Last Two Digits

We can use the binomial theorem to find the last two digits of $19^{39}$ . Writing $19$ as $(20 - 1)$ :

$(20 - 1)^{39} = \binom{39}{0} (-1)^{39} + \binom{39}{1} (-1)^{38} \times 20 + \binom{39}{2} (-1)^{37} \times 400 + \cdots$

Note that from third term onwards, all of them are divisible by 100 and do not affect the last two digits of $19^{39}$ .

$\begin{aligned} (20 - 1)^{39} & \equiv -1 + 39 \times 20 \pmod{100}\\ & \equiv 1 + 780 \\ & \equiv 79 \end{aligned}$

Hence, the last two digits of $19^{39}$ are $79$ . $_\square$

Let $19^{39}=x \pmod {100}$ where $0\le x \le 100$ and $\gcd(19,100)=1$ implies we can say $19^{\phi(100)}=1\pmod{100})$ and $\phi(100)=40$ .

So we need to solve $19x=1\pmod{100}$ which means we have to find the modular inverse of $19$ with respect to $100$ .

The restriction on the range in which $x$ lies, as well as the fact that it should end with $9$ , (an odd power of a number ending with $9$ always ends in $9$ , so $x$ must end with $9$ ), so we need to check the numbers { $19,29,39,49,59,69,79,89,99$ } and we will see that only $x=79$ satisfies the condition $19x=1\pmod{100}$ , so $79$ is our required answer.