A Problem (and Solution) from Bhaskaracharya's Lilavati

I was reading a book on ancient mathematics problems from Indian mathematicians. Here I wish to share one problem from Bhaskaracharya‘s famous creation Lilavati.

Problem

A beautiful maiden , with beaming eyes, asks of which is the number that multiplied by 3 , then increased by three-fourths of the product, divided by 7, diminished by one-third of the quotient, multiplied by itself, diminished by 52, the square root found, addition of 8, division by 10 gives the number 2 ?

Ahh.. Isn’t it very long sentenced problem? The solution is here:
The method of working out this problem is to reverse the whole process — Multiplying 2 by 10 (20), deducting 8 (12), squaring (144), adding 52 (196), ‘multiplied by itself’ means that 196 was found by multiplying 14 to itself.
Now, Let the number be $ n $ .

Then applying initial part of the problem on it. $ \dfrac {3n+3n \times \dfrac{3} {4} } {7} – \dfrac {1} {3} \times \dfrac {3n+3n \times \dfrac{3} {4} } {7} = 14 $

14 is what we got in first half of solution.
Now we get:
$ \dfrac {n} {2} = 14 $
Thus the number is
28 .

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9 Comments on A Problem (and Solution) from Bhaskaracharya's Lilavati

  • Vidhya Cgr -

    i was searching for the answer and i got it here. Thank you:)

    Reply
  • Loki gubbi -

    Waw, this is an interesting problem with beautiful soln. What a knok from bhaskara! I realy hats of u . What a great indian! Thanks

    Reply
  • prateek -

    what a problem it is?

    Reply
  • Srinath kesav -

    Thanks for the Answer

    Reply
  • ajay gole -

    fantastic try to give more examples please from indian ancient maths

    Reply

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