Understanding When the Sum of Powers is Divisible by 5

Introduction

Divisibility by 5 can be a fascinating topic in number theory, especially when dealing with powers of different numbers. This article will guide you through the process of determining for which values of ( n ) the expression ( 1^n 2^n 3^n 4^n ) is divisible by 5. We will explore this using modular arithmetic and explore the cyclical nature of the powers.

Evaluating the Expressions Modulo 5

First, let’s evaluate the individual terms modulo 5.

1n

Any power of 1 is always 1:

[1^n equiv 1 pmod{5}]

2n

2n modulo 5 has a repeating cycle of 4:

(n equiv 0 pmod{4} implies 2^n equiv 1 pmod{5}) (n equiv 1 pmod{4} implies 2^n equiv 2 pmod{5}) (n equiv 2 pmod{4} implies 2^n equiv 4 pmod{5}) (n equiv 3 pmod{4} implies 2^n equiv 3 pmod{5})

3n

3n modulo 5 also has a repeating cycle of 4:

(n equiv 0 pmod{4} implies 3^n equiv 1 pmod{5}) (n equiv 1 pmod{4} implies 3^n equiv 3 pmod{5}) (n equiv 2 pmod{4} implies 3^n equiv 4 pmod{5}) (n equiv 3 pmod{4} implies 3^n equiv 2 pmod{5})

4n

4n modulo 5 has a cycle of 2 since 4 is 2 modulo 5:

(n equiv 0 pmod{2} implies 4^n equiv 1 pmod{5}) (n equiv 1 pmod{2} implies 4^n equiv 4 pmod{5})

Multiplying the Results Based on ( n mod 4 )

Now, combining these results based on ( n mod 4 ), we can find when the entire expression is divisible by 5:

When ( n equiv 0 pmod{4} ): [1^n 2^n 3^n 4^n equiv 1 1 1 1 equiv 4 pmod{5}] When ( n equiv 1 pmod{4} ): [1^n 2^n 3^n 4^n equiv 1 2 3 4 equiv 10 equiv 0 pmod{5}] When ( n equiv 2 pmod{4} ): [1^n 2^n 3^n 4^n equiv 1 4 4 1 equiv 10 equiv 0 pmod{5}] When ( n equiv 3 pmod{4} ): [1^n 2^n 3^n 4^n equiv 1 3 2 4 equiv 10 equiv 0 pmod{5}]

Conclusion

Based on our calculations, we find that (1^n 2^n 3^n 4^n) is divisible by 5 for ( n equiv 1, 2, 3 mod 4 ).

Additional Insight

Additionally, we can check the cyclical nature of the powers directly:

Case 1: If ( n 4k ), where ( k ) is a natural number: (1^n equiv 1 pmod{5}) (2^n equiv 6 pmod{5} equiv 1 pmod{5}) (3^n equiv 1 pmod{5}) (4^n equiv 6 pmod{5} equiv 1 pmod{5}) The sum of this expression is (1 1 1 1 4 pmod{5}), which is not divisible by 5. Case 2: ( n 4k 1 ): (1^n equiv 1 pmod{5}) (2^n equiv 2 pmod{5}) (3^n equiv 3 pmod{5}) (4^n equiv 4 pmod{5}) The sum of this expression is (1 2 3 4 10 equiv 0 pmod{5}), which is divisible by 5. Case 3: ( n 4k 2 ): (1^n equiv 1 pmod{5}) (2^n equiv 4 pmod{5}) (3^n equiv 4 pmod{5}) (4^n equiv 1 pmod{5}) The sum of this expression is (1 4 4 1 10 equiv 0 pmod{5}), which is divisible by 5. Case 4: ( n 4k 3 ): (1^n equiv 1 pmod{5}) (2^n equiv 8 equiv 3 pmod{5}) (3^n equiv 2 pmod{5}) (4^n equiv 4 pmod{5}) The sum of this expression is (1 3 2 4 10 equiv 0 pmod{5}), which is divisible by 5.

In summary, the expression (1^n 2^n 3^n 4^n) is divisible by 5 for ( n equiv 1, 2, 3 mod 4 ). It is noteworthy that when ( n 4k ), the expression is not divisible by 5 due to the repeated modulo 5 results.