ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

PRELIMINARY OPERATIONS 227
number below (i.e. the penultimate) is multiplied by
the one just above it and then added by that just
below it. Divide the last number (obtained
by doing so repeatedly) by the divisor corres-
ponding to the smaller remainder; then multiply the
residue ty the divisor corresponding to the greater
remainder and add the greater remainder. (The result
will be) the number corresponding to the two divisors.
Āryabhaṭa’s problem may be enunciated thus :
To find a number (N) which being divided by two
given numbers (a, b) will leave two given remainders
(R₁, R₂).
This gives :
N=ax+R₁=by+R₂
(where R₁ is a greater remainder and R₂ lesser remainder, and a
is the divisor corresponding to greater remainder and b the
divisor corresponding to the lesser remainder.)
Denoting as before by c the difference between R₁, and
R₂, we get
(i) by=ax+c, if R₁>R₂
(ii) ax=by+c, if R₂>R₁
the equation being so written as to keep c always positive.
Hence the problem now reduces to making either
(ax+c)/b or (by+c)/a
according as R₁>R₂ or R₂>R₁, a positive integer. So Āryabhaṭa
says : Divide the divisor corresponding to the greater remainder
etc.”
Now we shall proceed with the details of the operation as
proposed by Datta and Singh in his History of Hindu Mathema-
tics, Part II. Algebra :
Suppose R₁>R₂; then the equation to be solved will be
ax+c=by ...(i)
a, b being prime to each other.