INDETERMINATE PROBLEMS. 
185 
2It I- 27y — 2000; and consequently x — 
2000 — 27.V 
= 95— y 
where, the fraction being in 
its least terms, and the numbers 6 and 21, at the same 
time, admitting of a common measure, a solution in 
whole numbers [by the note to the preceding lemma) is im 
possible. The reason of which depends on these two 
considerations; that, whatsoever number is divisible by 
a given number, must be divisible aiso by all the divi 
sors of it; and that any quantity which exactly mea* 
sures the whole and one part of another, must do the 
like by the remaining part. Thus, in the present case, 
the quantity Gy— 5, to have the result a whole num 
ber, ought to be divisible by 21, and therefore divisible by 
3, likewise (which is, here, a common measure of a and c): 
but Gy, the former part of 6y—5, is divisible by 3, there 
fore the latter part — 5 ought also to be divisible by 3 ; 
which is not the case, and shews the thing proposed to 
be impossible. 
PROBLEM V. 
A butcher bought a certain number o f sheep and oxen, 
for ichich he paid lOOl. \for the sheep he paid 17 shillings 
apiece,and for the oxen,one with another,he paid 7 pounds 
apiece, it is required to find how many he had of each sort. 
Let x be the number of sheep, and y that of the 
oxen; then, the conditions of the question being ex 
pressed in algebraic terms, vre shall have this equation 
viz. 17t 4- I40y — 2000; and consequently x — 
2000 — 140y 
17 
8y — —"i^- 1 - 1 .; which being 
a whole number, — — must therefore be a whole 
17 
riumber likewise: whence, by proceeding as above, we 
lind y ~ 7, and x — 60; and this is the only answer 
the question will admit of; for the greatest value of x 
cannot in this case be divided by the co-efficient of y t 
that is 140 cannot be had in 60; and therefore, ac