RESOLUTION OF INDETERMINATE PROBLEMS. ! SI 
Then multiply the last number thus found by the abso 
lute quantity fbJ in the numerator of the given fraction, 
and divide the product by the denominator; so shall the 
remainder be the true value of x, required; provided the 
number of terms in the upper line be even, and the sign o f 
b negative, or, if that number be odd and the sign o f b af 
firmative; but, if the number of terms be even, and the 
sign of b affirmative, or vice versa, then the difference 
between the said remainder and the denominator of the 
fraction ivill be the true answer.. 
In the general method here laid down a is supposed 
less than c, and that these two numbers are prime to 
each other: for, were they to admit of a common mea 
sure, whereby b is not divisible, the thing would be im 
possible, that is, no integer could be assigned for x t 
so as to give the value of ax an integer: the rea 
son of which, as w r ell as of the lemma itself, will be 
explained a little farther on : here it will be proper to 
put down an example or tw r o, to illustrate the use of 
w hat has been already delivered. 
Examp. 1. Let the given quantity be 
Then the operation will stand as follows : 
87 )j256 ( 2 
82 ) 87_( 1 
5 ) ¡82 ( 16 
2)5(2 
1 
2, 1, 16, 2 
1, 2, 3, 50, 103 
50 
256 ) 5150 ( 20 
~ 30 = X: 
Examp. 2. 
Given 
7\x -f 10 
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89