CONGRUENCE ¥

always, have to multiply the members also by this integer, in
which case the congruence will hold tru (§ 216).
For example, 5 =1 (mod 2); 5-=1(mod2 X 3)
but5 X 3=1X3(mod2 X 3)

However 8 =2 (mod 2) and 8 = 2 (mod 2 X 3)

Again, if we divide the members by any integer, in order
to be sure that the congruence holds we may, tho not always,
have to divide the modulus by the greatest common divisor
of the modulus and this integer, in which case it will hold tru
(§ 280).

For example, 6 = 0 (mod 6); 6 :2-=0:2(mod 6)

but 6:2 =0:2(mod6:(6(x2)).

However, — 6 = 6 (mod6)and — 6 :2 = 6 :2 (mod 6)

Also = —9(mod2)and3:3 = — 9 :3 (mod 2).

Here m || c.

283. Th. Ifa = b (mod m) and a = ¢ (mod n) and k and |
are two integers such that km + In = m Q n (§ 245),
then b = ¢ (mod m (X n)
and a = (kmc + Inb) : (m R n) (mod m X n);
conversely, if km 4+ In = m Xn, b =c (mod m R n),
and a = (kmc + Ind) : (m RQ n) (mod m X n),
then a = b (mod m) and a = ¢ (mod n).

For the direct theorem, that b = ¢ (mod m Q n) follows
easily from §§ 190, 185, 188.

From @ = b (mod m), multiplying the members and the
modulus by 7 : (m QR n), we get

an : (mRn) = bn: (m@n)(modmn : (mRn)) §216.
From this, multiplying the members by /, we get

aln : (mRQn) =Inb : (m R n)(modm X n)  §§ 212, 271.
Similarly, from ¢ = ¢ (mod 7), we get

akm : (m R n) = kmc : (m Q n)(mod m X n)
Adding these two results, we get
a(km 4 In) : (m R n) = (kmc + Inb) : (m @ n)(mod m X n)