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2 solutions
You need to show rigorously that the pattern you mentioned indeed holds for all integers, i.e., n 5 ≡ n ( m o d 1 5 ) ∀ n ∈ Z . Otherwise, your solution is incomplete and unjustified.
There's actually a very easy proof to this using Fermat's Little Theorem and basic modular arithmetic.
Claim: n 5 ≡ n ( m o d 1 5 ) ∀ n ∈ Z
Proof:
We have, by Fermat's Little Theorem that,
∀ n ∈ Z , n 5 ≡ n ( m o d 5 ) ∀ n ∈ Z , n 5 = n 3 × n 2 ≡ n × n 2 ≡ n 3 ≡ n ( m o d 3 )
Now, since g cd ( 5 , 3 ) = 1 and 1 5 = 3 × 5 , we have,
∀ n ∈ Z , n 5 ≡ n ( m o d 5 , 3 ) ⟺ n 5 ≡ n ( m o d 1 5 ) ∀ n ∈ Z
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Here's an alternative approach for proving n 5 ≡ n ( m o d 1 5 ) ∀ n ∈ Z See that n 5 − n = n ( n 4 − 1 ) = n ( n 2 − 1 ) ( n 2 + 1 ) = n ( n + 1 ) ( n − 1 ) [ ( n + 2 ) ( n − 2 ) + 5 ] = ( n − 2 ) ( n − 1 ) n ( n + 1 ) ( n + 2 ) + 5 ( n − 1 ) n ( n + 1 ) Obviously this is a multiple of 5 and is also a multiple of 3, since 3 and 5 are co-prime, hence n 5 − n is a multiple of 15, thus n 5 ≡ n ( m o d 1 5 ) ∀ n ∈ Z
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This is a really nice approach using elementary methods (even better than mine). +1 :)
A number to the power of 5 gives a number where the last digit is same as the original. The sum of numbers from 5 to 55 gives 1530 which means the last digit of the final number is 0. This means the possible solutions are 0 and 5. Testing both, we get that it is 0.
5 5 ≡ 5 ( m o d 1 5 )
6 5 ≡ 6 ( m o d 1 5 )
Similarly, 7 5 gives a remainder of 7 and 8 5 gives a remainder of 8 on division by 15 and so on...
Thus we get ∑ 5 5 5 which is = 1 5 3 0
Which in turn gives 0 as remainder