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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?

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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 5
When n = 1, the result is divisible by 5
When n = 2, the result is divisible by 5
When n = 3, the result is divisible by 5
When n = 4, the result is not divisible by 5


m = any positive integer

for n = 0 + 4m, the result is not divisible by 5

for n = 1 + 4m, the result is divisible by 5

for n = 2 + 4m, the result is divisible by 5

for 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


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