Home / Math / Puzzles / A Problem (and Solution) from Bhaskaracharya’s Lilavati
Bhaskara - Bhaskar II - Bhaskaracharya
Math Puzzles

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 already had in first half of solution.
Now as we have:
$$ \dfrac {n} {2} = 14 $$
Thus the number is 28 .

Also read:  Happy Holi! : The Village Tour

About the author

Gaurav Tiwari

Excellent Reader, Good Orator and Bad Writer. Talk about WordPress, PHP and Mathematics. I develop & design web applications that are visually brilliant and mass-friendly. I can also design stunning graphics to support my web-apps & amazing logos to amaze people. I have worked with several companies & individuals to deliver satisfactory results.

14 Comments

Click here to post a comment

Your email address will not be published. Required fields are marked *

Get your content published!

Subscribe

Blog & Earn