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 .

10 Comments

  1. Gajananda Swami March 4, 2012
  2. Vidhya Cgr May 24, 2012
  3. Loki gubbi June 10, 2012
  4. prateek March 28, 2013
  5. Srinath kesav June 7, 2013
  6. ajay gole December 10, 2013
  7. jayaprakash January 28, 2015