Determine the last digit of the following numbers:
(a) $3^{100}$ (b) $9^{100}$
(c) $2^{100}$ (d) $4^{100}$
The last digit of a number is congruent to the number modulo 10.
We also use Proposition (3.8).
If $a ≡ b \pmod {n}$ then $a^k ≡ b^k \pmod{n}$ where $k$ is a natural number.
(a) We have $3^2 = 9 \equiv -1 \pmod{10}$.
And so
$$ \begin{align} 3^{100} & = (3^2)^{50} = (9)^{50} \\ \\ & \equiv (-1)^{50} \pmod{10} \\ \\ & \equiv 1 \pmod{10} \end{align} $$
So the last digit is 1.
(b) We have $9^2 = 81 \equiv 1 \pmod{10}$.
And so
$$ \begin{align} 9^{100} & = (9^2)^{50} = (81)^{50} \\ \\ & \equiv (1)^{50} \pmod{10} \\ \\ & \equiv 1 \pmod{10} \end{align} $$
So the last digit is 1.
(c) We have $2^5 = 32 \equiv 2 \pmod{10}$.
And so
$$ \begin{align} 2^{100} & = (2^5)^{20} = (32)^{20} \\ \\ & \equiv (2)^{20} \pmod{10} \\ \\ & \equiv (2^5)^4 \pmod{10} \\ \\ & \equiv (2)^4 \pmod{10} \\ \\ & \equiv 16 \pmod{10} \\ \\ & \equiv 6 \pmod{10} \end{align} $$
So the last digit is 6.
(d) We already have $2^{100} \equiv 6 \pmod{10}$.
And so
$$ \begin{align} 4^{100} & = (2^2)^{100} \\ \\ & = (2^{100})^2 \\ \\ & \equiv 6^2 \pmod{10} \\ \\ & \equiv 36 \pmod {10} \\ \\ & \equiv 6 \pmod{10} \end{align} $$
So the last digit is 6.