How to find the last digit of $a^{b^{c^{d\ldots}}}$?

Question

To find last digit of $a^b$ I just need take the unit digit of $a$ and two last digits of $b$ and then follow patterns for different unit digits $0...9$. What if $b$ has own power that has another power and so on? For example how can I calculate $$123232^{694022^{140249}} \mod 10\:\:? $$

Answer 1

All that matters is the value of $a_1$ mod $10$, the value of $a_2$ mod $4$, and whether $a_3$ is even or odd, even if the tower goes higher.

Consider $a_1^{a_2^{\vdots^{a_n}}}$. I assume $n\geq3$ and all the $a_i$ are greater than $1$. For ease of notation, let $b=a_2^{a_3^{\vdots^{a_n}}}$.

• If $a_1\equiv0$ mod $10$, then $a_1^{a_2^{\vdots^{a_n}}}\equiv0$ mod $10$. So for all that follow, assume $10$ does not divide $a_1$.
• If $a_1\equiv5$ mod $10$, then any power of $a_1$ is also $5$ mod $10$. So for all that follow, assume $5$ does not divide $a_1$.
• If $a_1\equiv6$ mod $10$, then any power of $a_1$ is also $6$ mod $10$. So for all that follow, assume $a_1\not\equiv6$ mod $10$.
• If $a_1\equiv4$ or $a_1\equiv8$ mod $10$, then powers of $a_1$ alternate between $a_1$ and $6$ mod $10$. So all that matters is whether $a_2$ is even or odd. If $a_2$ is even, then $a_1^{a_2^{\vdots^{a_n}}}\equiv6$ mod $10$. Otherwise, $a_1^{a_2^{\vdots^{a_n}}}\equiv a_1$ mod $10$.
• If $a_1\equiv2$ mod $10$, then powers of $a_1$ cycle $(2,4,8,6)$ mod $10$. So you need to know what $b$ is mod $4$.
  • If $a_2\equiv0$ mod $4$, then $b\equiv0$ mod $4$, and $a_1^{a_2^{\vdots^{a_n}}}\equiv 6$ mod $10$.
  • If $a_2\equiv2$ mod $4$, then $a_3>1$ by assumption, and so $b\equiv0$ mod $4$. This means $a_1^{a_2^{\vdots^{a_n}}}\equiv 6$ mod $10$.
  • If $a_2\equiv1$ mod $4$, then any power of $a_2$ is $1$ mod $4$ as well, and $a_1^{a_2^{\vdots^{a_n}}}\equiv 2$ mod $10$.
  • If $a_2\equiv3$ mod $4$, then powers of $a_2$ alternate between $1$ and $3$ mod $4$. So if $a_3$ is even, $b\equiv 1$ mod $4$ and $a_1^{a_2^{\vdots^{a_n}}}\equiv 2$ mod $10$. And if $a_3$ is odd, $b\equiv 3$ mod $4$ and $a_1^{a_2^{\vdots^{a_n}}}\equiv 8$ mod $10$.

Now we may assume $a_1$ is relatively prime to $10$. It is $1$, $3$, $7$, or $9$ mod $10$. Now we have Euler's Theorem saying $$a_1^b\equiv a_1^{b\mod{\varphi(10)=4}}$$

So we care what is $b$ mod $4$. We have already studied that above.

• If $a_2\equiv0$ mod $4$, then $b\equiv0$ mod $4$, and $a_1^{a_2^{\vdots^{a_n}}}\equiv 1$ mod $10$.
• If $a_2\equiv2$ mod $4$, then $a_3>1$ by assumption, and so $b\equiv0$ mod $4$. This means $a_1^{a_2^{\vdots^{a_n}}}\equiv 1$ mod $10$.
• If $a_2\equiv1$ mod $4$, then any power of $a_2$ is $1$ mod $4$ as well, and $a_1^{a_2^{\vdots^{a_n}}}\equiv a_1$ mod $10$.
• If $a_2\equiv3$ mod $4$, then powers of $a_2$ alternate between $1$ and $3$ mod $4$. So if $a_3$ is even, $b\equiv 1$ mod $4$ and $a_1^{a_2^{\vdots^{a_n}}}\equiv a_1$ mod $10$. And if $a_3$ is odd, $b\equiv 3$ mod $4$ and $a_1^{a_2^{\vdots^{a_n}}}\equiv a_1^3$ mod $10$.

I think that covers all scenarios. For your example, $123232^{694022^{140249}}$ has $a_1\equiv2$ mod $10$. And it has $a_2\equiv2$ mod $4$. So the above outlines how the result is $6$ mod $10$.

Here is a table for the last digit of $a_1^{a_2^{\vdots^{a_n}}}$ based on the above.

$$ \begin{array}{c} a_3\text{ even}|a_3\text{ odd} \\ \begin{array}{cc} &a_2\mod 4\\ a_1\mod 10& \begin{array}{c|c} &\begin{array}{cccc} 1&2&3&4 \end{array}\\ \hline 1&\begin{array}{cccc} 1&1&1&1 \end{array}\\ 2&\begin{array}{cccc} 2&6&2|6&6 \end{array}\\ 3&\begin{array}{cccc} 3&1&3|7&1 \end{array}\\ 4&\begin{array}{cccc} 4&6&4&6 \end{array}\\ 5&\begin{array}{cccc} 5&5&5&5 \end{array}\\ 6&\begin{array}{cccc} 6&6&6&6 \end{array}\\ 7&\begin{array}{cccc} 7&1&7|3&1 \end{array}\\ 8&\begin{array}{cccc} 8&6&8&6 \end{array}\\ 9&\begin{array}{cccc} 9&1&9&1 \end{array}\\ 0&\begin{array}{cccc} 0&0&0&0 \end{array}\\ \end{array} \end{array} \end{array} $$