Finding powers of 2 and 3 in modular arithmetic 
Find all the powers of $2$ and $3$ modulo $17$.

How would you solve this question and explain the steps please!

Recall that:
$$2^m\equiv k \pmod{17} \implies 2^{m+1}\equiv 2k \pmod {17}$$
We can then list, remembering that if our $k$ is greater than $17$, we subtract $17$ from it.
$$1\equiv 1 \pmod {17}$$
$$2\equiv 2 \pmod {17}$$
$$4\equiv 4 \pmod {17}$$
$$8\equiv 8 \pmod {17}$$
$$16\equiv 16 \pmod {17}$$
$$32\equiv 15 \pmod {17}$$
(Because $16\cdot 2= 32$, but $32>17$ so we apply $32-17=15$)
$$64\equiv 13 \pmod {17}$$
(Because $15\cdot 2 = 30$, but $30>17$, so we apply $30-17=13$)
$$128\equiv 9 \pmod {17}$$
(same as above)
$$256\equiv 1 \pmod {17}$$
And now we have a cycle, so  $2^k\equiv 1,2,4,8,9,13,15,16\pmod{17}$
Now do the same for the powers of $3$. As a check, you should get $3^k\not\equiv 0 \pmod{17}$
