Let q be a positive whole number.

Determine whether or not 1^q + 2^q + 3^q + 4^q is always divisible by 10 whenever q is NOT divisible by 4.

Each decimal digit is periodic with a differing (though not distinct) period length. 

1 has a period of 1 {1}.
2 has a period of 4 {2, 4, 8, 6}.
3 has a period of 4 {3, 9, 7, 1}.
4 has a period of 2 {4, 6}.

Each period is a divisor of 4 and thus the sum of each term would be cyclic every four terms.

Sum of the nth terms of the period of the digits 1,2,3 and 4 are:
(1st) 1+2+3+4 = 10
(2nd) 1+4+9+6 = 20
(3rd) 1+8+7+4 = 20
(4th) 1+6+1+6 = 14.

As can be seen the first, second and third sums are divisible by 10 while the fourth is not.  Thus, 1^q + 2^q + 3^q + 4^q is always divisible by 10 whenever q is NOT divisible by 4.

Edited on May 14, 2007, 11:55 am

Posted by Dej Mar on 2007-05-14 11:52:46