will be distinguished by different numbers of those Cycles, until the last year, 7980, when the division by the prime numbers 28, 19, and 15, respectively, will leave no remainders; the numbers themselves then expressing the last years of each cycle. At the commencement of the Vulgar Era, the year of the current cycle of the Sun was 10, of the Moon 2, and of Indiction, 4; and from these characters, the corresponding year of the Julian Period may be found, by the help of the following analytical problem. To find a number, which being divided by three given integral numbers, 28, 19, and 15; shall leave given remainders, 10, 2, and 4, respectively. Let a be the integral number sought, which, divided by the two first divisors, will leave the remainders required. Then will and 2-2, be integers. Put =m, and 19 x-10 then x=28m+10; and x=19n+2; and consequently, n= an integer. From this, subtract the nearest integer, 28m +8 , 19 a+2=b. And consequently, a=9b-2. Substitute this value 9 of a, in the last found value of m; which will give m=19 b—3. Substitute this in the first original value of x, and then x=532b-74. Hence b will be affirmative; and substituting its least possible value, 1, then x=532-74-458. But this number, 458, will fulfil the conditions of the two first divisors; leaving the remainders 10 and 2, respectively. If now to this number, 458, we add the product of the two first divisors, 28 x 19-532, or any multiple thereof, 532 y: supposing y to denote any affirmative integer, the aggregate 532 y+458, will equally fulfil the conditions of the two first divisors. Let this aggregate next be supposed to answer the condition 532 y+454 of the third divisor, 15, also. Then will 532 y+458-4= 15 be an integer. From it subtract the nearest integer, and the remainder will be an integer also, 15 c-4 7y+4 15 From this, subtract the nearest in And consequently y= teger, 140, and the remainder will be an integer, ==d. And 7 7 consequently c=7 d+4. Substitute now this value of c, in the foregoing value of y, and y=15 d+8. Where d may be either any affirmative integer, or nothing; taking, therefore, its least value, 0, then 15 d, vanishing, y=8. Substitute now this value of y, in the aggregate numbers, it becomes 532 × 8+458= 4256+458=4714, which is the year of the Julian Period required*. But the problem may be proposed generally, as in Simpson's Algebra, Edit. 4. p. 191. "Supposing e, f, and g, to denote given integers; to find such a value of x, as that the quantities be all integers. 66 x-e x-f, and X-8, may 15 By making =y, we have x=28 y+e; X-e being substituted in our second expression, it becomes which as well as y, is to be a whole number: but by making b-e-f, will be = y + which value 28y+e-f, 19 28 y+e-f, 19 18 y +2 b, being both divisible by 19, their difference, y—2 b, must also be divisible by the same number; whence it is evident that one value of y, is 2 b; and that 2 b+19 z (supposing za whole number) will be a general value of y; and consequently, that x (=28 y+e) =532 z+56 b+e, is a general value of x, answering the two first conditions. * This easy and simple method of solving the problem, was originally given in my Analysis Equationum, 1784, p. 99. "Let this, therefore, be substituted in the remaining expres532 z+56b+e−g_ sion, 29; which, by that means, becomes 15 35 z+3b+ 72+ß 15 15 (supposing ẞ=11b+e-g=12e −11 ƒ—g.) Here 15 z and 14 z+2 ẞ being both divisible by 15, their difference, 2- -2 ß, must likewise be divisible by the same number; and therefore one value of z will be 2 ẞ, and the general value of z=2ẞ+15 w: from whence the general value of x (=532 z +56 b+e) is given=7980 w+1064 ẞ+56 b+e; which, by restoring the values of b and ẞ, becomes 7980 w +12825 e-11760 ƒ- 1064 g. "Now to have all the terms affirmative, and their co-efficients the least possible, let w be taken = −e+2 ƒ+g; whence there results, 4845 e +4200 ƒ + 6916 g, for a new value of x: from which, by expounding e, f, and g, by their given values, and dividing the whole by 7980; the least value of x, which is the remainder of the division, will be known." So far Simpson. From this ingenious and subtile analytical solution it appears, 1. That e, the given number of the Solar Cycle, is to be multiplied by 4845 (=19 × 15 × 17) or by the least multiple of the Lunar and Indiction Cycles, which, divided by the Solar 28, will leave 1 remainder. 2. That f, the given number of the lunar cycle, is to be multiplied by 4200 (= 28 × 15 × 10) or by the least multiple of the solar and indiction cycles, which, divided by the lunar 19, will leave 1. 