There are at least three methods of proving the results that
for odd values of $n$ , $9^{n} + 1^{n}$ , $8^{n} + 2^{n}$ , $7^{n} + 3^{n}$ and $6^{n} + 4^{n}$ are multiples of 10
whereas, if $n$ is even, then $9^{n} - 1^{n}$ , $8^{n} - 2^{n}$ , $7^{n} - 3^{n}$ and $6^{n} - 4^{n}$ are multiples of 10.

These results can be proved using the axiom of mathematical induction but a simpler approach is to use the periodicity of the values of the last digit for powers of integers. Emma from St Paul's Girls' School and Andrei from Tudor Vianu National College, Bucharest, Romania both used this method of solution. As Andrei tabulated his results his solution is shorter.

Proof using the periodicity of last digits of powers of integers

I shall determine the last digit of each of the numbers: $1^{n}, 2^{n}, 3^{n}, 4^{n}, 6^{n}, 7^{n}, 8^{n}, 9^{n}$ .

The last digit of $1^{n}$ is always 1, and for $6^{n}$ it is always 6.
For, 2, 3, 7 and 8 the last digit of their powers repeats with periodicity 4, as shown in the table:

n (mod 4) 0 1 2 3

2 6 2 4 8

3 1 3 9 7

7 1 7 9 3

8 6 8 4 2

For 4 and 9, the last digit of their powers repeats with periodicity 2:

n(mod 2)	0	1
4	6	4
9	1	9

After creating these tables, I shall prove the required results:

(1) If $n$ is odd, the last digit of $9^{n}$ is 9, and as the last digit of $1^{n}$ is always 1, so the the last digit of $9^{n} + 1^{n}$ is 0, and the number is a multiple of 10.

For $7^{n} + 3^{n}$ there are two possible cases:

$n = 1$ (mod 4). Here $7^{n} + 3^{n} = (7 + 3) (\mod 10) = 0 (\mod 10)$
$n = 3$ (mod 4). Here $7^{n} + 3^{n} = (3 + 7) (\mod 10) = 0 (\mod 10) .$

(2) If $n$ is even it is very easy to look in the table above to obtain the desired results. First I look at $8^{n} - 2^{n}$ for different values of $n$ - even.

If $n = 2$ (mod 4), then $8^{n} - 2^{n} = (4 - 4) (\mod 10) = 0 (\mod 10)$ .
If $n = 0$ (mod 4), $8^{n} - 2^{n} = (6 - 6) (\mod 10) = 0 (\mod 10)$ .

For, $6^{n} - 4^{n}$ , I also obtain (6-6) (mod 10) = 0 (mod 10).

(3) If $n$ is even: $9^{n} - 1^{n}$ is a multiple of 10. The proof derives evidently from the table. In a similar manner, $7^{n} - 3^{n}$ is also a multiple of 10.

(4) If $n$ odd: $8^{n} + 2^{n}$ and $6^{n} + 4^{n}$ are both multiples of 10.

Proof using the Binomial Theorem

$7^{n} + 3^{n} = 7^{n} + (10 - 7)^{n} = 7^{n} + 10^{n} + . . . . . + (- 7)^{n} .$

All the terms of the Binomial expansion involve powers of 10 apart from the term $(- 7)^{n}$ so the expression is a multiple of 10 when $n$ is odd but not when $n$ is even. The same method works for all other parts of the question.