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.

Who was Bhaskaracharya?

bhaskara ii with lilavati

Bhaskara II, who is popularly known as Bhaskaracharya, was an Indian mathematician and astronomer from the 12th century. He’s especially known for the discovery of the fundamentals of differential calculus and its application to astronomical problems and computations.

What is Lilavati?

Lilavati is his treatise on mathematics, written in 1150. Lilavati is the first volume of his main work, the Siddhānta Shiromani. Other volumes include the Bijaganita (algebra), the Grahaganita(astronomy) and the Golādhyāya(geodesy).

Lilavati contains thirteen chapters, mainly definitions, arithmetical terms, interest computation, arithmetical and geometrical progressions, plane geometry, solid geometry, the shadow of the gnomon, the Kuṭṭaka – a method to solve indeterminate equations, and combinations.

See: https://en.wikipedia.org/wiki/L%C4%ABl%C4%81vat%C4%AB

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, the addition of 8, division by 10 gives the number 2?

Ahh.. Isn’t it a very long sentence 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). Now ‘multiplied by itself’ means that 196 was found by multiplying 14 to itself.

We are done with this part, “multiplied by itself, diminished by 52, the square root found, the addition of 8, division by 10 gives the number 2”. Let’s try to solve the rest of the question.

Let’s suppose that the number (the end result) is $n$.

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, on simplification, as we have:

$$ \dfrac {n} {2} = 14 $$

Thus the number is 28.

You will eventually come with some problems that need the reverse process is the followed. Lilavati is an exceptional book and it contains even more awesome problems. If you wish to have a hand on Lilavati, you can buy one from Amazon here (for Indian readers) or from here (for Global readers).

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