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**.

Super Q and A

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

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

what a problem it is?

Thanks for the Answer

fantastic try to give more examples please from indian ancient maths

not good

Hmmm.

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