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**.

### Gaurav Tiwari

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**Tagged:** India, Lilavati, List of Indian mathematicians, Mathematical problem, Multiplication, Problem, Square root, Vedic Math

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

## Gajananda Swami -

Super Q and A

## Vidhya Cgr -

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

## 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

## prateek -

what a problem it is?

## Srinath kesav -

Thanks for the Answer

## ajay gole -

fantastic try to give more examples please from indian ancient maths

## sanjya -

not good

## Gaurav Tiwari -

Hmmm.

## Gaurav Tiwari -

I'll be providing more such content very soon. Better you try the Archives http://gauravtiwari.org/a/ for related posts.