(421) 
A remainder has been found greater than the smallest factof, 
namely 249 = 231 +18. So we can write 231 X 243 + 18, which 
causes the remainders to have smaller values. 
After a total of 82 operations all prime numbers (/ Gy have 
disappeared and G, proves to be indivisible. Without the pointing 
out of the non-divisors more than 260 operations would have been 
necessary. 
The factor G, requires about three times the number of operations 
as Go. 
As the calculation is performed with relatively small numbers, the 
method of remainders is to be most recommended for the investigation 
of great numbers. 
VI. Testing of divisors. 
Example G; — 898423 — 948% — 281. 
To find out whether 7 is a divisor the remainders must be 
determined that are left by 948 and 281 after division by 7. If the 
difference of the (remainder)? of 948 and the remainder of 281 is a 
sevenfold then 7 is a factor. 
So we must write: 
divisor quotient remainder (remainder)* remainder of 281 
7 135 3 2 1 
11 86 2 4 6 
13 73 —1 1 8 etc. 
To obtain smaller numbers the remainders can be taken negatively 
without any mconvenience. 
It is at once evident, that the divisor 73 will give a quotient 13 
with a remainder —1; likewise that the divisor 43 (half of 86) will 
give a quotient 2 X 11 with remainder 2; etc., so that often the 
result can be written down for two divisors at once. 
For larger numbers the operation would be: 
divisor quotient remainder (remainder) remainder of 281 
509} 439 192721 or 319 |} 
1 | 281 
‘neko 437 190969 or 366 J 
VIL. Calculation of the number of operations. 
If the operation in § V for the number 56151 is performed as 
