Determine the last two digits of the following by using modular arithmetic:
(a) $4 352 709 × 4 678 829$
(b) $4352783^5$
The last two digits are identified by using modulo 100.
(a) $$ \begin{align} 4 352 709 \times 4 678 829 & \equiv 9 \times 29 \pmod{100} \\ \\ & \equiv 261 \pmod{100} \\ \\ & \equiv 61 \pmod{100} \end{align} $$
So the last two digits are 61.
(b) We will use Proposition (3.8).
If $a ≡ b \pmod{n}$ then $a^k ≡ b^k \mod{n}$ where $k$ is a natural number.
$$ \begin{align} 4352783^5 & \equiv 83^5 \pmod{100} \\ \\ & \equiv 3939040643 \pmod{100} \\ \\ & \equiv 43 \pmod{100} \end{align} $$
So the last two digits are 43.
The author's solution uses a different approach which is educational.
We have
$$ 4352783 \equiv 83 \pmod{100} \equiv -17 \pmod{100} $$
And
$$ (-17)^2 \equiv 289 \pmod{100} \equiv 89 \pmod{100} $$
And so
$$ \begin{align} 4352783^5 & \equiv (-17)^5 \pmod{100} \\ \\ & \equiv \left((-17)^2\right)^2 \times (-17) \pmod{100} \\ \\ & \equiv 89^2 \times (-17) \pmod{100} \\ \\ & \equiv 7921 \times (-17) \pmod{100} \\ \\ & \equiv 21 \times (-17) \pmod{100} \\ \\ & \equiv -357 \pmod{100} \\ \\ & \equiv 43 \pmod{100} \end{align} $$
So the last two digits are 43.