FOR ANY COMPOSITE MODULUS, REAL OR COMPLEX. 243 
great; and consequently the saving of labour effected by the use of tables of indices 
for composite moduli proportionately so. 
As an example of the use of tables of indices for composite moduli let us use the 
I table given in the example to Proposition (24). 
Solve 
= 53 (mod If 2). 
53 = (4. 1. 1.) 
(written in this order the indices are referred to moduli 6. 4. 2), 
and, therefore, 
If, then, 
and, therefore, 
therefore 
Solve 
therefore 
therefore, if 
9 = (2. 2. 0), 
x" = (2. 3. 1). 
X = (rt. h. c) 
= (5«. 55. be), 
5a = 2 (mod 6), and, tlierefore, a = 4, 
55 = 3 (mod 4), and, therefore, 5 = 3, 
be = \ (mod 2), and, therefore, c = 1, 
X = (4. 3. 1), 
x= 109 (mod 112). 
47x® = 55 (mod 112). 
55 = (3. 0. 1), 
47 = (5. 2. 1), 
= (4. 2. 0), 
X = (a. 5. c) 
2a = 4 (mod 6), 25 = 2 (mod 4), 2c = 0 (mod 2), 
a = 2 (mod 3), 5 = 1 (mod 2), c = 0 or 1 
a = 2 or 5. 5 = 1 or 3. 
Hence there are eight solutions 
(2 1 0 ) = 
(■2 1 1 ) = 
(2 3 0) S 
• (2 3 1) = 
51 
(5 
1 
0) = 19 
37 
(5 
1 
V= 5 
107 
(5 
3 
0) = 75 
93 
(5 
3 
1) = 61, 
2 l 2 
