DATA
04-13-2002, 07:03 PM
LINEAR congruence
hi,
To solve 25 x(congruent)15 (mod 120)
d=(25,120)=5, d|15 the congruence has exactly 5 solutions modulo 120.
We know if (a,m)=d and d|b,the linear congruence
ax(congruent)b (mod m) has exactly d solution modulo m,given by
t,t+(m/d),t+2(m/d),...t+(d-1)m/d,where t is the solution ,unique modulo m/d,of the linear congruence
(a/d)x(congruent)b/d(mod(m/d))
Solution y =t +(km)/d for some k.
To solve 25 x(congruent)15 (mod 120)
it has 5 solutions modulo 120
To find them we divide by 5 and solve the congruence
5x(congruent)3(mod 24)
we use y=t +(km)/d
for(k=0,1,2,3,4) m/d is calculated
i.e y=t+(k*24) for k=0,1,2,3,4
How ever how is t actually computed and the 5 solutions actually found?
regards Data.
hi,
To solve 25 x(congruent)15 (mod 120)
d=(25,120)=5, d|15 the congruence has exactly 5 solutions modulo 120.
We know if (a,m)=d and d|b,the linear congruence
ax(congruent)b (mod m) has exactly d solution modulo m,given by
t,t+(m/d),t+2(m/d),...t+(d-1)m/d,where t is the solution ,unique modulo m/d,of the linear congruence
(a/d)x(congruent)b/d(mod(m/d))
Solution y =t +(km)/d for some k.
To solve 25 x(congruent)15 (mod 120)
it has 5 solutions modulo 120
To find them we divide by 5 and solve the congruence
5x(congruent)3(mod 24)
we use y=t +(km)/d
for(k=0,1,2,3,4) m/d is calculated
i.e y=t+(k*24) for k=0,1,2,3,4
How ever how is t actually computed and the 5 solutions actually found?
regards Data.