Question #139419
find the smallest positive solution of the system of linear congruences ? X=2 (mod3) , X=3(mod 5), X=2(mod 7)
1
Expert's answer
2020-10-21T15:57:32-0400

Since the numbers 3 5 7 are coprime, we use the Chinese remainder theorem

M0=m1m2m3=357=105{M_0} = {m_1}{m_2}{m_3}=3*5*7 =105

Mi=M0miM_i =\frac{M_0}{m_i}

M1=M0m1=1053=35M_1 =\frac{M_0}{m_1}=\frac{105}{3}=35

M2=M0m2=1055=21M_2 =\frac{M_0}{m_2}=\frac{105}{5}=21

M3=M0m3=1057=15M_3 =\frac{M_0}{m_3}=\frac{105}{7}=15

Miyi=ai(mod mi)M_iy_i =a_i(mod\ m_i)


35y1=2(mod 3)35y_1 =2(mod\ 3)

33y1+2y1=2(mod 3)33y_1+2y_1 =2(mod\ 3)

2y1=2(mod 3)2y_1 =2(mod\ 3)

y1=1y_1=1


21y2=3(mod 5)21y_2 =3(mod\ 5)

20y2+y2=3(mod 5)20y_2+y_2 =3(mod\ 5)

y2=3(mod 5)y_2 =3(mod\ 5)

y2=8y_2 =8


15y3=2(mod 7)15y_3=2(mod\ 7)

14y3+y3=2(mod 7)14y_3+y_3=2(mod\ 7)

y3=2(mod 7)y_3=2(mod\ 7)

y3=9y_3=9

x=(M1y1+M2y2+M3y3)(mod M0)x =( M_1y_1+M_2y_2+M_3y_3)(mod\ M_0)

x=(351+218+159)(mod 105)x =(35*1+21*8+15*9)(mod\ 105)

x=338(mod 105)x= 338(mod\ 105)

x=23x=23

Answer: x=23


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