38 PLANETARY PULVERISER Wednesday by one day, the residue of the week-cycle is unity. In other words, 4-1 will be whole numbers. If we assume A to be a multiple of 210389, we have simply to determine A such that A-1 may be completely divisible by 7. Let A 210389X. Then we have to solve the pulveriser 210389X - 1 =Y, 7 or (vide stanza 47) 576A 210389 and or (vide stanza 47) 4X-1 = 7 This is what the rule prescribes. Y', where Y= 30055 X+Y'. Evidently, a solution of (2) is X = 2, Y' = 1. The corresponding solution of (1) is X= 2, Y 30055×2 + 1 = 60111. The required ahargana is therefore 1000 + A, i.e., 1000+ 210389X i.e., 1000+ 210389x2 or 421778. 2, (ii) To find the ahargana for Friday. In this case, the residue of the week-cycle is 2. Let the required ahargana be 1000+ 210389X. Then we have to solve the pulveriser 210389X2 = Y, 7 4X-2 (1) (2) = Y', (3) (4) where Y30055X+Y'. Evidently, a solution of (4) is X= 4, Y' 2. The corresponding solution of (3) is X= 4, Y- 30055X+2= 120222. The required ahargana is therefore 842556. (iii) To find the ahargana for Wednesday. As before, let the ahargana be 1000+210389X. In this case, the residue of the week-cycle is 0 and we evidently have X-7, so that the required ahargana is 1473723. A rule for the solution of the so called vela-kuttakara (time- pulveriser): 49. First make the abraded dividend and the (new)
