पृष्ठम्:गणितसारसङ्ग्रहः॒रङ्गाचार्येणानूदितः॒१९१२.djvu/३१६

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GAŅITASĀRASAŃGRAHA.

the smaller group-value, so that a creeper-like chain of successive quotients may be obtained in this case also as before. Below the lowermost quotient in this chain, the optionally chosen multiplier of the last remainder in the odd position of order in this last successive division is to be put down; and below this again is to be put down the number which is obtained by adding the difference between the two group-values (already referred to) to the product (of the least remainder in the last odd position of order multiplied by the above optionally chosen multiplier thereof, and then by dividing the resulting eum by the last divisor in the

 

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The rationale of this process will be clear from the following considerations:- ${\text{We have (i) }}{\tfrac {B_{1}x+b_{1}}{A_{1}}}{\text{ is an integer; (ii)}}{\tfrac {B_{1}x+b_{2}}{A_{2}}}{\text{ is an integer; and}}$

(iii) ${\tfrac {B_{3}x+b_{3}}{A_{3}}}$ is an integer. In (i) Let the lowest value of $x=s_{1}$

In (ii)       ,,        ,,        ,,        $x=s_{2}$.

In (iii)      ,,        ,,        ,,        $x=s_{3}$.

(iv) When both (i) and (ii) are to be satisfied, $dA_{1}+s_{1}$ has to be euqal to $kA_{2}+s_{2}$, so that $s_{1}-s_{2}=kA_{2}-dA_{1}$. That is ${\tfrac {A_{1}d+(s_{1}-s_{2}}{A_{2}}}=k$.

From (iv), which is an indeterminate equation with the values o d and k unknown, we arrive, according to what has been already proved, at the lowest positive integral value of d/ This value of d multiplied by A₁, and then increased by s₁, gives the value of x which will satisfy (i) an (ii).

Let this be t₁; and let the next higher value of x which will satisfy both these equations be t₂.

(v) Now, $t_{1}+nA_{1}=t_{2}$;

(vi) and $t_{1}+mA_{2}=t_{2}$.

$\therefore {\tfrac {A_{1}}{A_{2}}}={\tfrac {m}{n}}.{\text{ Thus }}A_{1}=mp{\text{, and }}A_{2}=np{\text{ where p is the common factor between }}A_{1}{\text{ and }}A_{2}$.

$\therefore m={\tfrac {A_{1}}{p}}{\text{, and }}n={\tfrac {A_{2}}{p}}$.

Substituting in (v) and (vi), we have

$t_{1}+{\tfrac {A_{1}A_{2}}{p}}=t_{2}$.

From this it is obvious that the next higher value of x satisfying the two equations is obtained by adding the least common multiple of A₁ and A₂ to the lower value.

Now again, let v be the value of x which satisfies all the three equations.

Then $v=t_{1}+{\tfrac {A_{1}A_{?}}{p}}\times r$, (where r is a positive integer) =(say) $t_{1}+t_{2}$; and $v=s_{3}+cA_{3}=t_{1}+lr$.

$\therefore r={\tfrac {cA_{3}+s_{3}-t_{1}}{l}}$.