Finding the unit digit

Question

How can I find the unit digit of $2 ^{9^{100}}$. Is there any general method of finding the unit digit?

Answer 1

I'm not fully sure what the term "unit digit" translates to, as there is no wikipedia page for it. However, I will assume that this translates into the last digit of a number.

Let $f(n)$ denote the unit digit of $n$.

Note that $f(2^1)=2, f(2^2)=4, f(2^3)=8, f(2^4)=6$. Then note $f(2^5)=2, f(2^6)=4, f(2^7)=8, f(2^8)=6$.

As you can see, the unit digits repeat themselves with a period of $4$.

Then note $9 \equiv 1 \pmod 4$. We can conclude $9^{100}\equiv 1^{100} \pmod 4$.

Thus, since the remainder of $9^{100}$ divided by $4$ is $1$, we conclude the last digit is $2$.

Answer 2

If you look at $2^1, 2^2, 2^3 ...$ you will quickly notice the unit digits follow the pattern $2,4,8,6,2,4,8,6...$ to prove that this pattern will continue to repeat note that when we multiply a number by $2$ it is the unit digit alone that determines what the unit digit of the answer will be. Using this pattern we get that $2^9$ ends in $2$ (because the remainder when we divide $9$ by $4$ is $1$). But this means that the unit digits of ${2^9}^i$ follow the same pattern and powers of $2$ and hence ${2^9}^9$ will also end in $2$! By the same reasoning ${{2^9}^9}^9$ will also end in $2$ and so forth. And hence ${2^9}^{100}$ also ends in $2$

Answer 3

Note that the positive integer powers of $2$ go $2,4,8,16,32,...$. The unit digit goes in a cycle of length $4$.

So you need to determine $9^{100}\pmod 4$

Note that $ 9 = 8+1$. A simple application of binomial theorem will lead you to the answer.

Answer 4

The unit digit $\implies\pmod{10}$

Now as $(2,10)=2$ let us find $2^{9^n-1}\pmod{10/2}$

As $\phi(5)=4$ and $9\equiv1\pmod4$

$9^n\equiv1^n\equiv1\implies9^n-1\equiv0\pmod4$

$\implies2^{9^n-1}\equiv2^0\pmod5$

$\implies2^{9^n-1}\cdot2\equiv2^0\cdot2\pmod{5\cdot2}$

Answer 5

Yes, there's a general method: computing the number modulo $10$.

Observe the successive positive powers of $2\mod 10$ is periodic: $$ 2^n\equiv\begin{cases} 2 &\text{if }\;n\equiv 1\mod 4,\\ 4 &\text{if }\;n\equiv 2\mod 4, \\8 &\text{if }\;n\equiv 3\mod 4, \\ 6 &\text{if }\;n\equiv 0\mod 4, \end{cases} $$ Thus $\;2^{9^{100}}\equiv 2^{9^{100}\bmod 4}\mod 10 =2^{(9\bmod 4)^{100}}=2^{1}=2\mod 10 .$

Answer 6

Hint $ $ Examining $2^n$ mod $10$ gives evidence for the conjecture that it has period $\,4\,$ for $\,n\ge 1.\,$ ${\rm mod}\ 4,\,$ the power we seek is $\,9^{100}\equiv 1^{100} \equiv \color{#c00}{\bf 1}.\,$ Low powers hint $\,\color{#c00}{2^{4n+\bf 1}\!\equiv 2}\pmod{10}\,$ so we rigorously prove this conjecture by induction. Base case is $\,10\mid \color{#0a0}{2^{\large 5}-2},\,$ and inductive step

$\quad {\rm mod}\ 10\!:\,\ \color{#c00}{2^{\large 4n+1}\equiv 2}\,\Rightarrow\,2^{\large 4(n+1)+1}\!\equiv 2^{\large 4} \color{#c00}{2^{\large 4n+1}}\!\equiv 2^{\large 4}\color{#c00}2\equiv \color{#0a0}{2^{\large 5}\equiv 2}.\ \ $

Remark $\ $ If modular arithmetic is unfamilar you can eliminate it above to obtain $$ 2^{\large 4(n+1)+1}\!-2 = 2^4(\color{#c00}{2^{4n+1}\!-2}) + \color{#0a0}{2^5 - 2} $$

By our induction hypothesis we know that $\,10\mid \color{#c00}{2^{4n+1}-2}\,$ and $\,10\mid \color{#0a0}{2^{\large 5}-2}\,$ by the base case, hence $10$ divides the RHS above, so also the equal LHS, which completes the inductive proof.

Alternatively you can prove by induction that $\,2^4-1\mid 2^{4n}-1\,$ then scale that by $\,2,\,$ or you could instead make the substituttion $\,a=2^4\,$ in $\,a-1\mid a^n-1\ $ (Factor Theorem).

More generally a very similar proof shows that $\,30\mid a^{4n+1}-a\ $ for all $\,a.\ $