4 e — * 
: # 
Interesting Properties of Numbers. — 115 
If N=16, and P=13, the quotients will be 1, 3, 11; the re- 
mainders will be 3, 9, 1. . 
i Nx=70=2.5.7, andP= 32= =2°5, the quotients will be 2, 3, 
8, 52, 16; the remainders will be 6, 4, ‘24, 16,0. In this case the 
process Sininaten: 
If N=13, and P=11, the Gucecry will be§ 1 
2, 4, 8, 5, 10 
The remainders will be $9 9. 7, : 6. 1 
Now exchanging the values of N see P, that is, taking N=11, 
2,4, 9,5 
11, 10, 8, 3,7 
and P=13, we get the quotients, $ fa , o . ™ % the remain- 
| Bee Oe Megs sii | 
411,46, 3, 7742 
der ; 2.9.8 10,6, 1 
If N=509, and P=19, we find Tp = ynls=P- 1, therefore 
2 
the number of terms in the periods will be P—1. And since N 
and P are both primes, the one of the form 4n+1, and the other 
of the form 4n+3, it follows that if N=19, and P=509, the 
number of terms in the periods will be P—1=508. 
When N =10, our process resolves itself into the usual rule for 
converting the vulgar fraction 1 into its equivalent decimal. 
If rect N being supposed 10, we find the quotients to be 
1, - 
$8 ei ” the remainders are sa . = Hence }=0.142857 re- 
peated in endless succession. Now it is obvious that the same 
succession of figures must represent in decimals the value of any 
vulgar fraction whose denominator is 7 and numerator less than 
7; it is also evident that the period will commence with that quo- 
tient which follows the remainder which is equal to the numerator 
= the fraction ; ans’ 2 = (0.285714; 3=0.428571 ; 4=0.571428; 
=0.714285 ; ~ 0.857142. 0 35,2 
“r Pai 7, a quotients will be $9, rs i 7, 6, 4. ‘a and the 
‘ 16 
remainders will be ae 2 2 2 i 2 is t Therefore, 
77 = 0.0588235294117647 ; 3,=0. 1176470588235294 ; 
v7 =0.1764705882352941 ; 1, =0.2352941176470588 ; 
thus we could with the same period of figures represent in - 
mals, the fractions 5 Fn) 17) Ty SC. 
‘ a 
