Last Dight of Natural Numbers

Just for a different approach, the statement is equivalent to proving that $a^5-a$ is always a multiple of $10$ .

Factoring, $a^5-a=a\left( a^4-1\right)=a(a-1)(a+1)\left(a^2+1 \right)$

One of the factors $a,a-1$ is even, so the product is even too.

Now, $a^2+1$ is a multiple of $5$ if and only if $a^2-4$ is. But $a^2-4=(a-2)(a+2)$ , so that $a^5-a$ is a multiple of $5$ so long as $(a-2)(a-1)a(a+1)(a+2)$ is. But there's always one multiple of $5$ among any five consecutive integers.

So $a^5-a$ is always an even multiple of $5$ , which is to say a multiple of $10$ , and $a^5$ always has the same last digit as $a$ .

It can be reduced to the fact that all natural numbers either have a unit digit between $0$ to $9$ or they're simply one digit numbers. it's not difficult to see that all digits follow a cycle of unit digits when raised to some power, it turns out that from one to the fifth power a natural number with a unit digit $m$ will give that same unit digit with power 5 as with power one. So we only need to see the following pattern

$\text{1 1 1 1 1}$

$\text{2 4 8 6 2}$

$\text{3 9 7 1 3}$

$\text{4 6 4 6 4}$

$\text{5 5 5 5 5}$

$\text{6 6 6 6 6}$

$\text{7 9 3 1 7}$

$\text{8 4 2 6 8}$

$\text{9 1 9 1 9}$

$\text{0 0 0 0 0}$

Related Wiki: Euler's Totient Function , Euler’s Theorem

It’s true that the statement is equivalent to proving $a^5 \equiv a \pmod{10}$ . First, when $\gcd (a, 10) =1$ , using Euler’s theorem,

$a^{\phi(n)} \equiv 1 \pmod{n}$

When $n=10$ , we can see $10=2 \times 5$ , therefore $\phi(10)=10 \times \left( 1 - \dfrac12 \right) \left( 1 - \dfrac15 \right) =4$ . Substitute $\phi(10)=4$ , we can get $a^4 \equiv 1 \pmod{10}$ . Multiplying both side by $a$ , the equation can be written as $a^5 \equiv a \pmod{10}$ .

Anyway, $\gcd (a, 10)$ may equal to $2$ or $5$ . When $\gcd (a, 10)=5$ , obviously, $a^5$ always has the same last dight as $a$ . Also, $\gcd (a, 10)$ can be equal to $2$ . $a$ must be even, so we suppose $a=2m$ . Inevitably, $\gcd (m, 5)=1$ , therefore $\gcd (2m, 5) = \gcd (m, 5) \times \gcd (2, 5) =1$ . Now we can use Euler’s theorem,

$(2m)^4 \equiv 1 \pmod{5}$

In it, $\phi(5)=5 \times \left( 1- \dfrac15 \right) =4$ . From $2m \equiv 2m \pmod{2}$ , we can get $(2m)^5 \equiv 2m \pmod{10}$ .

All in all, the formula above satisfies all the natural numbers.

From the example they show that $a^5\equiv a\mod 10$ for $a=1,2,3,4\mod 10$ . since $(-1)^5=-1$ , we also have $a^5\equiv a\mod 10$ for $a=-1,-2,-3,-4\mod 10$ . Finally, $0^5\equiv 0 \mod 10$ and $5^5\equiv 5 \mod 10$ are trivial to check.

In my opinion, when it hits to the power of fifth, it will come back from the first. I got it from the sequence.