x^5 ~ 1 mod 10

How do I proved that x^5 always has a last digit 1?

why bother

Why your mother

See, I rhymed, therefore I'm awesome.

Fuck off now.

it doesn't.

good luck.

Fuck, I meant last digit x

>>5
You mean it has a last digit of the last digit of x?

>>6

IT'S LATE AND I'M TIRED, FUCK.

x^5 ~ x modulo 10

There, that should be correct.

consider case by case, it's not that hard
1^5 = 1
2^5 = 32 = 2
3^5 = 243 = 3
4^5 = 1024 = 4
5^5 = 3125 = 5
6^5 = 7776 = 6
7^5 = 16807 = 7
8^5 = 32768 = 8
9^5 = 59049 = 9

A proof is pretty easy though, basically euler-fermat formula states that a^phi(n) = 1 mod n where phi(n) is the number of integers coprime to n and less than n, if a is co-prime to n.

now phi(10) = 4, so a^4 = 1 for all invertible a, so a^5 = a for all invertible a.

Not sure how to show if for a not invertible, can't be arsed to think.

wow, that's not interesting.

have you heard of the mathematian, fermat? check out his little theorum.

If x is an integer, and in x^(p), p is a prime number, and x and p are coprime to eachother. x^(p)-x will be evenly divisible by p.

so x^p = x mod p .. coprime means x and p have no common factor other than 1 or -1. check out euler's totient function sometime.

anyhow you have mod 2p which is a tweak on fermat's thm.
basicly you'd be doing two divisions for x^(p)-x. once for your p, which divides evenly, then dividing by two, which is just 0/2. move the a over to the right again and you have:

x^(p)=x mod 2p   p=5 and x is coprime to 5 (ie. 2,3,13,91)

i hope that's correct enough, I don't know how to account for non-coprime x's and p's