Question

n为正整数,20n+2整除2003n+2002，求所有n的解

Answer 1

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.

Answer 2

20n+2整除2003n+2002