Simple Congruence Question Using Fermat's Little Theorem

Question

Find the units digit of $3^{100}$ by use of Fermat's theorem. Would it suffice to show that because 2,5 are both prime, we can use Fermat's little theorem to show that $3^{100}\cong1(\mod{2})$ and $3^{100}\cong1(\mod{5})$, then $10|(3^{100}-1)\Rightarrow3^{100}\cong1(\mod{10})$ thus the units digit is 1?

Answer 1

Fermat's Little Theorem gives us that $3 \equiv 1$ $(mod$ $2)$, and $3^{4} \equiv 1$ $(mod$ $5)$. So on $\mathbb{Z}_{2}$ over multiplication, $o(1) = 1$ (the order of the element); and on $\mathbb{Z}_{5}$ over multiplication, $o(3) = 4$. We have that if $o(a)|n$, then $a^{n} = e_{G}$, the identity element of the group. So $3^{100} \equiv 1$ $(mod$ $10)$.

If you're not familiar with group theory, then you can note that if $\phi(n)|x$ and $gcd(a, n) = 1$, then $a^{x} \equiv 1$ $(mod$ $n)$.

It may also be worth looking at the Euler-Fermat Theorem, which gives us that if $gcd(a, n) = 1$, then $a^{\phi(n)} \equiv 1$ $(mod$ $n)$.

Answer 2

Yes, slightly more generally, by little Fermat

$\quad {\rm mod}\,\ 5\!:\,\ a\not\equiv 0\,\Rightarrow\, a^4\equiv 1\,\Rightarrow\, a^{4n}\equiv 1$

$\quad {\rm mod}\,\ 2\!:\,\ a\not\equiv 0\,\Rightarrow\,\ a\equiv 1\,\Rightarrow\, a^{4n}\equiv 1$

So $\ 2,5\nmid a\,\Rightarrow\,2,5\mid a^{4n}\!-1\,\Rightarrow\,{\rm lcm}(2,5)=10\mid a^{4n}\!-1\,\Rightarrow\, a^{4n}\equiv 1 \pmod{10}$