# Divisible by 5 - Integral Values

by Andrew
(New York)

For what integral values of n is the expression (1^n) + (2^n) + (3^n) + (4^n) divisible by 5?

### Comments for Divisible by 5 - Integral Values

 Feb 07, 2012 Modular Arithmetic - Divisible by 5 - Integral Values by: Staff Question:by Andrew (New York)For what integral values of n is the expression (1^n) + (2^n) + (3^n) + (4^n) divisible by 5?Answer:For what integral values (of n is the expression (1^n) + (2^n) + (3^n) + (4^n) divisible by 5?Consider the following table: n     (1^n) + (2^n) + (3^n) + (4^n) 0      4   not div by 5 1      10 2      30 3      100 4      354    not div by 5 5      1300 6      4890 7      18700 8      72354   not div by 5 9      282340 10    108650 11    4373500 12    17312754   not div by 5 13    68711380 ...When n = 0, the result is not divisible by 5When n = 1, the result is divisible by 5When n = 2, the result is divisible by 5When n = 3, the result is divisible by 5When n = 4, the result is not divisible by 5m = any positive integerfor n = 0 + 4m, the result is not divisible by 5for n = 1 + 4m, the result is divisible by 5for n = 2 + 4m, the result is divisible by 5for n = 3 + 4m, the result is divisible by 5>>>The final answer is:The sum (1^n + 2^n + 3^n + 4^n ) is divisible by 5 if n = 1, 2, or 3 mod 4 Thanks for writing.Staff www.solving-math-problems.com