Find the units digit of $572^{42}$

Question

The idea of this exercise is that you use the modulus to get the right answer. What I did was:

$$572\equiv 2\pmod {10} \\ 572^2 \equiv 2^2 \equiv 4\pmod{10} \\ 572^3 \equiv 2^3 \equiv 8\pmod{10} \\ 572^4 \equiv 2^4 \equiv 6\pmod{10} \\ 572^5 \equiv 2^5 \equiv 2\pmod{10} \\ 572^6 \equiv 2^6 \equiv 4\pmod{10} \\ (...)$$

I can see that this goes 2,4,8,6 and then repeats. I remember that the gist of the exercise is to find the remainder based on this repetition. How do I do that? I know that $572^{42} \equiv 2^{42}\equiv ? \pmod {10}$. How do I simplify that 42 and answer this using that repetition?

Answer 1

Since you found $572 \equiv 2\pmod{10} $ you have to check only what is $$2^{42} \equiv ?\pmod{10} $$

Since $2^5 \equiv 2$ we have $$2^{42} = 2^2\cdot (\color{red}{2^5})^8 \equiv 2^2\cdot \color{red}{2}^8 \equiv 2^{10} \equiv (\color{blue}{2^5})^2 \equiv \color{blue}{2}^2\equiv 4 \pmod {10}$$

Answer 2

Another approach is:

The reminder is repeating between:

$$R=\{2,4,8,6\}$$

Solve the following, since $|R|=4$

$$42\equiv x\pmod{4}$$

Have $x=2$, which means the reminder is the second term in $R$ that is $4$

Therefore $$572^{42}\equiv4\pmod{10}$$

Answer 3

Consider $2^k$ modulo $10$.

You will notice that it enters a repeating pattern: $2,4,8,6,2,4,8,6,2,\dots$

Where $2^k\pmod {10} \equiv \begin{cases}2&\text{when}~k\equiv 1\pmod 4\\ 4&\text{when}~k\equiv 2\pmod 4\\8&\text{when}~k\equiv 3\pmod 4\\6&\text{when}~k\equiv 0\pmod 4\end{cases}$

Now, consider the exponent in this case. It is $42$. Since the pattern of $2^k$ depends on $k\pmod 4$, we see that $42\equiv2 \pmod 4 $. Therefore, $2^{42}\equiv 4 \pmod {10}$ which implies that ${572}^{42}\equiv 4 \pmod {10}$.

Answer 4

Applying $\ xy\bmod xz\, =\, x(y\bmod z)\ = $ mod Distributive Law to factor out $\,x = 2\,$ below

$\ \ \ \ \begin{align}(2a)^{\large 2+4N}\!\bmod 10 &\,=\, 2(2\, \color{#c00}2^{\large\color{#c00} 4N} a^{\large 2} \color{#c00}a^{\large\color{#c00} 4N}\bmod 5)\\[.2em] &\,=\, 2(2a^{\large 2}\bmod 5)\ \ {\rm by}\ \ 5\nmid a\,\Rightarrow\ \color{#c00}{a^{\large 4}\equiv 1}\!\!\!\pmod{5}\\[.3em] &\,=\, 4\ \ \ {\rm if}\ a\equiv \pm1\!\!\!\pmod{\!5}\end{align}$

Answer 5

You want $2^{42} \mod 10$. You found that $2^4\equiv6\mod 10$. Now $6^n\equiv 6$ for all $n\in\mathbb N$ (you could prove that by induction), so $2^{42}\equiv2^{40}2^2\equiv(2^4)^{10}2^2\equiv6^{10}\times4\equiv6\times4\equiv4\mod10$.

Answer 6

Notice the pattern in the indices, the indices that get mapped to each of $2,4,8$ or $6.$

Note that the indices $1,5,9,\cdots, 1+4n$ get mapped to $2.$ The indices $2,6,10,\cdots,2+4n$ to $4,$ the indices $3,7,11,\cdots 3+4n$ get mapped to $8,$ and finally all multiples of $4,$ of the form $4n$ get mapped to $6,$ where $n$ is a nonnegative integer.

Then you only need reduce $42$ modulo $4,$ which shows you that it is of the second form. Thus, you have that to the $42$nd power, your number is $4\pmod {10}.$