EXAMPLES Eleven examples on the pulveriser (kuttākāra): 13. The signs, etc., up to the thirds of the Sun's (mean) longitude have all been carried away by the strong wind; the residue of thirds is known to me to be 101. Tell (me) the Sun's (mean) longitude and also the ahargana.¹ 14. The minutes together with the signs and degrees of the Moon's (mean) longitude have been destroyed being rubbed out by the hands of a child; twenty-five seconds are seen to remain (undestroyed). Calculate from them, O you of noble descent, the ahargana and the (mean) longitude of the Moon. The following is the solution of this example: Revolution-number of the Moon reduced to minutes 68167872 civil days in a yuga 86225 By the usual process³, we get x = 70091, y = 55412633'. Multiplying 25 by 86225 and dividing the product by 60, the quotient is 35927. This is the residue of the minutes (Kalašeṣa).² We have, therefore, to solve the pulveriser 68167872 x 35927 86225 = y, where x denotes the required ahargana and y the Moon's mean longitude in terms of minutes. Hence the required ahargana longitude 4 signs 23° 53′ 25". 219 = H 70091 days, and the Moon's mean 15. The signs and degrees (of the mean longitude of Mars) have been carried away by the hurricane; the residue of the ¹ Answer: Ahargana -106141; Sun's mean longitude 52'23"11". For complete solution see supra, pp. 35-36. 2 Vide supra, chapter I, stanza 46(ii), p. 33. 8 Vide supra, chapter I, stanzas 42-44, p. 30. 3 signs 32°
