What is the unit number of $17^{17}$? What is the unit number of $17^{17}$? The question concerns students between the ages of 14 and 16.
The possible solution are
(A) $1$, (B) $3$, (C) $5$, (D) $7$, (E) $9$.
I have searched by induction by trying the first powers of $17^{17}$ to understand what the rule was:
$17^2=28\color{red}{9}$
$17^3=491\color{red}{3}$
$17^4=8352\color{red}{1}$
$17^5=141985\color{red}{7}$
$17^6=2413756\color{red}{9}$
$17^7=41033867\color{red}{3}$
$17^8= 697575744\color{red}{1}$
I have seen that the last digits have a sequence as $\color{magenta}{9317-9317\ldots}$ but I don't understand how to get $9$ which is the last digit of  $17^{17}$ from this sequence. There is an unusual fact that the solution provides this:
The digit of the units of a product is determined only by the product of the digits of the units
of the factors. Thus:
$17^2$ has the same digit of units as $7 × 7$ i.e. $9$;
$17^3$ has the same digit of the units as $9 × 7$ i.e. $3$;
$17^4$ has the same digit as the units of $3 × 7$ i.e. $1$;
$17^5$ has the same unit digit as $1 × 7$ i.e. $7$, which is the same unit digit as $17^1$.
The digit of units of the powers of $17$ thus repeats every $4$ consecutive values of the exponent. Since $17 = 4 \cdot 4 + 1$ the digit of units of $17^{17}$ is the same as that of $17^1$, that is, $7$. And this is false,
How, therefore, do I get $9$?
 A: Your solution is correct but calculating the entirety of these powers is unnecessary.
Think of a number with digits:
$a_1a_2a_3\cdot\cdot\cdot a_{n-1}a_n$ as: $$10 \cdot a_1a_2a_3\cdot\cdot\cdot a_{n-1} + a_n$$
We can see that if we multiply through by some number ($17$ in this case) then the first n-1 digits, multiplied by this number will still be a multiple of $10$ and, hence, will have no effect on the last digit of this product. We can conclude that the last digits of:
$$a_1a_2a_3\cdot\cdot\cdot a_{n-1}a_n \cdot 17$$ is the last digit of: $$17a_n$$
We can also use the same reasoning to notice this is equal to the last digit of:
$$7a_n$$
Using this, last digits of $17^n$:
$$17^1 \rightarrow 7$$
$$17^2 \rightarrow 7\cdot7 \rightarrow 9$$
$$17^3 \rightarrow 7\cdot9 \rightarrow 3$$
$$17^4 \rightarrow 7\cdot3 \rightarrow 1$$
$$17^5 \rightarrow 7\cdot1 \rightarrow 7$$
We could keep doing this up to $19$ or we could notice that this sequence is going to loop at intervals of $4$ to deduce:
$$17^{4k} \rightarrow 1 \implies 17^{16} \rightarrow 1 \implies 17^{17} \rightarrow 7$$
This is a shoddy example of something called modular arithmetic. I'd recommend researching this and once you have a very basic understanding of it, this question will be easy to do by considering $17^n\equiv7^n\mod{10}$
