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100 CONJUNCTION OF A PLANET AND A STAR [CH. VIII where (3=2, 3, 4, .... neglecting the case in which * or y is zero. 1 Putting j3=2, 3, 4, we see that *=40 and ^=24 is the least solution. Assuming now that the residues of the revolution of Saturn and Mars are 24 and 40 respectively, we have to obtain the ahargana and the revolu- tions performed by Saturn and Mars. To obtain the ahargana and the revolutions performed in the case of Saturn, we have to solve the equation 3664 1«-24 _ #i) 394479375 where u and v denote the ahargana and the revolutions performed respectively. Applying the rules given in the Maha-BKaskarlya (i. 41-45), the general solution of the above equation is found to be a = 394479375/4/346688814, and 3664U4- 32202, where t = 0, 1, 2, ... . The least solution corresponds to t = 0. To obtain the ahargena and the revolutions performed in the ease of Mars, the equation to be solved is 191402^-40 {9 . 131493125 Z and w denoting the ahargana and the revolutions performed by Mars respec- tively. The general solution of this equation is

  • =131493125*+ 118076020,

«;=191402j+-171872, 1 This solution was given by the Hindu Mathematician Narayaria (1356 A. D.). See GK, i. 47. The Hindu mathematician Brahraagupta (628 A.D.), who was a contemporary of Bhaskara I, had given the following solution ; y = A(p-y% i A (P 8 4-r a )-HP*-r 2 ) wn rc A ~ iHiF+y 2 )-iP-yw ' which reduces to Narayana's solution by taking y =*1. The commentator Udayadivakara has given a unique method for solving the above multiple equations. His method has been discussed by me in a paper entitled •"Acarya Jayadeva, the mathematician". Sec Canita, Vol. 5, No. 1, June 1954, pp. 18-19.