3. That g, the given number of the indiction, is to be multiplied by 6916 (28 × 19 × 13) or by the least multiple of the solar and lunar cycles, which, divided by the cycle of indiction 15, will leave 1. 4. That the aggregate of these three terms is to be divided by the Julian period 7980, and the remainder will be the year required. Hence is derived Beverege's Arithmetical Rule, p. 192, 380. I. To find the year of the Julian Period corresponding to certain given years of the Cycles of the Sun, Moon, and Indiction. Multiply the given year of the cycle of the sun by 4845; of the moon by 4200, and of indiction by 6916; and divide the sum of the products by 7980: the remainder will be the year of the Julian Period required. Thus, if we repeat the foregoing example of the given years of the three cycles, A.D. 1. II. To find the respective years of the Cycles of the Sun, Moon, and Indiction, corresponding to a given year of the Julian Period. Divide the given year by the numbers 28, 19, 15, successively; the quotients will shew the number of revolutions of each cycle till that time, and the remainders the respective years of each current cycle. If there be no remainders, then the divisors themselves will be the last years of the cycles. Thus, if the same year 4714, be divided by 28, 19, 15, sucsessively, the quotients 168, 248, 314, will express the number of revolutions of each cycle, from the beginning of the period till that time; and the remainders, 10, 2, 4, the years of the current cycles respectively. Hence it appears, that the Julian Period began B.C. 4714, and will end A.D. 3266. It does not therefore precede the creation, and include the commencement of historical time, as was imagined by its inventor, and the followers of the shorter Jewish system of chronology. As an historical period, therefore, it is considerably inferior to the Vulgar Christian Era, which running infinitely, backwards and forwards, from a well-known fixed point, is immediately commensurate to the whole of duration, past, present, and future. ἡμεραι EPACTS. Epacts, or nμɛpaι ɛπαктαι, are “additional days,” requisite to find out the moon's age. Since the lunar year of 354 days is deficient from the solar of 365 days, by 11 days, this deficiency will run through every year of the lunar cycle. Thus the epact of the first year of the cycle is 11, because 11 days are to be added to the lunar, in order to complete the solar year; the epact of the second is 22; the epact of the third 33-30=3, because the moon's age cannot exceed 30 days; the epact of the fourth, 14; and so on till the last year of the cycle, whose epact is 29; and the epact of the first year of the next cycle, 11, as before. The following Rules will shew the use of Epacts: RULE I. To find the year of the Lunar Cycle, or the Golden Number, in any given year of our Lord. Add 1 to the given year, then divide the sum by 19, the remainder, if any, is the golden number; if there be no remainder, then 19 is the golden number. Thus 1808 divided by 19 leaves 3, which is the golden number of the year 1807. The reason of the addition of unit, is, because the Vulgar Christian Era began in the second year of the lunar cycle, as shewn before. RULE II. To find the Epact in any given year. If the year precede the alteration of the style, A.D. 1752 ; First find the golden number of that year; multiply it by 11; if the product be less than 30, it will be the epact, but if greater, divide it by 30, and the remainder will be the epact. But if the year follow A.D. 1752, because eleven days were then struck out of the calendar, the epact so found will require correction. If it be greater than 11, subtract 11 from it; if less, add to it 30, and subtract 11 from the sum: the remainder, in either case, will give the epact. Thus the golden number of the year 1807, namely 3, multiplied by 11, and the product divided by 30, left 3 for the epact; this again multiplied by 11 gives 33, from which subtracting 11, the remainder 22 gives the corrected epact. RULE III. To find the Moon's Age on any given day of the year. Add together the epact of the given year, the number of months from March inclusive, and the proposed day of the month; if the sum be less than 30, it will be the moon's age, but if greater, its remainder, when divided by 30, will be the moon's age. Thus, if it be required to find the moon's age on November 15, |