Limiting final digits

Here is another, perhaps more systematic, approach. Consider the following alternate definition of $d$ : $d(m,n)$ is equal to the least positive number that is $\equiv m^n \mod 10$ . This is because taking the congruence mod 10 is known for giving us the last digit. With this in mind I will prove the following equivalence:

A number $m$ has the sought property if and only if $m^2 \equiv m \mod 10$ . First I will prove this in the reverse direction. If we have that $m^2 \equiv m \mod 10$ then $d(m,2) = d(m,1)$ . And then $m^3 \equiv m \times m^2 \equiv m \times m \equiv m^2 \equiv m \mod 10$ so $d(m,3) = d(m,2)$ , and so on. So we will have a constant sequence and as it is known, constant sequences converge.

In the other direction let's argue by contradiction. Suppose that $m^2 \not \equiv m \mod 10$ . Then $d(m,2) \neq d(m,1)$ . Thinking about this, consider the following lemma: For all $m, d(m,5) = d(m,1)$ .

Proof: We just need to show that the congruence $x^5 \equiv x \mod 10$ holds for all integers. This may be assumed but to prove it simply use the Chinese Remainder Theorem to break it down into $x^5 \equiv x \mod 5$ and $x^5 \equiv x \mod 2$ . Here Fermat's little theorem proves all $x$ work for either congruence and then via the Chinese Remainder Theorem this implies all $x$ have the property $x^5 \equiv x \mod 10$ .

So then we would have $d(m,6) \neq d(m,5)$ . But then we would also have $d(m,25) = d(m,5)$ so $d(m,26) \neq d(m,25)$ and so on. And this will occur infinitely often at every power of $5$ and because $d$ always takes integer values this means that the difference between $d(m,5^n)$ and $d(m,5^n + 1)$ never closes so the limit cannot exist.

So the limit will exist for a given $m$ if and only if $m^2 \equiv m \mod 10$ so let's solve this congruence. We really just have the polynomial congruence $m^2 - m \equiv 0 \mod 10$ so let's solve the congruences:

$x^2 - x \equiv 0 \mod 2 \implies x \equiv 0,1 \mod 2$

$x^2 - x \equiv 0 \mod 5 \implies x \equiv 0,1 \mod 5$ .

If $x \equiv 0 \mod 2$ and $x \equiv 0 \mod 5$ we get $x = 10$ which isn't single digit so it is not counted. If $x \equiv 0 \mod 2, x \equiv 1 \mod 5$ we get $x = 6$ . If $x \equiv 1 \mod 2, x \equiv 0 \mod 5$ we get $x = 5$ and if $x \equiv 1 \mod 2, x \equiv 1 \mod 5$ we get $x = 1$ .

This means the answer to the problem is $1 + 5 + 6 = 12$ .

Let's think about what it means for a limit to infinity $\displaystyle \lim_{x \ \to \ \infty} f(x) = L$ to exist: as $f(x)$ approaches $\infty$ , its values converge to a single, limiting value. This means that the limits to infinity of cyclical or periodic functions, like $\sin x$ , fail to exist, because they oscillate infinitely.

Now let's consider the function $d(m, n)$ and look at the set of last digits of powers:

$\displaystyle \begin{array} {|c|c|} \hline \text{The last digit of powers of 1 is always} & 1\\\hline \text{The last digits of powers of 2 repeat in a cycle of} & 4,8,6,2\\\hline \text{The last digits of powers of 3 repeat in a cycle of} & 9,7,1,3\\\hline \text{The last digits of powers of 4 repeat in a cycle of} & 6,4\\\hline \text{The last digit of powers of 5 is always} & 5\\\hline \text{The last digit of powers of 6 is always} & 6\\\hline \text{The last digits of powers of 7 repeat in a cycle of} & 9,3,1,7\\\hline \text{The last digits of powers of 8 repeat in a cycle of} & 4,2,6,8\\\hline \text{The last digits of powers of 9 repeat in a cycle of} &1,9\\\hline \end{array}$

The set of the last digits of powers of $2, 3, 4, 7, 8, 9$ all form periodic sequences. This means that $\displaystyle \lim_{n \ \to \ \infty} d(m, n)$ fails to exist for $m = 2, 3, 4, 7, 8, 9$ . Only the last digits of powers of $1$ , $5$ , and $6$ are always the same - they are the respective powers themselves, in fact. Thus, for $m = 1, 5, 6$ , $\displaystyle \lim_{n \ \to \ \infty} d(m, n)$ exists, because their value stays constant.

Hence, our answer is $1 + 5 + 6 = \boxed{12}$