Equation Of Arjunas Arrows
Equation Of Arjunas Arrows
Arjuna’s Arrows and Algebra (Beejganit)!
Bhāskarāchārya (1114 -1185 CE), has been called the greatest mathematician of medieval India. And is considered the progenitor of Differential Calculus. He was 500 years before Newton and Leibniz!
Bhāskarāchārya wrote at least four mathematical treatises in Sanskrit. One of them, titled Leelavati, contains many algebra-related teasers, which have become the subject of significant research by scholars. These teasers are in the form of shlokas which pose the problems. The shlokas need to be interpreted correctly to decipher the meaning in order to find the solution.
In the 70th Shloka of Leelavati, we see equations related to Arjuna’s Arrow.
पार्थ: कर्णवधाय मार्गणगणं क्रुद्धो रणे संदधे
तस्यार्धेन निवार्य तच्छरगणं मूलैश्चतुभिर्हयान् |
शल्यं षड्भिरथेषुभिस्त्रिभिरपि च्छत्रं ध्वजं कार्मुकम्
चिच्छेदास्य शिरः शरेण कति ते यानर्जुनः संदधे || 70 ||
The direct meaning of this shloka is a question formulated as follows:
During the battle between Arjuna and Karna in the Mahabharata, Arjuna released some arrows, out of the released arrows:
- Half were consumed in stopping the arrows coming from Karna.
- 4 times the square root of the arrows were consumed to control the horses of Karna’s chariot
- Six were for gaining control over Shalya, the charioteer of Karna. (Shalya was the maternal uncle of Nakula and Sahadeva)
- Three were used to take on the umbrella and flag of the chariot and the bow of Karna.
- Finally, Karna was killed by a single arrow. So how many arrows were released by Arjuna in the battle?
If we formulate this equation correctly, then the basic algebra easily yields the answer to this question,
X = X/2 + 4√X + 6 + 3 + 1
Solving the above, we get the value of X=100 for the total number of arrows shot by Arjuna.