220 degrees is 73. Say what is the (mean) longitude of Mars and also what is the ahargana. The following is the solution of this example: Revolution-number of Mars reduced to degrees civil days in a yuga EXAMPLES The pulveriser to be solved is therefore 13780944x73 26298625 = y, where x denotes the required ahargana and y the mean longitude of Mars in terms of degrees. Solving the pulveriser by the usual process, we get x= 17420617 y = 9128711. Hence the required ahargana-17420617 days, and the mean longitude of Mars 9128711°, i.e., 25357 revs., 6 signs, 11 degrees. 16. The (mean) longitude of Mercury is 3 signs, 15 degrees, and 5 minutes. Considering this give out the days elapsed (i.e., the ahargana) and also the revolutions performed by him. The following is the solution of this example: Longitude of Mercury- 3 signs 15° 5' = 6305'. Now 13780944 26298625* revolution-number of Mercury civil days in a yuga 896851 78895875* Therefore multiplying 6305 by 78895875 and then dividing the pro- duct by 21600, we get 23029559 as the quotient. This is the residue of the revolutions. Thus we have to solve the pulveriser 896851x23029559 78895875 where x denotes the ahargana and y the required revolutions. = y,
