IX DETER MIN ATE PROBLEM S. 
183 
sought; then, since this number, when 13 is subtracted 
from it, is divisible by 26, it is manifest that 
^ 1 Till- or —must be a whole number: 
26 26 
whence, by proceeding according to the lemma, x will 
be found = 8; and consequently 17# + 7 — 143, the 
number required. See the operation 
17)26(1 
17 
6 
18 
26 
9 
8 )_9 ( 1 
1 
PROBLEM II. 
Supposing Qx 4- 13// — 2000, it is required to find all 
the possible values of x and y in whole positive numbers. 
By transposing 13//, and dividing the whole equation 
2000— 13y 
9 
by 9, we have x — 
which, as x is a whole positive number,by the question, 
must also be a whole positive number, and so likewise 
2 
; from which the least value of y, in Avhole 
9 
numbers, will come out 5 ; and consequently the 
corresponding value of x — 215. From whence the 
rest of the answers, which are 16 in number, will be 
found, by adding 9, continually to the last value of //, 
and subtracting 13 from that of x, as in the annexed 
table, which exhibits all the possible answers in whole 
numbers. 
x =21512021189117611631150] 15711241111198185] 72! 591 46| SSI 201 7 
b| 14| 2ô\ 52) 4l| 501 59l 681 77(361951104J1 is| 122! 13111149 
In the same manner, the least value of y, and the 
greatest of x being found, in any other case, the rest of 
the answers will be obtained, by only adding the co 
efficient of x, in the given equation, to the last value of//, 
continually, and subtracting the co-efficientof y from the 
N 4