What an Odd Fact(or)

Thanks to Steve of Bayridge Secondary School for the following solution. Solutions were also recieved from Hin-Tai from Bourne Grammar School, Junwei from BHASVIC, Josh of Lawrence High School, Stephanie and Stephen of Knowles Hill School, Andrei of School 205, Bucharest and Robert and Josh from Highgate school.

Well done to all of you but there is more. What powers and what numbers will this pattern work for? Charlotte of The Mount School, York thinks it works for all powers.

Show that $1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99}$ is divisible by 5.

For each base (1, 2, 3, 4 or 5), there exists a pattern between the value of the exponent, and the first place value.
For $n ^ 1$, the first place value is always 1.

n	$1 ^ n$	$2 ^ n$	$3 ^ n$	$4 ^ n$	$5 ^ n$
1	1	2	3	4	5
2	1	4	9	16	25
3	1	8	27	64	125
4	1	16	81	256	625
5	1	32	243	...4	...5
6	1	64	729	...6	...5
7	1	128	...7	...4	...5
8	1	256	...1	...6	...5

if (n mod 4) = 0 then $2 ^ n$ mod 10 = 6

if (n mod 4) = 1 then $3 ^ n$ mod 10 = 3
if (n mod 4) = 2 then $3 ^ n$ mod 10 = 9
if (n mod 4) = 3 then $3 ^ n$ mod 10 = 7

$4 ^ n$

if (n mod 2) = 0 then $4 ^ n$ mod 10 = 6
if (n mod 2) = 1 then $4 ^ n$ mod 10 = 4

$5 ^ n$

$5 ^ n$ mod 10 = 5

Let n = 99

n mod 4 = 3
n mod 2 = 1

$1^{99}$= 1
$2^{99}$ = ...8
$3^{99}$ = ...7
$4^{99}$ = ...4
$5^{99}$ = ...5
-------------
...15

If the first place value of any natural number greater than 0 is equal to 0 or 5, then the number is a multiple of 5.

Let $m$ be a natural number greater than, or equal to 5.

$\{m \ \mid \ m \geq 5, m \in N \}$

if m mod 10 = 0, 5, then m is divisible by 5

Example:

5 mod 5 = 0
90 mod 5 = 0
5 mod 10 = 5
90 mod 10 = 0

$1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99} mod 10 = 5$

therefore $1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99}$ is divisible by 5.