in the chosen year. On a separate sheet of paper is attached a “pocket calendar for 200 years” (from 1800 to 2000) for distributing the numbers of a given year by day of the week. The “Public Perpetual Calendar” we considered was compiled in relation to the data of Mr. G.M. (although, perhaps, independently of it) and completely satisfies its goals. An even greater simplification is represented by a witty and original method proposed in 1891 on the pages of the Moscow illustrated magazine “Science and Life”. This is a cross-shaped table; in the upper part - numbers of centuries of the Julian and Gregorian calendars, below - tens and units, on the right - months, on the left - numbers. In the middle there is a movable circle indicating the days of the week. You just have to make three turns of this circle, and based on the date of the month and year, you can easily find the corresponding day of the week. This calendar answers only one question, but its advantages are that it is simplified to a minimum and takes into account the Gregorian calendar, which is not the case in others. On the same principle of possible simplicity, but on a larger scale, G. Ioffe’s “Complete public wall calendar of the old and new styles with the eternal Easter of the Orthodox Church,” released in 1891 in Moscow, was built (price with postage 1 rub. 20 kopecks). This is a large thick parchment sheet on which the “Eternal Calendar”, “Eternal Paschal of the Orthodox Church” with an “auxiliary table” and a table of “moving holidays, continuous weeks, fasts and days of remembrance depending on Easter” are clearly printed in several colors. The necessary explanations and “scientific notes” are placed in the margins and on the reverse side. To use it, you need to move the corresponding strips, plates and circles in the same way. Using these simple manipulations, the “Perpetual Calendar” indicates: 1) days of the week by year and date; And 2) years - by day on a known date of a given month. “Eternal Easter” with a table shows Easter and passing holidays, and according to the day of Easter of the given year - the date and day of the event before or after it; using them, you can find back the years for well-known combinations of holidays, for example, when Kyrio-Easter (i.e., Bright Resurrection) coincides with the Annunciation, March 25, or when the Ascension falls on May 9, St. Nicholas the Wonderworker, etc.? You can also put a “calendar” at the beginning of the year, and it will serve throughout it for all the issues noted. From this it can be seen that Joffe’s work “gives calendar information and answers to all sorts of questions, direct and inverse, regarding permanent and temporary holidays, memorial days, service days, etc. in the past, present and future tense.” All this information, with a little skill, can be obtained easily, quickly and conveniently without any calculations, and therefore Ioffe’s “Calendar” is suitable for all literate people and can be useful for scientific and practical purposes. Its scientific merit was recognized by the physics and mathematics commission, which gave an approving review. (See volume 4 of the “Proceedings of the Department of Physical Sciences of the Imperial Society of Lovers of Natural History”, Moscow 1891). For our part, we could add that to complete the matter, it is not superfluous to know the Paschal in the Gregorian new style, which is not in any of the analyzed manuals. For many, and not infrequently, this is necessary. For this purpose, the excellent arguments of Mr. Perevoshchikov (“Rules for the calculation of time adopted by the Orthodox Church,” Moscow 1850) and Archpriest P.S. Delitsyn, professor of the Moscow Theological Academy (“A way to find the day of Holy Easter in a given year among Christians, both Orthodox and Western” in “Readings in the Moscow Society of Lovers of Spiritual Enlightenment” for 1865), who also give an overview of Orthodox timekeeping in general; We used them at the beginning of the article. All chronological issues are discussed in even more detail in the book published in St. Petersburg in 1879 by the late professor of Novorossiysk University V.I. Lapshin entitled “Lunar current and different ways of determining Orthodox and Western Easter” (XVI+83 pages, price 25 kopecks with postage instead of the nominal 75 kopecks), although the presentation is sometimes somewhat confusing. As can be seen from here, the author focuses on a particular issue, but explores it quite thoroughly both from a scientific point of view and from a practical point of view. The essay is divided into six sections. The first (§§ 1–15) provides general information about Easter, years, the movement of the Moon and other necessary data, and then reports in detail on the “lunar flow device” adapted by Mr. Lapshin to determine Bright Resurrection and the days depending on it; it is printed on a special sheet and can easily be adapted for practical use. The second part describes various ancient “lunniki” based on the book by V. Sreznevsky “Northern Carved Calendar”, Magnusson Description of a norwegian clog Calender (Cambrige, 1879), the Followed Psalter of 1686, “The Hand of the Theologians” (Moscow, 1787) and the manuscript of the Imperial Public Library No. 199 (§§ 16–21); The author's reasoning is illustrated here with precise drawings in the appendix. In sections 3, 4 and 5 (§§ 22–38) various ways of determining Easter are given, for example, through indefinite equations with two unknowns, according to the formula of Prof. A. I. Savich, etc., and the doctrine of Easter elements (epakta, foundation, vrutseleto, key of boundaries), which indicates “the application of reckoning to the verification of chronological data in Russian chronicles.” At the end (§§ 39–43) the question of Easter according to the Gregorian calendar is discussed; it is analyzed in less detail, but so satisfactorily that the essence of the matter is clear, especially in view of the fact that the presentation is adapted to Orthodox Easter terms. On the relationship of the accepted “Easter” elements to the astronomical ones and on the need to retain the former even if they do not correspond to the latter, see the article by Prof. Moscow Theological Academy D. F. Golubinsky “On the time of celebrating Easter among Christians of the East and West” in the April 4 issue of “Theological Bulletin” for 1892, pp. 73–88. For the 19th century (editor's note). According to the old style (editor's note). If we take into account only those cells where numbers are located (editor's note).