求出所有的整数n，使得20n+2能整除2003n+2002.解 因为20n+2是偶数,且能整除2003n+2002.所以2003n+2002为偶数.从而n为偶数.令2m=n,m为整数,则(2003n+2002)/(20n+2)∈Z;<==>(2003m+1001)/(20m+1)∈Z;<===>[100(20m+1)+3m+901]/(20m+1)∈Z;<===>(3m+901)/(20m+1)∈Z;<===>[20(3m+901)]/(20m+1)∈Z;<===>[3(20m+1)+18017]/(20m+1)∈Z;<==>18017/(20m+1)∈Z;<===>(43*419)/(20m+1)∈Z;故20m+1=±1,±43,±419,±18017得m=0,-21.故所求n=0或n=-42.