Find the least positive four-digit solution to the following system of congruences.
7x = 21 (mod 14)
2x + 18 = 16 (mod 9)
Find the least positive four-digit solution to the following system of congruences.
7x = 21 (mod 14)
2x + 18 = 16 (mod 9)
7x=21 mod14
x=3 mod14
x=14p+3
2x+18=16 mod9
2x=-2 mod9
x=-1 mod9
x=9q-1
9q-1 = 14p+3
9q - 14p = 4
-------------------------
9=-1*(-14) - 5
-14= 3(-5)+1
-14 - 3(-5) = 1
-14 - 3 [9+1*(-14)] = 1
(-14) - 3*9 -3(-14) = 1
-2(-14) -3(9) = 1
9(-3) -14(-2)=1
9(-12) -14(-8)=4
9(-12+14g) - 14(-8+9g) = 4
so
q =-12+14g and p =-8+9g
x = 9q-1 and x= 14p+3
x= 9(-12+14g)-1 and x= 14(-8+9g)+3
x= -109 +9*14g and x = -109 + 14*9g they are the same so it is right.
x= -109 +126g
I want the smallest 4 digit value of x
-109+126g>1000
126g > 1109
g > 8.8
g=9
x = -109+126*9
x= 1025