Question

若m为正整数,且m不能被4整除,试说明1^m+2^m+3^m+4^m+…+9^m一定是5的倍数

Answer 1

1^m+2^m+3^m+4^m+…+9^m | 5

= (1^m+2^m+3^m+4^m)*2+0^m | 5

= 2 + (2^m+3^m+4^m)*2 | 5

当M被4除余1时

2 + (2^m+3^m+4^m)*2 | 5

= 2+（2+3+4）*2 |5

= 20 |5

= 0

当M被4除余2时

2 + (2^m+3^m+4^m)*2 | 5

= 2+（4+9+16）*2 |5

= 60 |5

= 0

当M被4除余3时

2 + (2^m+3^m+4^m)*2 | 5

= 2+（8+27+64）*2 |5

= 200 |5

= 0

综上得证.