INDETERMINATE PROBLEMS» 
PROBLEM XII. 
if Ì7# + 19}/ 4- 21 z rr 400; it is proposed to find 
all the possible values of x, y> and z, in whole positive 
numbers. 
When the co-efficients of the indeterminate quanti 
ties x, a, and z, are nearly equal, as in this equation, it 
will be convenient to substitute for the sum of those 
quantities. Thus, let x -f y -f z be put r= m; then 
by subtracting 17 times this last equation from the pre 
ceding one, we shall have 2y + 4z — 400 — 17m; and 
by subtracting the given equation from 21 times the 
assumed one x + y + z — m, there will remain 
4r -h 2y — 21 m — 400. Therefore, since y and 2 can 
have no values less than unity, it is plain, from the first 
of these two equations, that 400 — 17m cannot be less 
than 6, and therefore m not greater than , or 
23 : also, because by the second of the two last equa 
tions, 21m — 400 cannot be less than 6, it is obvious 
that m cannot be less than , or 19 : therefore 
19 and 23 are the limits of m, in this case. These be 
ing determined, let 4;r be transposed in the last equation 
and the whole be divided by 2, and we shall have 
y — lOm— 200— itx + — : which being a whole 
number, by the question, must likewise be a whole 
number, and consequently m an even number ; which, 
as the limits of m are 19 and 23, can only be 20, or 22 : 
let, therefore, m be first taken — 20, then y will be 
come 10 — 2x and z [m — x — y) — 10 + x ; 
wherein x being taken equal to 1, 2, 3, and 4, suc 
cessively, we shall have y equal to 8, 6, 4, 2, and z 
equal to 11, 12, 13, 14, respectively, which are four 
of the answers required. Again, let m be taken — 22 ; 
then will y — 31 — 2x, and 2 — x — 9 ; wherein, let 
x be interpreted by 10, 11, 12, 13, 14, and 15, succes 
sively, whence ?/ will come out 11, 9, 7, 5, 3, and 1 ; 
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