How to find the last digit of a number in base b?

Question

For a number $a^{x^{...^n}} $. To find its last digit in a base b,

Imagine that I have this number $a^{x^{y}}$ to simplify the problem. Then I calculate $a^{x} \equiv c \pmod b$ and after that $c^{y} \equiv d \pmod b$ which means that d is the last digit. Am i right?

Thanks !

Be careful! Exponentiation is not associative. The order of operations you propose disagrees with the conventional understanding. Perhaps an easy example will help you see the point. — hardmath, Oct 27 '19 at 14:08
Oh I see! I didn't think about that. Then, I should solve $x^y \equiv c \pmod b$ first and $a^c \equiv \pmod b$. Right? — Moh18, Oct 27 '19 at 15:00
@hardmath but for $a^{x^{...^n}}$ is there a general method? — Moh18, Oct 27 '19 at 15:03
Yes, if you know Euler's generalization of Fermat's Little Theorem, you can take the "tower of exponents" down one by one. But one works from the top of the tower downward, not from the base upward as you were trying to do. Again, a smallish example should show you how to proceed. — hardmath, Oct 27 '19 at 15:06
Possible duplicate of How do I compute $a^b,\bmod c$ by hand? — , Oct 27 '19 at 15:17
@RoddyMacPhee: While that would clearly form a subproblem to this Question, the issue of how to deal with a tower of exponents seems especially important here. — hardmath, Oct 27 '19 at 16:43
it's the same basic method, unless you want to also find quotients... — , Oct 27 '19 at 16:44

score 1 · Answer 1 · 2019-10-27T14:36:31.600

1

The procedure you suggest does not work since the conventional meaning of $a^{x^{y}}$ is that it means $a^{(x^{y})}$ whereas you were finding $(a^x)^{y}$.

Depending on the problem you wish to solve, it is likely that the best method for you would be to find a power $N$ of $a$ which is 1 modulo $b$ and then consider $x^y$ modulo $N$.

edited Oct 27 '19 at 14:36

answered Oct 27 '19 at 14:07

$2^{(2^{3})}=256$ whereas $(2^2)^{3}=64$. They do not have the same last digit in base 10 – Oct 27 '19 at 14:28
As I said it depends on the problem - it is usually the best method – Oct 27 '19 at 14:49
what would be a method that would work in all cases? Would Euler's Theorem be useful? So according to Euler's theorem, if $a$ and $b$ are coprime, there will always be an $N=\varphi(b)$ such that $a^N\equiv 1\pmod b.$ So we have $x^y \equiv b \pmod N\Leftrightarrow a^{x^y}\equiv a^b\pmod N.$ – Simon Fraser Oct 27 '19 at 14:59
@S. Dolan It's $x^y \equiv c \pmod b$ first and $a^c \equiv \pmod b$. As Simon Fraser said, is there a general mmethod ? – Moh18 Oct 27 '19 at 15:05
it's $x^y\equiv c\bmod N \implies a^{x^y}\equiv a^c\bmod b$ – Oct 27 '19 at 15:13
@Moh18, as I mentioned in my above comment, you can use Euler's Theorem if $a$ and $b$ are coprime. If not, then there are two cases: 1) $a$ and $b$ have the same prime factors. In that case, there exists an $N$ such that $a^N\equiv 0\pmod b.$ This $N$ is simply the least common multiple of all the powers of the prime factors of $b.$ This is easy to prove. 2) $a$ and $b$ do not have the same prime factors. Then – Simon Fraser Oct 27 '19 at 15:14
correction: I should've written $x^y \equiv c\pmod N \Leftrightarrow a^{x^y}\equiv a^c\pmod b$ – Simon Fraser Oct 27 '19 at 15:19
Also @Moh18 it is very easy to find the last digit of a power. It's the same as the last digit of its units digit raised to the same power. So you just need to find the period of the units digit and take the exponent $x^y$ modulo this period. – Oct 27 '19 at 15:31

How to find the last digit of a number in base b?

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