INDETERMINATE PROBLEMS. 
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Then, making b — 651, the whole process will stand 
as follows: 
From ,, ,, 
... 12354 
sub. 
987# 4- b 
1. rem 
. . .. 2484’ — b 
l. rem. x 3 
7444 — 3b 
2. rein. 
243x + 4b 
3. rem . .. . 
54 — 5 b 
3. rem. x 48 .. .. 
..... 2404 — 240?» 
4. rem 
34 ~h 244') 
5. rem 
6. rem. 
x -f 493 ô: 
where, x being without a co-efficient, let 4936 or its 
equal 320943 be now divided by 1235, the common 
measure to all those quantities, and the remainder will 
be found 1078 ; therefore x +■ 1078 is likewise divi 
sible.by 1235 ; and consequently the least value of x 
{— 1235 — 1078) — 157. The manner of working, 
according to this method, may be a little varied; it 
being to the same eifect, whether the last remainder, or 
a multiple of it, be subtracted from the preceding one, 
or the preceding one, from some greater multiple of the 
last. Thus, in the example before us, the quantity 
2434— b, in the third line, might have been multi 
plied by 4, and the preceding one subtracted from the 
product; which would have given 5x — 5b (as in the 
sixth line) by one step less. If the manner of proceed 
ing in these two examples be compared with the process 
for finding the same values, according to the lemma, the 
grounds of this will appear obvious. 
PROBLEM X. 
Supposing e,f, and g to denote given integers, to deter* 
x—e x—f 
mine the value of x, such that the quantities 
■ may all of them be integers, 
2S 
19 * 
and