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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?

H5159
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2 Answers2

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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 $a^{\phi(n)} \equiv 1$ $(mod$ $n)$.

ml0105
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  • Right, I left out the work to get to $3^{100}$ for both 2 and 5. I am not familiar with those as of yet, this is my first course in number theory. My question on the whole was if that the logic that I used was correct. – H5159 Mar 17 '14 at 02:49
  • You should hopefully be learning about the order of an element pretty soon then, even if you haven't had algebra yet. An rules of exponents argument should actually work, as long as you cite Fermat's Little Theorem on the factoring (ie., choose your exponents to get $1$, then exponentiate $1$). – ml0105 Mar 17 '14 at 02:54
  • I am currently taking abstract/modern algebra at the same time. So yes I am learning about it somewhat. Albeit I'm behind in all my classes :'( – H5159 Mar 17 '14 at 02:55
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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 $\ 2,5\nmid a\,\Rightarrow\,2,5\mid a^{4n}\!-1\,\Rightarrow\,{\rm lcm}(2,5)=10\mid a^{4n}\!-1\,\Rightarrow\, a^{4n}\equiv 1 \pmod{10}$

Bill Dubuque
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