设f(x)=1+a(x-1)+b(x-1)(x-2)+(x-1)(x-2)(x-3)g(x)
那么f(2)=2,f(3)=3,可得
a=1,b=0,于是
f(x)=x+(x-1)(x-2)(x-3)g(x)
即知f(x)除以(x-1)(x-2)(x-3)的余式为x.