THE RESOLUTION OF 
158 
PROBLEM IX. 
Supposing 87* + 256?/ = 15410; to determine the least 
value of x, and the greatest of y, in whole positive 
numbers. 
By transposition and division, we have 
V — 
15410—87* 
256 
_ 87* — 50 . - 
CO —— : where the frac- 
2?6 
tion being the same with that in Ex amp. l. to the pre 
mised lemma, the required value of* will he given from 
thence z= 30; from thence that of y will likewise be 
known. But I shall in this place shew the manner of 
deducing these values, independent of all previous con 
siderations, by a method on which the demonstration 
of the lemma itself depends. 
In order to this, it is evident, as the quantity 87* — h 
(supposing b — 50) is divisible by 256, that its double 
174* — 26 must be likewise divisible by 256. But 
256r is plainly divisible by 256; and if from this the 
quantity in the preceding line be subtracted, the re 
mainder, 82* 4- 2b will be likewise divisible by the 
same number; since whatsoever number measures the 
whole, and one part of another, must do the like by the re 
maining part: for which reason, if the quantity last 
found be subtracted from the first,the remainder 5*—3b 
will also be divisible by 256: and, if this new remain 
der multiplied by 16, be subtracted from the preceding 
one (in order to farther diminish the co-efficient of *), 
the difference 2* 4 50b must be still divisible by the 
same number. In like manner, the double of the last 
line, or remainder, being subtracted from the preceding 
one, we have *— 1036, a quantity, still, divisible by 
256 : but 
1036 
256 
, 30 
= 20 xi 
256 
therefore * — 30 must 
be divisible by 256; and consequently * be either equal 
to 30, or to 30 increased by some multiple of 256; but 
30, being the least value, is that required. 
It may not be amiss to add here another Example, to 
illustrate the way of proceeding by this last method: 
wherein let us suppose the quantity given tobe 
987*4051 
1235