पाटीगणित (श्रीधराचार्य - प्राचीन संस्कृत टीका एवं प्रो. कृपाशंकर शुक्ल सविस्तार व्याख्या)

62 DETERMINATIONS
The second part of the rule relates to the solution of the simple
equation of the type
x − (a/b) x = d′,
where a/b is the fraction and d′ the visible quantity. The solution is
correctly stated as
x = d′ / (1 − a/b).
Ex. 99. A number is diminished by its square root,
what remains is diminished by its one-sixth, what remains
after that is diminished by its square root, what remains after
that is diminished by its one-fifth, and what remains after
that is diminished by twice the square root of itself; the
residue now left is 8. (Find the number).¹
This problem reduces to the solution of the following equation :
x − √x − ⅙ (x − √x) − √[x − √x − ⅙ (x − √x)] − ⅕ {x − √x − ⅙ (x − √x)
− √[x − √x − ⅙ (x − √x)]} − 2 [x − √x − ⅙ (x − √x) − √[x − √x − ⅙ (x − √x)]
− ⅕ {x − √x − ⅙ (x − √x) − √[x − √x − ⅙ (x − √x)]}]^½ = 8,
which can be written as
x − √x = y, (1)
where y − ⅙ y = z, (2)
z − √z = u, (3)
u − ⅕ u = v, (4)
v − 2 √v = 8. (5)
Solving these equations in the inverse order, we get x = 36.
Hindu Method of Solution. Writing down the square roots and frac-
tions, which have been successively subtracted from the original number,
in order and then the visible quantity, we get
Square root, 1/6 part of the remainder, square root of the remainder, 1/5 part of the
remainder, 2 times the square root of the remainder, visible quantity 8.
──────────────────────────────────────────────────────────
¹ For other examples see, GSS, iv. 41, 42-45, 46; GT, p. 49, lines
7-10; p. 50, lines 2-5.