520 PHILOSOPHICAL TRAN3ACTIOKS. [aNNO I768. 



what number is there, between 1800 and IQOO, which divided by 28 leaves 

 1, 7, 18, or 24; and being divided by 19 leaves 14 ? 



Solution. Here then in the general theorem -r- X (d — e) -{• d, is a == 28, 

 b= IQ,D = \, 7, 18, or 24; E = 14, 1 = 1. To find* 19)28( 1 

 r, the quotients are * 1, 2, 9; drop the first and last, be- 9)19(2 



cause their number is odd; then the series required will be 1)9(9 



0, 1 , 2 ; therefore r = 2. 



Note, if to any year of Christ be added 9, and the sum divided by 28 ; the re- 

 mainder, or 28, if O remain, will be the cycle of the sun for that year; and if 1 

 be added to any year of Christ, and the sum divided by 19; the remainder, or 

 19, if O remains, is the cycle of the moon. Hence, if any year of Christ be 

 severally divided by 28 and 1 9, and the remainders be d and e respectively ; then 

 rf + 9, or d-\- g — 28 = d; and e -)- 1, or e -f 1 — J 9 = e. In the present 

 case, taking d = 1, d -\- g = 1 cannot be; because </ would be negative: but d 

 must not only be affirmative, but also greater than e; therefore make d -\- g — 

 28 = D = 1 ; therefore d= 1 + 28 — 9 = 25 ; and e -f 1 = e = 14, there- 

 fore e = 13. Here d and e are the cyclar numbers, and d and e are the anno 

 domini numbers suited to the theorem. 



Here then («r X d — e) + d=(28 X 2 x 7) + 20 = 412; and412 -|- 532 

 4- 532 + 532 = 2008. The first answer therefore, a. d. 412, is too little, and 

 when increased by 3 dionysian periods, or multiples of 28 and 1 9, is too great, 

 going beyond the century required. So, when this solar cycle is 1, it will not do. 



Let D = 7, the rest as before. Then d -{■ g — 28 = 7 ; therefore d= 26. 

 Here then {ra X d — e) + d= {28 X 2 X 13) -j- 26 = 754; and 754 -f- 533 

 -|- 532 = 1818. So A. D. J 818 will answer the question. 



Let D = 18, the rest as before. Then d -\- g — 28 = 11; therefore d= 37. 

 Here (ra x d — e) + d = (28 X 2 X 24) -{- 37 = 1381 ; and 1381 -f- 532 

 5= 1913. This goes beyond the century required ; so will not do. 



Let D = 24, the rest as before. Then d-\- g = 24; therefore d = 15. Here 

 (ra X ^- e) -f (/= (28 X 2 X 2) -}- 15 = 127; 127 + 532 X 4 = 2255; 

 which goes beyond the century required. 



So there is but one year in the 19th century, viz. 1818, that will have the 

 conditions required. The cycle of the sun will then be 7 ; the cycle of the moon 

 14; and the Sunday letter d ; and Easter-day the 22d of March. 



XVI. A Determination of the Solar Parallax attempted by a peculiar Method, 

 from the Observations of the last Transit of Venus. By Andretv Planman, of 

 the Acad, of Sciences at Stockholm, &c. From the Latin, p. I07. 

 § I. From the centre of the earth x (fig. 7, pi. 15), conceive lines to be 



