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.
$$\dfrac {3n+3n \times \dfrac{3} {4} } {7} – \dfrac {1} {3} \times \dfrac {3n+3n \times \dfrac{3} {4} } {7} = 14$$
Now as we have:
$$\dfrac {n} {2} = 14$$
Thus the number is 28 .

A designer by profession, a mathematician by education but a Blogger by hobby. Loves reading and writing. Just that.

1. Super Q and A

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

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

4. what a problem it is?

6. fantastic try to give more examples please from indian ancient maths

7. I want learn lelawathi ganith

8. Its a very good question but i didn’t understand how u got the answer? The first half is clear but didn’t understand the second part which is $$dfrac {3n+3n times dfrac{3} {4} } {7} – dfrac {1} {3} times dfrac {3n+3n times dfrac{3} {4} } {7} = 14$$
Now as we have:
$$dfrac {n} {2} = 14$$
Thus the number is 28 .

9. Its a very good question but i didn’t understand how u got the answer? The first half is clear but didn’t understand the second half. If u can please show it in a better way. .

10. What a nice problem?