Modular arithmetic: 5th powers (mod 11) Doing a question about modular arithmetic and number theory and I have a mark scheme that doesn't really explain very well. 
It say " The 5th powers (mod 11) are 0,1 and 10 ". I can find examples to support this, such as $11^5$ is equal to 0(mod 11), $2^5$ is equal to 10(mod 11).
How do we know that these are the only values that it can take? 

Well you can write any number as $11n,11n+1,11n+2,11n+3,...11n+10$
This basically means that a number can leave a remainder from  0 to 10  upon division by 11
So if you know the remainders for the fifth powers of  all the numbers upto 11 , you can know which remainders all the numbers will leave upon division by 11.
So ,
$1^5 \equiv 1$ mod $11$
$2^5 \equiv 10$ mod $11$
$3^5 \equiv 1$ mod $11$
$4^5 \equiv 1$ mod $11$
$5^5 \equiv 1$ mod $11$
$6^5 \equiv 10$ mod $11$
$7^5 \equiv 10$ mod $11$
$8^5 \equiv 10$ mod $11$
$9^5 \equiv 1$ mod $11$
$10^5 \equiv 10$ mod $11$
So , as you can see the remainders are only 0 , 1 , 10 , so any number will only leave 0,1,10 as remainder when their fifth power is divided by 11.
Eg) $51^5$ mod $11$
$51 = 44 + 7$ mod $11$
So $51 ^5 $ will leave the remainder of $ 7^ 5$ upon division by 11 ( i.e 10 )
I hope this helps you .
