Problem: Find the last digit of $3^{1999}$.

My answer is $3$, but the answer sheet says $7$.

Here is what I did:

  • $3^{1999}=(3^9)^{222}\cdot3$
  • Using Fermat's Little Theorem: $3^9\equiv1\pmod{10}$
  • Therefore, $3^{1999}\equiv(3^9)^{222}\cdot3\equiv1^{222}\cdot3\equiv3\pmod{10}$
  • Therefore, the last digit should be $3$

Where did I go wrong?

  • 5
    $\begingroup$ $\phi(10)=4$ not 9 $\endgroup$ Jun 18 '16 at 4:13
  • 7
    $\begingroup$ This requires Euler's theorem, not Fermat's, because Fermats requires a prime modulus. $\endgroup$ Jun 18 '16 at 4:14

Chill! If we evaluate powers of 3 a few times, we see that the last digit will be a sequence of $3$, then $9$, then $7$, then $1$. We don't need to worry about the next few digits because it doesn't ask about it. We know that $3^{1999} = (3^4)^{499} * 3^3$ and since $3^4$ has a last digit of 1, we can negate that entire term. $3^3 = 27$, which has a last digit of 7.

Also, this can be modelled by the sequence $3,9,7,1,3,9....$ and you just need to find the $1999^{th}$ term.


Here's a straightforward alternative that does not require Euler's or Fermat's, and only requires noticing that

$$3^2 \equiv -1 \pmod {10}$$ so that $$\begin{align}3^{1999} &= (3^2)^{999}\cdot3\\&\equiv (-1)^{999}\cdot3\pmod{10}\\&\equiv-3\pmod{10}\\&\equiv{7}\pmod{10}\end{align}$$


By Euler's theorem,

$3^4 \equiv 1 \pmod{10}$

$\implies 3^{1996} \equiv 1 \pmod{10}$ [Raising to the power 499]

$\implies 3^{1999} \equiv 27 \pmod{10} \equiv 7 \pmod{10}$

Hence the result.

  • $\begingroup$ I felt very confident in changing your $=>$ to $\implies$. But I stopped short of changing your $27 \pmod{10}$ to $27 \pmod{100}$. $\endgroup$ Jan 8 '17 at 2:19

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