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The congruence axā”b(modm)ax\equiv b\pmod{m}axā”b(modm) where d=gcd(a,m)d=gcd(a,m)d=gcd(a,m) and dā£bd|bdā£b has a solution xā”a1ā1b1(modm/d)x\equiv a_{1}^{-1}b_{1}\pmod{m/d}xā”a1ā1āb1ā(modm/d)where a1=ad,b1=bda_{1}=\frac{a}{d},b_{1}=\frac{b}{d}a1ā=daā,b1ā=dbā.